# Matrices

## Class 12 NCERT

### NCERT

1   In the matrix $A=\begin{bmatrix} 2 & 5 & 19 & -7 \\[0.3em] 35 & -2 & \frac{5}{2} & 12 \\[0.3em] \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$,write :$\\$ (i) The order of the matrix $\\$ (ii) The number of elements,$\\$ (iii) Write the elements $a _{13} , a_ {21} , a_ {33} , a_ {24} , a_ {23}$

##### Solution :

(i) In the given matrix, the number of rows is 3 and the number of columns is $4$. Therefore, the order of the matrix is $3 * 4$ $\\$. (ii) Since the order of the matrix is $3 * 4$, there are $3 * 4 = 12$ elements in it.$\\$ (iii)$a _{13} = 19, a_{ 21 }= 35, a_{ 33 }= - 5, a_{ 24 }= 12, a_{ 23}=\dfrac{5}{2}$

2   If a matrix has $24$ elements, what are the possible order it can have? What, if it has $13$ elements?

##### Solution :

We know that if a matrix is of the order $m * n$, it has mn elements.$\\$ Thus, to find all the possible orders of a matrix having $24$ elements, we have to find all the ordered pairs of natural numbers whose product is $24.$$\\ The ordered pairs are: ( 1, 24 ) , ( 24,1 ) , ( 2,12 ) , (12, 2 ) , ( 3,8 ) , ( 8,3 ) , ( 4,6 ) , and ( 6, 4 )$$\\$ Hence, the possible orders of a matrix having $24$ elements are:$\\$ $1 *24, 24 * 1, 2 * 12,12 * 2,\\ 3 * 8,8 * 3, 4 * 6$ and $6 * 4\\ ( 1,13 )$ and $( 13,1 )$ are the ordered pairs of natural numbers whose product is $13$.$\\$ Hence the possible orders of matrix having $13$ elements are (1* 13) and (13 * 1 ).

3   If a matrix has $18$ elements, what are the possible orders it can have? What, if it has $5$ elements?

We know that if a matrix is of the order $m * n$ , it has mn elements.$\\$ Thus, to find all the possible orders of a matrix having $18$ elements, we have to find all the ordered pairs of natural numbers whose products is $18$.$\\$ The ordered pairs are:$( 1,18 ) , ( 18,1 ) , ( 2,9 ) , ( 9, 2 ) , ( 3,6 ) , and ( 6,3 )$$\\ Hence, the possible orders of a matrix having 18 elements are :\\ 1 * 18,18 * 1, 2 * 9,9 * 2,3 * 6 and 6 * 3\\ ( 1,5 ) and ( 5,1 ) are the ordered pairs of natural numbers whose product is 5.\\ Hence, the possible orders of a matrix having 5 elements are 1 * 5 and 5 * 1 . 4 Construct a 3 x 4 matrix, whose elements are given by\\ (i) a_{ij} =\dfrac{1}{2}|-3i+j|$$\\$ $(ii) a_{ij} =2i-j$$\\ ##### Solution : a_{13}=\dfrac{1}{2}|-3*1+3|=\dfrac{1}{2}|-3+3|=0\\ a_{23}=\dfrac{1}{2}|-3*2+3|=\dfrac{1}{2}|-6+3|=\dfrac{1}{2}|-3|=\dfrac{3}{2}\\ a_{33}=\dfrac{1}{2}|-3*3+3|=\dfrac{1}{2}|-9+3|=\dfrac{1}{2}|-6|=\dfrac{6}{2}=3\\ a_{14}=\dfrac{1}{2}|-3*1+4|=\dfrac{1}{2}|-3+4|=\dfrac{1}{2}|1|=\dfrac{1}{2}\\ a_{24 } =\dfrac{1}{2}|-3*2+4|=\dfrac{1}{2}|-6+4|=\dfrac{1}{2}|-2|=\dfrac{2}{2}=1\\ a_{34} =\dfrac{1}{2}|-3*3+4|=\dfrac{1}{2}|-9+4|=\dfrac{1}{2}|-5|=\dfrac{5}{2}$$\\$ Therefore , the required matrix is $A=\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\[0.3em] \frac{5}{2} & 2 & \dfrac{3}{2} & 1 \\[0.3em] 4 & \frac{7}{2} & 3 &\frac{5}{2} \end{bmatrix}$

In general, a $3 x 4$ matrix is given by $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\[0.3em] a_{21} & a_{22} & a_{23} & a_{24}\\[0.3em] a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$$\\ (i) Given : a_{ij} =\dfrac{1}{2}|-3i+j|,i=1,2,3 and j=1,2,3,4$$\\$ Thus , we have $\\$ $a_{11}=\dfrac{1}{2}|-3*1+1|=\dfrac{1}{2}|-3+1|=\dfrac{1}{2}|-2|=\dfrac{2}{2} =1$$\\ a_{21}=\dfrac{1}{2}|-3*2+1|=\dfrac{1}{2}|-6+1|=\dfrac{1}{2}|-5|=\dfrac{5}{2}$$\\$ $a_{31}=\dfrac{1}{2}|-3*3+1|=\dfrac{1}{2}|-9+1|=\dfrac{1}{2}|-8|=4$$\\ a_{12}=\dfrac{1}{2}|-3*1+2|=\dfrac{1}{2}|-3+2|=\dfrac{1}{2}|-1|=\dfrac{1}{2}$$\\$ $a_{22}=\dfrac{1}{2}|-3*2+2|=\dfrac{1}{2}|-6+2|=\dfrac{1}{2}|-4|=\dfrac{4}{2} =2$$\\ a_{32}=\dfrac{1}{2}|-3*3+2|=\dfrac{1}{2}|-9+2|=\dfrac{1}{2}|-7|=\dfrac{7}{2}$$\\$ $a_{13}=\dfrac{1}{2}|-3*1+3|=\dfrac{1}{2}|-3+3|=0\\ a_{23}=\dfrac{1}{2}|-3*2+3|=\dfrac{1}{2}|-6+3|=\dfrac{1}{2}|-3|=\dfrac{3}{2}\\ a_{33}=\dfrac{1}{2}|-3*3+3|=\dfrac{1}{2}|-9+3|=\dfrac{1}{2}|-6|=\dfrac{6}{2}=3\\ a_{14}=\dfrac{1}{2}|-3*1+4|=\dfrac{1}{2}|-3+4|=\dfrac{1}{2}|1|=\dfrac{1}{2}\\ a_{24 } =\dfrac{1}{2}|-3*2+4|=\dfrac{1}{2}|-6+4|=\dfrac{1}{2}|-2|=\dfrac{2}{2}=1\\ a_{34} =\dfrac{1}{2}|-3*3+4|=\dfrac{1}{2}|-9+4|=\dfrac{1}{2}|-5|=\dfrac{5}{2}$$\\ Therefore , the required matrix is A=\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\[0.3em] \frac{5}{2} & 2 & \dfrac{3}{2} & 1 \\[0.3em] 4 & \frac{7}{2} & 3 &\frac{5}{2} \end{bmatrix} (ii) Given :\\ a _{ij }= 2 i - j , i = 1, 2,3 and j = 1, 2,3, 44\\ Thus, we have\\ \bullet a _{11} = 2 * 1 - 1= 2 - 1=1\\ \bullet a_{21}=2*2-1=4-1=3\\ \\ \bullet a_{31}= 2*3-1=6-1=5\\ \\ \bullet a_{12} =2*1-2=2-2=0\\ \\ \bullet a_{22} =2*2-2=4-2=2\\ \\ \bullet a_{32}=2*3-2=6-2=4\\ \\ \bullet a_{13}=2*1-3=2-3=-1\\ \\ \bullet a_{23} =2*2-3=4-3=1\\ \\ \bullet a_{33}=2*3-3=6-3=3\\ \\ \bullet a_{14}=2*1-4=2-4=-2\\ \\ \bullet a_{24}=2*2-4=4-4=0\\ \\ \bullet a_{34} =2*3-4=6-4=2$$\\$ Therefore ,the required matrix is $A=\begin{bmatrix} 1 &0&-1&-2 \\[0.3em] 3&2&1&0 \\[0.3em] 5&4&3&2 \end{bmatrix}$

5   Find the value of x, y and z from the following equation:$\\$ $(i) \begin{bmatrix} 4& 3 \\[0.3em] x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\[0.3em] 1& 5 \end{bmatrix}$ $(ii) \begin{bmatrix} x+y &2 \\[0.3em] 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 &2 \\[0.3em] 5 & 8 \end{bmatrix}$ $(iii) \begin{bmatrix} x+y +z \\[0.3em] x+z\\[0.3em] y+z \end{bmatrix}=\begin{bmatrix} 9 \\[0.3em] 5\\[0.3em] 7 \end{bmatrix}$

$(ii) \begin{bmatrix} x+y &2 \\[0.3em] 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 &2 \\[0.3em] 5 & 8 \end{bmatrix}$$\\ As the given matrices are equal, their corresponding elements are also equal.\\ Comparing the corresponding elements, we get:\\ x + y = 6, xy = 8, 5 +z = 5 Now, \\ 5 + z = 5 \implies z = 0$$\\$ Using $( x - y )^2=(x+y)^2- 4 xy$ , we get$\\$ $\implies (x-y)^2=36-32=4\\ \implies x-y=\pm 2$$\\ • When x - y = 2 and x + y =6 ,we get x = 4 and y = 2\\ • When x - y =- 2 and x + y = 6 we get x = 2 and y = 4\\ \therefore x =4, y = 2, and z = 0 or x = 2 , y = 4 , and z = 0 (i) Given : \begin{bmatrix} 4& 3 \\[0.3em] x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\[0.3em] 1& 5 \end{bmatrix}$$\\$ As the given matrices are equal, their corresponding elements are also equal.Comparing the corresponding elements, we get:$\\$$x = 1, y = 4 and z = 3 (iii) \begin{bmatrix} x+y +z \\[0.3em] x+z\\[0.3em] y+z \end{bmatrix}=\begin{bmatrix} 9 \\[0.3em] 5\\[0.3em] 7 \end{bmatrix} As the two matrices are equal, their corresponding elements are also equal.\\ Comparing the corresponding elements, we get:\\ x + y + z = 9 ...(1)\\ x + z = 5 ...(2)\\ y + z = 7 ...(3)$$\\$ From (1) and (2), we have:$\\$ $y + 5 = 9 \implies y = 4$$\\ From (3), we have:\\ 4 + z = 7 \implies z = 3\\ \therefore x + z = 5 \implies x = 2\\ \therefore x = 2, y = 4, and z = 3 6 Find the value of a, b , c , and d from the equation:\\ \begin{bmatrix} a-b & 2a+c \\[0.3em] 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\[0.3em] 0 & 13 \end{bmatrix} ##### Solution : \begin{bmatrix} a-b & 2a+c \\[0.3em] 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\[0.3em] 0 & 13 \end{bmatrix}$$\\$ As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:$\\$ $a - b =- 1 ........(1)\\ 2 a -b = 0........(2)\\ 2 a + c = 5.......(3)\\ 3 c + d = 13$$\\ From (2), we have:\\ b = 2 a$$\\$ Then, from (1), we have:$\\$ $a - 2 a =- 1\\ \implies a = 1\\ \implies b = 2$$\\ Now, from (3), we have:\\ 2 * 1 = c - 5\\ \implies c = 3$$\\$ From (4) we have:$\\$ $3 * 3 + d = 13\\ \implies 9 + d = 13 \implies d = 4 \\ \therefore a = 1, b = 2, c = 3,$ and $d = 4$

7   $A=\begin{bmatrix} a_{ij} \end{bmatrix}_{m*n}$ is a square matrix, if

The correct answer is $C.$$\\ It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.\\ Therefore, A =\begin{bmatrix} a_{ i j} \end{bmatrix}_{m*n} is a square matrix, if m = n. 8 Which of the given values of x and y make the following pair of matrices equal\\ \begin{bmatrix} 3x+7 & 5 \\[0.3em] y+1 & 2-3x \end{bmatrix} =\begin{bmatrix} 0 & y-2 \\[0.3em] 8 & 4 \end{bmatrix} ##### Solution : The Correct answer is B. \\ It is given that \begin{bmatrix} 3x+7 & 5 \\[0.3em] y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \\[0.3em] 8 & 4 \end{bmatrix} Equating the corresponding elements, we get:\\ 3x+7 = 0 \implies x=\dfrac{-7}{3}\\ 5=y-2\implies y=7\\ y+1=8 \implies y=7\\ 2-3x =4\implies x=-\dfrac{2}{3}$$\\$ We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.$\\$ Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

9   The number of all possible matrices of order $3 * 3$ with each entry $0$ or $1$ is:

##### Solution :

The correct answer is $D.$ The given matrix of the order $3 * 3$ has $9$ elements and each of these elements can be either $0$ or $1.$$\\ Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 2 ^9 = 512 10 Let A = \begin{bmatrix} 2 & 4 \\[0.3em] 3&2 \end{bmatrix} ,B=\begin{bmatrix} 1&3 \\[0.3em] -2 & 5 \end{bmatrix}, C=\begin{bmatrix} -2 & 5\\[0.3em] 3 & 4 \end{bmatrix}$$\\$ Find each of the following$\\$ $(i) A + B \ \ \ \ \ \ \ \ \ \ \ \ (ii) A - B \\ (iii) 3A -C \ \ \ \ \ \ \ (iv) AB \\(v) BA$

##### Solution :

$(i) A+B =\begin{bmatrix} 2 & 4 \\[0.3em] 3&2 \end{bmatrix}+\begin{bmatrix} 1 & 3 \\[0.3em] -2&5 \end{bmatrix}=\begin{bmatrix} 2+1 & 4+3 \\[0.3em] 3-2&2+5 \end{bmatrix}=\begin{bmatrix} 3 & 7 \\[0.3em] 1&7\end{bmatrix}$ $\\$ $(i) A-B =\begin{bmatrix} 2 & 4 \\[0.3em] 3&2 \end{bmatrix}-\begin{bmatrix} 1 & 3 \\[0.3em] -2&5 \end{bmatrix}=\begin{bmatrix} 2-1 & 4-3 \\[0.3em] 3-(-2)&2-5 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\[0.3em] 5&-3\end{bmatrix}$ $\\$ $(iii)3A-C=3\begin{bmatrix} 2 & 4 \\[0.3em] 3&2 \end{bmatrix}-\begin{bmatrix} -2 & 5\\[0.3em] 3 & 4 \end{bmatrix} =\begin{bmatrix} 3*2 & 3*4 \\[0.3em] 3*3&3*2 \end{bmatrix}-\begin{bmatrix} -2 & 5\\[0.3em] 3 & 4 \end{bmatrix}$$\\ =\begin{bmatrix} 6& 12\\[0.3em] 9 & 6 \end{bmatrix}-\begin{bmatrix} -2 & 5\\[0.3em] 3 & 4 \end{bmatrix}$$\\$ $\begin{bmatrix} 6+2 & 12-5\\[0.3em] 9-3 & 6-4 \end{bmatrix}$$\\ \begin{bmatrix} 8 & 7\\[0.3em] 6 & 2 \end{bmatrix}$$\\$ $\\$ (iv) Matrix $A$ has $2$ columns. This number is equal to the number of rows in matrix $B.$ Therefore, $AB$ is defined as:$\\$ $AB=\begin{bmatrix} 2 & 4 \\[0.3em] 3&2 \end{bmatrix}\begin{bmatrix} 1&3 \\[0.3em] -2 & 5 \end{bmatrix}$$\\ \begin{bmatrix} 2(1)+4(-2)&2(3)+4(5) \\[0.3em] 3(1)+2(-2) &3(3)+2( 5) \end{bmatrix}\\ \begin{bmatrix} 2-8&6+20 \\[0.3em] 3-4 & 9+10 \end{bmatrix}$$\\$ $\begin{bmatrix} -6&26 \\[0.3em] -1 & 19 \end{bmatrix}$ $\\$ $\\$ (v) matrix $B$ has $2$ columns. This number is equal to the number of rows in matrix $A$. Therefore, $BA$ is defined as:$\\$ $BA=\begin{bmatrix} 1&3 \\[0.3em] -2 & 5 \end{bmatrix}\begin{bmatrix} 2 & 4 \\[0.3em] 3&2 \end{bmatrix}$$\\ =\begin{bmatrix} 1(2)+3(3)&1(4)+3(2) \\[0.3em] -2(2)+5(3) & -2(4)+5(2) \end{bmatrix}$$\\$ $=\begin{bmatrix} 2+9&4+6 \\[0.3em] -4+15 & -8+10 \end{bmatrix}=\begin{bmatrix} 11&10 \\[0.3em] 11 & 2 \end{bmatrix}$

11   Compute the following:$\\$ $(i)\begin{bmatrix} a & b \\[0.3em] -b & a \end{bmatrix}+\begin{bmatrix} a & b \\[0.3em] b & a \end{bmatrix}$$\\ (ii)\begin{bmatrix} a^2+b^2 & b^2+ c^2 \\[0.3em] a^2+c^2 & a^2+b^2 \end{bmatrix}+\begin{bmatrix} 2ab & 2bc \\[0.3em] -2ac & -2ab \end{bmatrix}$$\\$ $(iii) \begin{bmatrix} -1 & 4 &-6 \\[0.3em] 8 & 5 & 16 \\[0.3em] 2 & 8 &5 \end{bmatrix}+\begin{bmatrix} 12 & 7 & 6 \\[0.3em] 8 & 0 & 5 \\[0.3em] 3 & 2 & 4 \end{bmatrix}$$\\ (iv)\begin{bmatrix} \cos^2 x & \sin^2 x \\[0.3em] \sin^2 x & \cos^2 x \end{bmatrix}+\begin{bmatrix} \sin^2 x & \cos^2 x \\[0.3em] \cos^2 x & \sin^2 x \end{bmatrix} ##### Solution : (ii)\begin{bmatrix} a^2+b^2 & b^2+ c^2 \\[0.3em] a^2+c^2 & a^2+b^2 \end{bmatrix}+\begin{bmatrix} 2ab & 2bc \\[0.3em] -2ac & -2ab \end{bmatrix}$$\\$ $=\begin{bmatrix} a^2+b^2+ 2ab & b^2+c^2+2bc \\[0.3em] a^2+c^2 -2ac & a^2+b^2-2ab \end{bmatrix}$$\\ =\begin{bmatrix} (a+b)^2 & (b+c)^2 \\[0.3em] (a-c)^2 & (a-b)^2 \end{bmatrix}$$\\$

$(iv)\begin{bmatrix} \cos^2 x & \sin^2 x \\[0.3em] \sin^2 x & \cos^2 x \end{bmatrix}+\begin{bmatrix} \sin^2 x & \cos^2 x \\[0.3em] \cos^2 x & \sin^2 x \end{bmatrix}$$\\ =\begin{bmatrix} \cos^2 x+ \sin^2 x & \sin^2 x+\cos^2 x \\[0.3em] \sin^2 x+\cos^2 x & \cos^2 x + \sin^2 x \end{bmatrix}$$\\$ $=\begin{bmatrix} 1 &1 \\[0.3em] 1 &1 \end{bmatrix}$$\ \ \ \ \ \$$ ( \therefore \cos^2 x=1)$

$(i)\begin{bmatrix} a & b \\[0.3em] -b & a \end{bmatrix}+\begin{bmatrix} a & b \\[0.3em] b & a \end{bmatrix}=\begin{bmatrix} a+a & b+b \\[0.3em] -b+b & a+a \end{bmatrix}=\begin{bmatrix} 2a & 2b \\[0.3em] 0 & 2a \end{bmatrix}$$\\ (ii)\begin{bmatrix} a^2+b^2 & b^2+ c^2 \\[0.3em] a^2+c^2 & a^2+b^2 \end{bmatrix}+\begin{bmatrix} 2ab & 2bc \\[0.3em] -2ac & -2ab \end{bmatrix}$$\\$ $=\begin{bmatrix} a^2+b^2+ 2ab & b^2+c^2+2bc \\[0.3em] a^2+c^2 -2ac & a^2+b^2-2ab \end{bmatrix}$$\\ =\begin{bmatrix} (a+b)^2 & (b+c)^2 \\[0.3em] (a-c)^2 & (a-b)^2 \end{bmatrix}$$\\$ $(iii) \begin{bmatrix} -1 & 4 &-6 \\[0.3em] 8 & 5 & 16 \\[0.3em] 2 & 8 &5 \end{bmatrix}+\begin{bmatrix} 12 & 7 & 6 \\[0.3em] 8 & 0 & 5 \\[0.3em] 3 & 2 & 4 \end{bmatrix}$$\\ = \begin{bmatrix} -1+12 & 4+7 &-6+6 \\[0.3em] 8+8 & 5+0 & 16+5 \\[0.3em] 2+3 & 8+2 &5+4 \end{bmatrix}$$\\$ $= \begin{bmatrix} 11 & 11 &0 \\[0.3em] 16&5&21 \\[0.3em] 5 & 10 &9 \end{bmatrix}$$\\ (iv)\begin{bmatrix} \cos^2 x & \sin^2 x \\[0.3em] \sin^2 x & \cos^2 x \end{bmatrix}+\begin{bmatrix} \sin^2 x & \cos^2 x \\[0.3em] \cos^2 x & \sin^2 x \end{bmatrix}$$\\$ $=\begin{bmatrix} \cos^2 x+ \sin^2 x & \sin^2 x+\cos^2 x \\[0.3em] \sin^2 x+\cos^2 x & \cos^2 x + \sin^2 x \end{bmatrix}$$\\ =\begin{bmatrix} 1 &1 \\[0.3em] 1 &1 \end{bmatrix}$$\ \ \ \ \ \ $$( \therefore \cos^2 x=1) (iii) \begin{bmatrix} -1 & 4 &-6 \\[0.3em] 8 & 5 & 16 \\[0.3em] 2 & 8 &5 \end{bmatrix}+\begin{bmatrix} 12 & 7 & 6 \\[0.3em] 8 & 0 & 5 \\[0.3em] 3 & 2 & 4 \end{bmatrix}$$\\$ $= \begin{bmatrix} -1+12 & 4+7 &-6+6 \\[0.3em] 8+8 & 5+0 & 16+5 \\[0.3em] 2+3 & 8+2 &5+4 \end{bmatrix}$$\\ = \begin{bmatrix} 11 & 11 &0 \\[0.3em] 16&5&21 \\[0.3em] 5 & 10 &9 \end{bmatrix}$$\\$

12   Compute the indicated products$\\$ $(i)\begin{bmatrix} a &b \\[0.3em] -b & a \end{bmatrix}\begin{bmatrix} a & -b \\[0.3em] b & a \end{bmatrix}$$\\ (ii)\begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\begin{bmatrix} 2 & 3 &4 \end{bmatrix}$$\\$ $(iii)\begin{bmatrix} 1 & -2 \\[0.3em] 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 2 & 3 & 1 \end{bmatrix}$$\\ (iv) \begin{bmatrix} 2 &3 &4 \\[0.3em] 3 & 4 &5 \\[0.3em] 4 & 5& 6 \end{bmatrix}\begin{bmatrix} 1 & -3 &5 \\[0.3em] 0 & 2 & 4 \\[0.3em] 3 & 0 &5 \end{bmatrix}$$\\$ $(v)\begin{bmatrix} 2 & 1 \\[0.3em] 3 & 2\\[0.3em] -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\[0.3em] -1 & 2 & 1 \end{bmatrix}$$\\ (vi)\begin{bmatrix} 3 & -1 & 3 \\[0.3em] -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\[0.3em] 1 & 0 \\[0.3em] 3 & 1 \end{bmatrix} ##### Solution : (iii) \begin{bmatrix} 1 & -2 \\[0.3em] 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 2 & 3 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\[0.3em] 2(1)+3(2) & 2(2)+ 3(3) & 2(3)+3(1) \end{bmatrix}$$\\ =\begin{bmatrix} 1-4 & 2-6 & 3-2 \\[0.3em] 2+6 & 4+9 & 6+3 \end{bmatrix} \\ = \begin{bmatrix} -3 & -4 & 1 \\[0.3em] 8 & 13 & 9 \end{bmatrix} (v)\begin{bmatrix} 2 & 1 \\[0.3em] 3 & 2\\[0.3em] -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\[0.3em] -1 & 2 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\[0.3em] 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1)\\[0.3em] -1(1)+1(-1) & -1(0)+1(2) & -(1)+1(1) \end{bmatrix}$$\\ =\begin{bmatrix} 2-1 & 0+2 & 2+1 \\[0.3em] 3-2 & 0+4 & 3+2\\[0.3em] -1-1 & 0+2 & -1+1 \end{bmatrix}=\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 1 & 4 & 5\\[0.3em] -2 & 2&0 \end{bmatrix}$$\\$ $\\$ $(vi)\begin{bmatrix} 3 & -1 & 3 \\[0.3em] -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\[0.3em] 1 & 0 \\[0.3em] 3 & 1 \end{bmatrix}$$\\ =\begin{bmatrix} 3(2)-1(1)+3(3) & 3(-3) -1(0) +3(1) \\[0.3em] -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{bmatrix}$$\\$ $=\begin{bmatrix} 6-1+9 & -9-0+3 \\[0.3em] -2+0+6 & 3+ 0+2 \end{bmatrix}=\begin{bmatrix} 14 & -6 \\[0.3em] 4 & 5 \end{bmatrix}$

$(i)\begin{bmatrix} a &b \\[0.3em] -b & a \end{bmatrix}\begin{bmatrix} a & -b \\[0.3em] b & a \end{bmatrix}$$\\ =\begin{bmatrix} a(a)+b(b)& a(-b)+b(a) \\[0.3em] -b(a)+a(b) & -b(-b) + a(a) \end{bmatrix}$$\\$ $=\begin{bmatrix} a^2+b^2 & -ab+ ab \\[0.3em] -ab+ab & b^2+a^2 \end{bmatrix}$$\\ =\begin{bmatrix} a^2+b^2 & 0 \\[0.3em] 0 & a^2 + b^2 \end{bmatrix}$$\\$ $(ii)\begin{bmatrix} 1 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}\begin{bmatrix} 2 & 3 &4 \end{bmatrix}$$\\ =\begin{bmatrix} 1 (2) & 1(3) & 1(4) \\[0.3em] 2(2) & 2(3 ) & 2(4) \\[0.3em] 3(2) & 3(3) & 3(4) \end{bmatrix}=\begin{bmatrix} 2 & 3 &4 \\[0.3em] 4 & 6 & 8 \\[0.3em] 6 & 9 & 12 \end{bmatrix}$$\\$ $(iii) \begin{bmatrix} 1 & -2 \\[0.3em] 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 2 & 3 & 1 \end{bmatrix}$$\\ =\begin{bmatrix} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\[0.3em] 2(1)+3(2) & 2(2)+ 3(3) & 2(3)+3(1) \end{bmatrix}$$\\$ $=\begin{bmatrix} 1-4 & 2-6 & 3-2 \\[0.3em] 2+6 & 4+9 & 6+3 \end{bmatrix}$ $\\$ $= \begin{bmatrix} -3 & -4 & 1 \\[0.3em] 8 & 13 & 9 \end{bmatrix}$

$(iv) \begin{bmatrix} 2 &3 &4 \\[0.3em] 3 & 4 &5 \\[0.3em] 4 & 5& 6 \end{bmatrix}\begin{bmatrix} 1 & -3 &5 \\[0.3em] 0 & 2 & 4 \\[0.3em] 3 & 0 &5 \end{bmatrix}$$\\ = \begin{bmatrix} 2(1)+3(0)+4(3 ) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\[0.3em] 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5)\\[0.3em] 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5) \end{bmatrix}$$\\$ $=\begin{bmatrix} 2+0+12 & -6+6+0 & 10 +12+20 \\[0.3em] 3+0+15 & -9+8+0 & 15 + 16+25\\[0.3em] 4+0+18 & -12 +10+0 & 20 +20 +30 \end{bmatrix}$$\\ =\begin{bmatrix} 14 & 0 & 42 \\[0.3em] 18 & -1 & 56\\[0.3em] 22 &-2 & 70 \end{bmatrix}$$\\$ $(v)\begin{bmatrix} 2 & 1 \\[0.3em] 3 & 2\\[0.3em] -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\[0.3em] -1 & 2 & 1 \end{bmatrix}$$\\ =\begin{bmatrix} 2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\[0.3em] 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1)\\[0.3em] -1(1)+1(-1) & -1(0)+1(2) & -(1)+1(1) \end{bmatrix}$$\\$ $=\begin{bmatrix} 2-1 & 0+2 & 2+1 \\[0.3em] 3-2 & 0+4 & 3+2\\[0.3em] -1-1 & 0+2 & -1+1 \end{bmatrix}=\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 1 & 4 & 5\\[0.3em] -2 & 2&0 \end{bmatrix}$$\\ \\ (vi)\begin{bmatrix} 3 & -1 & 3 \\[0.3em] -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\[0.3em] 1 & 0 \\[0.3em] 3 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 3(2)-1(1)+3(3) & 3(-3) -1(0) +3(1) \\[0.3em] -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{bmatrix}$$\\ =\begin{bmatrix} 6-1+9 & -9-0+3 \\[0.3em] -2+0+6 & 3+ 0+2 \end{bmatrix}=\begin{bmatrix} 14 & -6 \\[0.3em] 4 & 5 \end{bmatrix} 13 If A=\begin{bmatrix} 1 & 2 &-3 \\[0.3em] 5 & 0 & 2\\[0.3em] 1 & -1 & 1 \end{bmatrix} ,B =\begin{bmatrix} 3 & -1 & 2 \\[0.3em] 4 & 2 & 5\\[0.3em] 2 & 0 & 3 \end{bmatrix},and C=\begin{bmatrix} 4 & 1 & 2 \\[0.3em] 0 & 3 & 2\\[0.3em] 1 & -2 & 3 \end{bmatrix} ,then \\ Compute (A+B) and (B-C). Also , verify that A+(B-C) =(A+B)-C ##### Solution : A+B=\begin{bmatrix} 1 & 2 &-3 \\[0.3em] 5 & 0 & 2\\[0.3em] 1 & -1 & 1 \end{bmatrix}+\begin{bmatrix} 3 & -1 & 2 \\[0.3em] 4 & 2 & 5\\[0.3em] 2 & 0 & 3 \end{bmatrix}$$\\$ $=\begin{bmatrix} 1+3 & 2-1 & -3+2 \\[0.3em] 5+4 & 0+2 & 2+5\\[0.3em] 1+2 & -1+0 & 1+3 \end{bmatrix}=\begin{bmatrix} 4 & 1 & -1 \\[0.3em] 9& 2 & 7\\[0.3em] 3 & -1 & 4 \end{bmatrix}$$\\ B-C =\begin{bmatrix} 3 & -1 & 2 \\[0.3em] 4 & 2 & 5\\[0.3em] 2 & 0 & 3 \end{bmatrix}-\begin{bmatrix} 4 & 1 & 2 \\[0.3em] 0 & 3 & 2\\[0.3em] 1 & -2 & 3 \end{bmatrix}$$\\$ $A+(B-C)=\begin{bmatrix} 1 & 2 &-3 \\[0.3em] 5 & 0 & 2\\[0.3em] 1 & -1 & 1 \end{bmatrix}+\begin{bmatrix} -1 & -2 &0 \\[0.3em] 4 & -1 & 3\\[0.3em] 1 & 2 & 0 \end{bmatrix}$$\\ \begin{bmatrix} 1+(-1) & 2+(-2) & -3+0 \\[0.3em] 5+4 & 0+(-1) & 2+3\\[0.3em] 1+1 & -1+2 & 1+0 \end{bmatrix}=\begin{bmatrix} 0 & 0 & -3 \\[0.3em] 9 & -1 & 5\\[0.3em] 2 & 1 & 1 \end{bmatrix}$$\\$ $(A+B)-C=\begin{bmatrix} 4 & 1 & -1 \\[0.3em] 9 & 2 & 7\\[0.3em] 3 & -1 & 4 \end{bmatrix}-\begin{bmatrix} 4 & 1 & 2 \\[0.3em] 0 & 3 & 2\\[0.3em] 1 & -2 & 3 \end{bmatrix}$$\\ =\begin{bmatrix} 4-4 & 1-1 & -1-2 \\[0.3em] 9-0 & 2-3 & 7-2\\[0.3em] 3-1 & -1-(-2) & 4-3 \end{bmatrix}$$\\$ $=\begin{bmatrix} 0&0&-3 \\[0.3em] 9& -1 & 5\\[0.3em] 2 & 1 & 1 \end{bmatrix}$$\\ Hence, we have verified that A +( B - C )=( A +B )- C . 14 IfA=\begin{bmatrix} \frac{2}{3} &1& \frac{5}{3} \\[0.3em] \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\[0.3em] \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} and B=\begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\[0.3em] \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\[0.3em] \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} then compute 3 A - 5 B ##### Solution : 3 A -5 B =3\begin{bmatrix} \frac{2}{3} &1& \frac{5}{3} \\[0.3em] \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\[0.3em] \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}-5\begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\[0.3em] \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\[0.3em] \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$$\\$ $=\begin{bmatrix} 2 & 3 & 5 \\[0.3em] 1 & 2 & 4 \\[0.3em] 7 & 6 & 2 \end{bmatrix}-\begin{bmatrix} 2 & 3 & 5 \\[0.3em] 1 & 2 & 4\\[0.3em] 7 & 6 & 2 \end{bmatrix}$ $\\$ $=\begin{bmatrix} 0& 0& 0 \\[0.3em] 0& 0 &0 \\[0.3em] 0 & 0 &0 \end{bmatrix}$

15   Simplify $\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\[0.3 em] -\sin \theta & \cos \theta \end{bmatrix}+\sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\[0.3em] \cos \theta & \sin \theta \end{bmatrix}$

##### Solution :

$\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\[0.3em] -\sin \theta & \cos \theta \end{bmatrix}+\sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\[0.3em] \cos \theta & \sin \theta \end{bmatrix}$$\\ =\begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\[0.3em] -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix}+\begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\[0.3em] \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$$\\$ $=\begin{bmatrix} \cos^2 \theta +\sin^2 \theta & \cos \theta \sin \theta -\sin \theta \cos \theta \\[0.3 em] -\sin \theta \cos \theta + \sin \theta \cos \theta & \cos^2 \theta +\sin ^ 2 \theta \end{bmatrix}$$\\ =\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix} \ \ \ \ \ \ \ \ \ \ \ (\therefore \sin^2 \theta =1) 16 Find X and Y ,if\\ (i)X+Y=\begin{bmatrix} 7 & 0 \\[0.3em] 2 & 5 \end{bmatrix} and X-Y = \begin{bmatrix} 3 & 0\\[0.3em] 0 & 3 \end{bmatrix}$$\\$ $(ii) 2 X + 3Y=\begin{bmatrix} 2 & 3 \\[0.3em] 4 & 0\end{bmatrix}$ and $3X+2Y=\begin{bmatrix} 2 & -2 \\[0.3em] -1 & 5 \end{bmatrix}$

##### Solution :

$(ii) 2X+3Y =\begin{bmatrix} 2 & 3\\[0.3em] 4 & 0\end{bmatrix}$.......(3) $3X+2Y=\begin{bmatrix} 2 & -2\\[0.3em] -1 & 5\end{bmatrix}$.....(4) Multiplying equation (3) with (2), we get$\\$ $2(2X+3Y)=2\begin{bmatrix} 2 & 3\\[0.3em] 4 & 0\end{bmatrix}\\ \implies 4X+6Y=\begin{bmatrix} 4 & 6\\[0.3em] 8 & 0\end{bmatrix}....(5)$$\\ Multiplying equation (4) with (3), we get\\ 3(3X+2Y)=3 \begin{bmatrix} 2 & -2\\[0.3em] -1 & 5\end{bmatrix}\\ \implies 9X+6Y=\begin{bmatrix} 6 & -6\\[0.3em] -3 & 15\end{bmatrix}.....(6)$$\\$ From(5) and (6),we have$\\$ $(4X+6Y)-(9X+6Y) =\begin{bmatrix} 4 & 6\\[0.3em] 8& 0\end{bmatrix}-\begin{bmatrix} 6 & -6\\[0.3em] -3 & 15\end{bmatrix}\\ \implies -5X=\begin{bmatrix} 4-6 & 6-(-6)\\[0.3em] 8-(-3) & 0-15\end{bmatrix}=\begin{bmatrix} -2 & 12\\[0.3em] 11 & -15\end{bmatrix}\\ \therefore X =-\dfrac{1}{5}\begin{bmatrix} -2 & 12\\[0.3em] 11 & -15\end{bmatrix}=\begin{bmatrix} \frac{2}{5} & -\frac{12}{5}\\[0.3em] -\frac{11}{5} & 3 \end{bmatrix}\\ 2X+3Y=\begin{bmatrix} 2 & 3\\[0.3em] 4 & 0\end{bmatrix}$$\\ Now ,\\ (i)X+Y=\begin{bmatrix} 7 & 0 \\[0.3em] 2 & 5 \end{bmatrix} ......(1) X-Y = \begin{bmatrix} 3 & 0\\[0.3em] 0 & 3 \end{bmatrix}.......(2)\\ Adding equations(1) and (2) , we get:\\ 2X=\begin{bmatrix} 7 & 0 \\[0.3em] 2 & 5 \end{bmatrix}+X-Y = \begin{bmatrix} 3 & 0\\[0.3em] 0 & 3 \end{bmatrix}$$\\$ $\begin{bmatrix} 7+3 & 0+0\\[0.3em] 2+0 & 5+3 \end{bmatrix} =\begin{bmatrix} 10 & 0\\[0.3em] 2 & 8 \end{bmatrix}$$\\ \therefore X=\dfrac{1}{2}\begin{bmatrix} 10 & 0\\[0.3em] 2 & 8 \end{bmatrix}=\begin{bmatrix} 5 & 0\\[0.3em] 1&4 \end{bmatrix}$$\\$ Now, $X+Y=\begin{bmatrix} 7 & 0\\[0.3em] 2 & 5\end{bmatrix}\\ \implies \begin{bmatrix} 5 & 0\\[0.3em] 1&4 \end{bmatrix}+Y=\begin{bmatrix} 7 & 0\\[0.3em] 2 & 5\end{bmatrix}\\ Y=\begin{bmatrix} 7 & 0\\[0.3em] 2 & 5\end{bmatrix}-\begin{bmatrix} 5 & 0\\[0.3em] 1&4 \end{bmatrix}\\ \implies Y=\begin{bmatrix} 7-5 & 0-0\\[0.3em] 2-1 & 5-4\end{bmatrix}\\ \therefore Y=\begin{bmatrix} 2 & 0\\[0.3em] 1 & 1\end{bmatrix}$

$\implies 2\begin{bmatrix} \frac{2}{5} & -\frac{12}{5}\\[0.3em] -\frac{11}{5} & 3 \end{bmatrix}+3Y=\begin{bmatrix} 2 & 3\\[0.3em] 4 & 0\end{bmatrix}$$\\ \implies \begin{bmatrix} \frac{4}{5} &-\frac{24}{5}\\[0.3em] -\frac{22}{5} & 6 \end{bmatrix}+3Y = \begin{bmatrix} 2 & 3\\[0.3em] 4 & 0\end{bmatrix}$$\\$ $\implies 3Y =\begin{bmatrix} 2& 3\\[0.3em] 4 &0\end{bmatrix}-\begin{bmatrix} \frac{4}{5} &-\frac{24}{5}\\[0.3em] -\frac{22}{5} & 6 \end{bmatrix}$$\\ 3Y=\begin{bmatrix} 2-\frac{4}{5} & 3+\frac{24}{5}\\[0.3em] 4+\frac{22}{5} & 0-6\end{bmatrix} =\begin{bmatrix} \frac{6}{5} & \frac{39}{5}\\[0.3em] \frac{42}{5} & -6\end{bmatrix}$$\\$ $\therefore Y=\dfrac{1}{3}\begin{bmatrix} \frac{6}{5} & \frac{13}{5}\\[0.3em] \frac{42}{5} & -6\end{bmatrix}=\begin{bmatrix} \frac{2}{5} & \frac{13}{5}\\[0.3em] \frac{14}{5} & -2\end{bmatrix}$

17   Find $X ,$if $Y=\begin{bmatrix} 3 & 2\\[0.3em] 1 & 4 \end{bmatrix}$ and $2X+Y=\begin{bmatrix} 1 & 0\\[0.3em] -3 & 2\end{bmatrix}$

##### Solution :

$2X+Y=\begin{bmatrix} 1 & 0\\[0.3em] -3 & 2 \end{bmatrix}$$\\ \implies 2X+\begin{bmatrix} 3 & 2\\[0.3em] 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 &0\\[0.3em] -3 & 2 \end{bmatrix}$$\\$ $\implies 2X=\begin{bmatrix} 1 & 0\\[0.3em] -3 &2 \end{bmatrix}-\begin{bmatrix} 3 & 2\\[0.3em] 1 & 4 \end{bmatrix}=\begin{bmatrix} 1-3 & 0-2\\[0.3em] -3-1 & 2-4 \end{bmatrix}$$\\ \implies 2X=\begin{bmatrix} -2 &-2\\[0.3em] -4 &-2 \end{bmatrix}$$\\$ $\therefore X=\dfrac{1}{2}\begin{bmatrix} -2 &-2\\[0.3em] -4 & -2 \end{bmatrix}= \begin{bmatrix} -1 & -1\\[0.3em] -2 & -1 \end{bmatrix}$

18   Find $X$ and $Y$,if $2 \begin{bmatrix} 1 & 3\\[0.3em] 0 & x \end{bmatrix}+\begin{bmatrix} y & 0\\[0.3em] 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6\\[0.3em] 1 & 8 \end{bmatrix}$

$2 \begin{bmatrix} 1 & 3\\[0.3em] 0 & x \end{bmatrix}+\begin{bmatrix} y & 0\\[0.3em] 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6\\[0.3em] 1 & 8 \end{bmatrix}$$\\ \implies \begin{bmatrix} 2 & 6\\[0.3em] 0 & 2x \end{bmatrix}+\begin{bmatrix} y & 0\\[0.3em] 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6\\[0.3em] 1 & 8 \end{bmatrix}$$\\$ $\begin{bmatrix} 2+y & 6\\[0.3em] 1 &2x+ 2 \end{bmatrix}=\begin{bmatrix} 5 & 6\\[0.3em] 1 & 8 \end{bmatrix}$$\\ Comparing the corresponding elements of these two matrices, we have:\\ 2 + y = 5 \implies y = 3\\ 2 x +2 = 8 \implies x = 3\\ \therefore x = 3 \text{and} \ y =3 19 Solve the equation for x, y, z and t if\\ 2\begin{bmatrix} x & z \\[0.3em] y & t \end{bmatrix}+3 \begin{bmatrix} 1 & -1 \\[0.3em] 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\[0.3em] 4 &6 \end{bmatrix} ##### Solution : 2\begin{bmatrix} x & z \\[0.3em] y & t \end{bmatrix}+3 \begin{bmatrix} 1 & -1 \\[0.3em] 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\[0.3em] 4 &6 \end{bmatrix}$$\\$ $\implies \begin{bmatrix} 2x & 2z \\[0.3em] 2y & 2t\end{bmatrix}+\begin{bmatrix} 3 & -3 \\[0.3em] 0 & 6\end{bmatrix}=\begin{bmatrix} 9 & 15 \\[0.3em] 12 & 18\end{bmatrix}$$\\ \implies \begin{bmatrix} 2x+3 & 2z-3 \\[0.3em] 2y & 2t+6 \end{bmatrix}=\begin{bmatrix} 9 & 15 \\[0.3em] 12 & 18 \end{bmatrix}$$\\$ Comparing the corresponding elements of these two matrices, we get: $\\$ $2 x + 3 = 9\\ \implies 2 x = 6\\ \implies x = 3\\ \\ 2 y = 12\\ \implies y = 6\\ \\ 2 z - 3 = 15\\ \implies 2 z = 18\\ \implies z = 9 $$\\ 2 t + 6 = 18\\ \implies 2 t = 12 \implies t =6 \therefore x = 3, y = 6, z = 9, \text{and } \ t =6 20 If x \begin{bmatrix} 2 \\[0.3em] 3 \end{bmatrix}+y \begin{bmatrix} -1 \\[0.3em] 1 \end{bmatrix}= \begin{bmatrix} 10 \\[0.3em] 5 \end{bmatrix}, find values of x and y ##### Solution : x \begin{bmatrix} 2 \\[0.3em] 3 \end{bmatrix}+y \begin{bmatrix} -1 \\[0.3em] 1\end{bmatrix}= \begin{bmatrix} 10 \\[0.3em] 5 \end{bmatrix}$$\\$ $\implies \begin{bmatrix} 2x \\[0.3em] 3x \end{bmatrix}+ \begin{bmatrix} -y \\[0.3em] y\end{bmatrix}= \begin{bmatrix} 10 \\[0.3em] 5 \end{bmatrix}$$\\ \implies \begin{bmatrix} 2x-y \\[0.3em] 3x+y \end{bmatrix}+ \begin{bmatrix} 10 \\[0.3em] 5 \end{bmatrix}$$\\$Comparing the corresponding elements of these two matrices, we get:$\\$ $2 x - y = 10$ and $3 x + y = 5$$\\ Adding these two equations, we have:\\ 5 x = 15 \implies x = 3$$\\$ Now, $3 x + y = 5\\ \implies y = 5 - 3 x\\ \implies y = 5 - 9 =- 4\\ \therefore x = 3 \text{and} \ y =- 4$

21   Given $3 \begin{bmatrix} x & y \\[0.3em] z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\[0.3em] -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\[0.3em] z+w &3 \end{bmatrix},$ find the values of x,y,z and w.

$(ii) \begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}$$\\ \begin{bmatrix} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1) + 3(4) \\[0.3em] 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\[0.3em] 1(-1)+1(0)+0(2) & 1(1)+1(-1) +0(3) & 1(0)+1(1)+0(4) \end{bmatrix}$$\\$ $\begin{bmatrix} 5 & 8 & 14 \\[0.3em] 0 & -1 & 1 \\[0.3em] -1 & 0 & 1 \end{bmatrix}$ $\\$ $\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}$$\\ \begin{bmatrix} -1(1)+1(0)+0(1) & -1(2)+1(-1)+0(1) & -1(3)+ 1(0) + 0(0) \\[0.3em] 0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+(-1)(0)+1(0) \\[0.3em] 2(1)+3(0)+4(1) & 2(2)+ 3(1)+4(1)& 2(3) + 3(0) + 4(0) \end{bmatrix}$$\\$ $\begin{bmatrix} -1 & -1 & -3 \\[0.3em] 1 & 0 & 0 \\[0.3em] 6 & 11 & 6 \end{bmatrix}$$\\ \therefore \begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\neq \begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 5 & -1 \\[0.3em] 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1\\[0.3em] 3 & 4 \end{bmatrix}$$\\$ $\begin{bmatrix} 5(2)-1(3) &5(1)-1(4)\\[0.3em] 6(2) + 7(3) & 6(1)+7(4) \end{bmatrix}$$\\ \begin{bmatrix} 10-3 & 5-4\\[0.3em] 12+21 & 6+28\end{bmatrix}=\begin{bmatrix} 7 & 1\\[0.3em] 33& 34 \end{bmatrix}$$\\$ $\begin{bmatrix} 2 & 1\\[0.3em] 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1\\[0.3em] 6 & 7 \end{bmatrix}$$\\ =\begin{bmatrix} 2(5)+1(6) & 2(-1) + 1(7)\\[0.3em] 3(5)+4(6) & 3(-1) + 4 (7)\end{bmatrix}$$\\$ $=\begin{bmatrix} 10+6 & -2+7\\[0.3em] 15+24 & -3+28 \end{bmatrix}=\begin{bmatrix} 16 & 5\\[0.3em] 39 & 25\end{bmatrix}$$\\ \therefore \begin{bmatrix} 5 & -1\\[0.3em] 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1\\[0.3em] 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1 \\[0.3em] 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\[0.3em] 6 & 7 \end{bmatrix} 24 Find A^2 -5 A + 6 I if A = \begin{bmatrix} 2 & 0 & 1 \\[0.3em] 2 & 1 & 3 \\[0.3em] 1 & -1 & 0 \end{bmatrix} ##### Solution : We have A^2 = A * A$$\\$ $A^2 = AA \begin{bmatrix} 2 & 0 & 1 \\[0.3em] 2 & 1 & 3 \\[0.3em] 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\[0.3em] 2 & 1 & 3 \\[0.3em] 1 & -1 & 0 \end{bmatrix}$$\\ \begin{bmatrix} 2(2)+0(2)+1(1) & 2(0) + 0(1)+ 1(-1) & 2(1)+ 0(3)+1(0) \\[0.3em] 2(2) + 1(2) +3(1) & 2(0)+ 1(1)+3(-1) & 2(1) + 1(3) + 3(0) \\[0.3em] 1(2)+ (-1)(2)+ 0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0) \end{bmatrix}$$\\$ $\begin{bmatrix} 4+0+1 & 0+0-1 & 2+0+0 \\[0.3em] 4+2+3 & 0+1-3 & 2+ 3+0 \\[0.3em] 2-2+0 & 0-1+0 & 1-3+0\end{bmatrix}$$\\ =\begin{bmatrix} 5 & -1 & 2 \\[0.3em] 9 & -2 & 5 \\[0.3em] 0 & -1 & -2 \end{bmatrix}$$\\$ Substituting the matrices in the given equation :$A ^2 - 5 A + 6 I$$\\ =\begin{bmatrix} 5 & -1 & 2 \\[0.3em] 9 & -2 & 5 \\[0.3em] 0 & -1 & -2 \end{bmatrix}-5 \begin{bmatrix} 2 & 0 & 1 \\[0.3em] 2 & 1 & 3 \\[0.3em] 1 & -1 & 0 \end{bmatrix}+6 \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0\\[0.3em] 0& 0& 1\end{bmatrix}$$\\$ $= \begin{bmatrix} 5 & -1 & 2 \\[0.3em] 9 & -2 & 5 \\[0.3em] 0 & -1 & -2 \end{bmatrix}-\begin{bmatrix} 10 & 0 & 5 \\[0.3em] 10 & 5 & 15\\[0.3em] 5 & -5 & 0 \end{bmatrix}+ \begin{bmatrix} 6 & 0 & 0 \\[0.3em] 0 & 6 & 0\\[0.3em] 0& 0& 6\end{bmatrix}$$\\ = \begin{bmatrix} 5-10 & -1-0 & 2-5\\[0.3em] 9-10 & -2-5 & 5-15\\[0.3em] 0-5 & -1+5 & -2-0\end{bmatrix}+\begin{bmatrix} 6 & 0 & 0 \\[0.3em] 0 & 6 & 0\\[0.3em] 0& 0& 6\end{bmatrix}$$\\$ $=\begin{bmatrix} -5 & -1 & -3 \\[0.3em] -1 & -7 & -10\\[0.3em] -5 & 4 & -2\end{bmatrix}+\begin{bmatrix} 6 & 0 & 0 \\[0.3em] 0 & 6 & 0\\[0.3em] 0& 0& 6\end{bmatrix}$$\\ =\begin{bmatrix} -5+6 & -1+0 & -3+0 \\[0.3em] -1+0 & -7+6 & -10+0\\[0.3em] -5+0& 4+0& -2+6\end{bmatrix}$$\\$ $=\begin{bmatrix} 1 &-1 & -3 \\[0.3em] -1 & -1 & -10\\[0.3em] -5 & 4 & 4\end{bmatrix}$

25   If $A=\begin{bmatrix} 1 & 0 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 2 & 0 & 3 \end{bmatrix},$ prove that $A^3 - 6 A^2-6A^2+7 A + 2I =0$

##### Solution :

Substituting the matrices in the given equation $A^ 3 - 6 A^ 2 + 7 A + 2 I$$\\ =\begin{bmatrix} 21 & 0 & 34 \\[0.3em] 12 & 8 & 23 \\[0.3em] 34 & 0 & 55 \end{bmatrix}-6\begin{bmatrix} 5 & 0 & 8 \\[0.3em] 2 & 4 & 5 \\[0.3em] 8 & 0 & 13 \end{bmatrix}+\begin{bmatrix} 1 & 0 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 2 & 0 & 3 \end{bmatrix}+2\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 21 & 0 & 34 \\[0.3em] 12 & 8 & 23 \\[0.3em] 34 & 0 & 55 \end{bmatrix}-\begin{bmatrix} 30 & 0 & 48 \\[0.3em] 12 & 24 & 30\\[0.3em] 48 & 0 & 78\end{bmatrix}+\begin{bmatrix} 7 & 0 & 14 \\[0.3em] 0 & 14 & 7\\[0.3em] 14 & 0 & 21 \end{bmatrix}+\begin{bmatrix} 2 & 0 & 0\\[0.3em] 0 & 2 & 0\\[0.3em] 0 & 0 &2 \end{bmatrix}$$\\ =\begin{bmatrix} 21+7+2 & 0+0+0 & 34+14+0 \\[0.3em] 12+0+0 & 8+14+2 & 23+7 +0 \\[0.3em] 34+14+0 & 0+0+0 & 55+21+2 \end{bmatrix}-\begin{bmatrix} 30 & 0 & 48 \\[0.3em] 12 & 24 & 30\\[0.3em] 48 & 0 & 78\end{bmatrix}$$\\$ $= \begin{bmatrix} 30 & 0 & 48 \\[0.3em] 12 & 24 & 30\\[0.3em] 48 & 0 & 78\end{bmatrix}-\begin{bmatrix} 30 & 0 & 48 \\[0.3em] 12 & 24 & 30\\[0.3em] 48 & 0 & 78\end{bmatrix}$$\\ =\begin{bmatrix} 0 & 0 & 0\\[0.3em] 0 & 0 & 0\\[0.3em] 0 & 0 & 0\end{bmatrix} =0$$\\$ $\therefore A^3-6 A^2 + 7 A+2I=0$

$A^2 =AA=\begin{bmatrix} 1 & 0 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 2 & 0 & 3 \end{bmatrix}$ $\\$ $= \begin{bmatrix} 1+0+4 & 0+0+0 & 2+0+6 \\[0.3em] 0+0+2 & 0+4+0 & 0+2+3 \\[0.3em] 2+0+6 & 0+0+0 & 4+0+9 \end{bmatrix}$$\\ =\begin{bmatrix} 5 & 0 & 8 \\[0.3em] 2 & 4 & 5 \\[0.3em] 8 & 0 & 13 \end{bmatrix}$$\\$ Now, $A^3= A^2.A\\ \begin{bmatrix} 5 & 0 & 8 \\[0.3em] 2 & 4 & 5 \\[0.3em] 8 & 0 & 13 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 2 & 0 & 3 \end{bmatrix}$$\\ \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\[0.3em] 2 +0+10 & 0+8+0 & 4+4+15 \\[0.3em] 8+0+ 26 & 0+0+0 & 16+0+39 \end{bmatrix}$$\\$ $= \begin{bmatrix} 21 & 0 & 34 \\[0.3em] 12 & 8 & 23 \\[0.3em] 34 & 0 & 55\end{bmatrix}$

26   If $A=\begin{bmatrix} 3 & -2 \\[0.3em] 4 & -2 \end{bmatrix}$ and $I=\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1\end{bmatrix},$ find $k$ so that $A^2=kA-2I$

##### Solution :

$A^2=A.A=\begin{bmatrix} 3 & -2 \\[0.3em] 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\[0.3em] 4 & -2 \end{bmatrix}$$\\ =\begin{bmatrix} 3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\[0.3em] 4(3)+(-2)(4) & 4(-2)+(-2)(-2) \end{bmatrix}=\begin{bmatrix} 1 & -2 \\[0.3em] 4 & -4 \end{bmatrix}$$\\$ Now $A^2=kA-2I$$\\ \implies \begin{bmatrix} 1 & -2 \\[0.3em] 4 & -4 \end{bmatrix}=k\begin{bmatrix} 3 & -2 \\[0.3em] 4 & -2 \end{bmatrix}-2\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}$$\\$ $\implies \begin{bmatrix} 1 & -2 \\[0.3em] 4 & -4 \end{bmatrix}=\begin{bmatrix} 3k & -2k \\[0.3em] 4k & -2k \end{bmatrix}-\begin{bmatrix} 2 & 0 \\[0.3em] 0 & 2 \end{bmatrix}$$\\ = \begin{bmatrix} 1 & -2 \\[0.3em] 4 & -4 \end{bmatrix}= \begin{bmatrix} 3k-2 & -2k \\[0.3em] 4k & -2k-2 \end{bmatrix}$$\\$ Comparing the corresponding elements, we have:$\\$ $3 k - 2 = 1\\ \implies 3 k =3\\ \implies k = 1$ Thus, the value of $k$ is $1.$

27   If $A=\begin{bmatrix} 0 & -\tan\dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order $2$,show that $I+A=(I-A)$ $\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$

##### Solution :

$=\left(\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}-\begin{bmatrix} 0 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 0 \end{bmatrix}\right)$ $\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$$\\ =\begin{bmatrix} 1 & \tan \dfrac{\alpha}{2} \\[0.3em] -\tan \dfrac{\alpha}{2} & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$$\\$ $=\begin{bmatrix} \cos \alpha +\sin \alpha \tan \dfrac{\alpha}{2} & -\sin \alpha +\cos \alpha \tan \dfrac{\alpha}{2} \\[0.3em] -\cos \alpha \tan \dfrac{\alpha }{2}+\sin \alpha & \sin \alpha \tan \dfrac{\alpha}{2}+\cos \alpha \end{bmatrix}$....(2)

$=\begin{bmatrix} 1-2\sin^2 \dfrac{\alpha}{2} +2\sin^2\dfrac{\alpha}{2} & -2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} +2 \sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} -\tan \dfrac {\alpha}{2} \\[0.3em] -2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} + \tan \dfrac{\alpha}{2} +2 \sin\dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} & 2 \sin^2 \dfrac{\alpha}{2} +1-2\sin^2\dfrac{\alpha}{2} \end{bmatrix}$ $\\$ $=\begin{bmatrix} 1 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 1 \end{bmatrix}$ $\\$ Thus, from (1) and (2), we get $L.H.S. = R.H.S.$

$=\begin{bmatrix} 1-2\sin^2\dfrac{\alpha}{2} +2\sin\dfrac{\alpha}{2} -\cos\dfrac{\alpha}{2}\tan \dfrac{\alpha}{2} & -2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} +(2 \cos^2 \dfrac{alpha}{2}-1)\tan \dfrac{\alpha}{2} \\[0.3em] -(2 \cos^2 \dfrac{\alpha}{2}-1)\tan \dfrac{alpha}{2} +2 \sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} & 2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2}\tan\dfrac{\alpha}{2} +1 -2\sin^2 \dfrac{\alpha}{2} \end{bmatrix}$ $\\$ $=\begin{bmatrix} 1-2\sin^2 \dfrac{\alpha}{2} +2\sin^2\dfrac{\alpha}{2} & -2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} +2 \sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} -\tan \dfrac {\alpha}{2} \\[0.3em] -2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} + \tan \dfrac{\alpha}{2} +2 \sin\dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} & 2 \sin^2 \dfrac{\alpha}{2} +1-2\sin^2\dfrac{\alpha}{2} \end{bmatrix}$ $\\$ $=\begin{bmatrix} 1 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 1 \end{bmatrix}$ $\\$ Thus, from (1) and (2), we get $L.H.S. = R.H.S.$

$L.H.S.$ $\\$ $I+A \\ =\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 0 \end{bmatrix} \\ \begin{bmatrix} 1 & -\tan \dfrac{\alpha}{2} \\[0.3em] \tan \dfrac{\alpha}{2} & 1 \end{bmatrix} ....(1)\\ R.H.S. \\ (I-A)\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$$\\ =\left(\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}-\begin{bmatrix} 0 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 0 \end{bmatrix}\right) \begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$$\\$ $=\begin{bmatrix} 1 & \tan \dfrac{\alpha}{2} \\[0.3em] -\tan \dfrac{\alpha}{2} & 1 \end{bmatrix}$ $\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$$\\ =\begin{bmatrix} \cos \alpha +\sin \alpha \tan \dfrac{\alpha}{2} & -\sin \alpha +\cos \alpha \tan \dfrac{\alpha}{2} \\[0.3em] -\cos \alpha \tan \dfrac{\alpha }{2}+\sin \alpha & \sin \alpha \tan \dfrac{\alpha}{2}+\cos \alpha \end{bmatrix}....(2)\\ =\begin{bmatrix} 1-2\sin^2\dfrac{\alpha}{2} +2\sin\dfrac{\alpha}{2} -\cos\dfrac{\alpha}{2}\tan \dfrac{\alpha}{2} & -2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} +(2 \cos^2 \dfrac{alpha}{2}-1)\tan \dfrac{\alpha}{2} \\[0.3em] -(2 \cos^2 \dfrac{\alpha}{2}-1)\tan \dfrac{alpha}{2} +2 \sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} & 2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2}\tan\dfrac{\alpha}{2} +1 -2\sin^2 \dfrac{\alpha}{2} \end{bmatrix} \\ =\begin{bmatrix} 1-2\sin^2 \dfrac{\alpha}{2} +2\sin^2\dfrac{\alpha}{2} & -2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} +2 \sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} -\tan \dfrac {\alpha}{2} \\[0.3em] -2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} + \tan \dfrac{\alpha}{2} +2 \sin\dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} & 2 \sin^2 \dfrac{\alpha}{2} +1-2\sin^2\dfrac{\alpha}{2} \end{bmatrix} \\ =\begin{bmatrix} 1 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 1 \end{bmatrix} \\ Thus, from (1) and (2), we get L.H.S. = R.H.S. 28 A trust fund has Rs 30,000 that must be invested in two different types of bonds. \\ The first bond pays 5 % interest per year, and the second bond pays 7% interest per year.\\ Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds.\\ If the trust fund must obtain an annual total interest of : (a) Rs 1,800$$\\$ (b) $Rs 2,000$

##### Solution :

(b) Let Rs $x$ be invested in the first bond. Then,$\\$ the sum of money invested in the second bond will be $Rs (30000 - x ) $$\\. Therefore, in order to obtain an annual total interest of Rs 2000,$$\\$ we have :$\\$ $[x(30000-x)]\begin{bmatrix} \\[0.3em] \dfrac{5}{100}\\[0.3em] \dfrac{7}{100} \end{bmatrix} =2000$$\\ \implies \dfrac{5x}{100}+\dfrac{7(30000-x)}{100} \\ =2000\\ \implies 5x+210000-7x\\ =200000\\ \implies 210000-2x=200000\\ \implies 2x=210000-200000\\ \implies 2x=10000\\ \implies x=5000$$\\$ Thus, in order to obtain an annual total interest of $Rs 2000,$$\\ the trust fund should invest Rs 5000 in the first bond \\ and the remaining Rs 25000 in the second bond. a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second\\ bond pays Rs (30000 - x ) .$$\\$ It is given that the first bond pays $5$% interest per year $\\$and the second bond pay $7$% interest per year.$\\$ Therefore, in order to obtain an annual total interest of $Rs 1800,$ $\\$ we have:$\\$

$[x(30000-x)]\begin{bmatrix} \\[0.3em] \dfrac{5}{100}\\[0.3em] \dfrac{7}{100} \end{bmatrix} [S.I. \text{for} 1 \text{year} = \dfrac{\text{Principal * Rate}}{100}]$$\\ \implies \dfrac{5x}{100}+\dfrac{7(30000-x)}{100}=1800\\ \implies 5x+210000-7x=180000\\ \implies 210000-2x=180000\\ 2x=210000-180000\\ 2x=30000\\ x=15000$$\\$ Thus, in order to obtain an annual total interest of $Rs 1800,$$\\ the trust fund should invest Rs 15000 in the first bond \\ and the remaining Rs 15000 in the second bond.\\ (b) Let Rs x be invested in the first bond. Then,\\ the sum of money invested in the second bond will be Rs (30000 - x )$$\\$. Therefore, in order to obtain an annual total interest of $Rs 2000,$$\\ we have :\\ [x(30000-x)]\begin{bmatrix} \\[0.3em] \dfrac{5}{100}\\[0.3em] \dfrac{7}{100} \end{bmatrix} =2000$$\\$ $\implies \dfrac{5x}{100}+\dfrac{7(30000-x)}{100} \\ =2000\\ \implies 5x+210000-7x\\ =200000\\ \implies 210000-2x=200000\\ \implies 2x=210000-200000\\ \implies 2x=10000\\ \implies x=5000$$\\ Thus, in order to obtain an annual total interest of Rs 2000,$$\\$ the trust fund should invest $Rs 5000$ in the first bond $\\$ and the remaining $Rs 25000$ in the second bond.

29   The bookshop of a particular school has $10$ dozen chemistry books, $8$ dozen physics books, $10$ dozen economics books.$\\$ Their selling prices are $Rs 80, Rs 60$ and $Rs 40$ each respectively. $\\$ Find the total amount the bookshop will receive from selling all the books using matrix algebra.

##### Solution :

The bookshop has $10$ dozen chemistry books,$\\$ $8$ dozen physics books, and $10$ dozen economics books.$\\$ The selling prices of a chemistry book, a physics book, $\\$and an economics book are respectively given as $Rs 80, Rs 60$ and $Rs 40.$$\\ The total amount of money that will be received from the \\sale of all these books can be represented in the form of a matrix as :\\ 12 \begin{bmatrix} 10 & 8 & 10 \end{bmatrix} \begin{bmatrix} 80 \\[0.3em] 60 \\[0.3em] 40\end{bmatrix}$$\\$ $=12[10*80+8*60+10*40]\\ =12(800+480+400)\\ =12(1680)\\ 20160$$\\ Thus, the bookshop will receive Rs 20160 from the sale of all these books. 30 Assume X , Y , Z , W and P are the matrices of order \\ 2 * n , 3 * k , 2 *p , n * 3 and p * k respectively.\\ The restriction on n , k and p so that PY + WY will be defined are:\\ A. k = 3, p = n$$\\$ B. $k$ is arbitrary, $p = 2$$\\ C. p is arbitrary, k = 3$$\\$ D.$k = 2, p = 3$

##### Solution :

Matrices $P$ and $Y$ are of the orders$\\$ $p * k$ and $3 *k$ respectively.$\\$ Therefore, matrix $PY$ will be defined if $k *3.$$\\ Consequently, PY will be of the order p * k .$$\\$ Matrices $W$ and $Y$ are of the orders $\\$ $n * 3$ and $3 *k$ respectively.$\\$ Since the number of columns in $W$ is equal to the number of rows in $Y$ , $\\$ matrix $WY$ is well- defined and is of the order $n * k .$$\\ Matrices PY and WY can be added only when their orders are the same.\\ However, PY is of the order p * k$$\\$ and $WY$ is of the order $n *k$ .$\\$Therefore. we must have $p * n .$$\\ Thus, k * 3 and p * n . are the restrictions on n , k , and p$$\\$ so that $PY + WY$ will be defined

31   Assume $X , Y , Z , W$ and $P$ are matrices of order $2 * n , 3 * k , 2 * p , n * 3$ and $p * k$ respectively.$\\$ If $n * p ,$ then the order of the matrix $7 X - 5 Z$ is$\\$ $A. p * 2\\ B. 2 * n\\ C. n * 3\\ D. p * n$

33   If$A=\begin{bmatrix} -1 & 2 & 3 \\[0.3em] 5 & 7 & 9 \\[0.3em] -2 & 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} -4& 1 & -5 \\[0.3em] 1 & 2 & 0 \\[0.3em] 1 & 3 & 1 \end{bmatrix},$ then verify that $\\$ (i) $(A+B)'=A'+B'$$\\ (ii)(A-B)'=A'-B' ##### Solution : (ii)A-B=\begin{bmatrix} -1 & 2 & 3\\[0.3em] 5 & 7 & 9 \\[0.3em] -2 & 1 & 1 \end{bmatrix}-\begin{bmatrix} -4 & 1 &-5 \\[0.3em] 1 & 2 & 0 \\[0.3em] 1 & 3 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 3 & 1 & 8\\[0.3em] 4 & 5 & 9 \\[0.3em] -3 & -2 & 0 \end{bmatrix}$$\\ \therefore (A-B)'=\begin{bmatrix} 3 & 4 & -3\\[0.3em] 1 & 5 & -2 \\[0.3em] 8 & 9 & 0 \end{bmatrix}$$\\$ $A'-B'=\begin{bmatrix} -1 & 5 & -2\\[0.3em] 2 & 7 & 1\\[0.3em] 3 & 9 & 1 \end{bmatrix}-\begin{bmatrix} -4 & 1 & 1\\[0.3em] 1 & 2 & 3\\[0.3em] -5 & 0 & 1 \end{bmatrix}$$\\ =\begin{bmatrix} 3 & 4 & -3\\[0.3em] 1 & 5 & -2\\[0.3em] 8 & 9 & 0 \end{bmatrix}$$\\$ Hence, we have verified that $(A-B)'=A'-B'.$

We have:$\\$ $A'=\begin{bmatrix} -1 & 5 & -2 \\[0.3em] 2 & 7 & 1 \\[0.3em] 3 & 9 & 1 \end{bmatrix},$ $B'=\begin{bmatrix} -4 & 1 & 1 \\[0.3em] 1 & 2 & 3 \\[0.3em] -5 & 0 & 1 \end{bmatrix}$$\\ (i)A+B=\begin{bmatrix} -1 & 2 & 3 \\[0.3em] 5 & 7 & 9 \\[0.3em] -2 & 1 & 1 \end{bmatrix}+\begin{bmatrix} -4& 1 & -5 \\[0.3em] 1 & 2 & 0 \\[0.3em] 1 & 3 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} -5 & 3 & -2 \\[0.3em] 6 & 9 & 9 \\[0.3em] -1 & 4 & 2 \end{bmatrix}$$\\ \therefore (A+B)'=\begin{bmatrix} -5 & 6 & -1 \\[0.3em] 3 & 9 & 4 \\[0.3em] -2& 9 & 2 \end{bmatrix}$$\\$ $A'+B'=\begin{bmatrix} -1 & 5 & -2 \\[0.3em] 2 & 7 & 1 \\[0.3em] 3 & 9 & 1 \end{bmatrix}+\begin{bmatrix} -4 & 1 & 1 \\[0.3em] 1 & 2 & 3 \\[0.3em] -5 & 0 & 1 \end{bmatrix}$$\\ =\begin{bmatrix} -5 & 6 & -1 \\[0.3em] 3 & 9 & 4 \\[0.3em] -2 & 9 & 2 \end{bmatrix}$$\\$ Hence, we have verified that$(A+B)'=A'+B'$

34   If $A'=\begin{bmatrix} 3 & 4 \\[0.3em] -1 & 2\\[0.3em] 0 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} -1 & 2 & 1 \\[0.3em] 1 & 2 & 3 \end{bmatrix},$ then verify that $\\$ (i)$(A+B)'=A'+B'$$\\ (ii)(A-B)'=A'-B'$$\\$

##### Solution :

$=\begin{bmatrix} (\cos \alpha)(\cos \alpha)+(-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\cos \alpha)+(-\sin \alpha)(\cos \alpha)\\[0.3em] (\sin \alpha )(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha)+(\cos \alpha)(\cos \alpha) \end{bmatrix}$$\\ \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\[0.3em] \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & \sin^2 \alpha +\cos ^2 \alpha \end{bmatrix}$$\\$ $=\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1\end{bmatrix} =I$$\\ Hence, we have verified that A'A=1. =\begin{bmatrix} (\sin \alpha)(\sin \alpha )+(-\cos \alpha )(-\cos \alpha ) & (\sin \alpha )(\cos \alpha )+(-\cos \alpha )(\sin \alpha ) \\[0.3em] (\cos \alpha )(\sin \alpha )+(\sin \alpha )(-\cos \alpha )& (\cos \alpha )(\cos \alpha )+(\sin \alpha)(\sin \alpha ) \end{bmatrix}$$\\$ $=\begin{bmatrix} \sin^2 \alpha + \cos ^2 \alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\[0.3em] \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & \cos ^2 \alpha+\sin^2 \alpha \end{bmatrix}$$\\ =\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}=I$$\\$ Hence, we have verified that $A'A=1.$

$A=\begin{bmatrix} \cos \alpha & \sin \alpha \\[0.3em] -\sin \alpha & \cos \alpha \end{bmatrix}$$\\ \therefore A'=\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}$$\\$ $A'A=\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha & -\sin \alpha \\[0.3em] -\sin \alpha & \cos \alpha \end{bmatrix}$$\\ =\begin{bmatrix} (\cos \alpha)(\cos \alpha)+(-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\cos \alpha)+(-\sin \alpha)(\cos \alpha)\\[0.3em] (\sin \alpha )(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha)+(\cos \alpha)(\cos \alpha) \end{bmatrix}$$\\$ $\begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\[0.3em] \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & \sin^2 \alpha +\cos ^2 \alpha \end{bmatrix}$$\\ =\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1\end{bmatrix} =I$$\\$ Hence, we have verified that $A'A=1.$

(ii)$A=\begin{bmatrix} \sin \alpha & \cos \alpha \\[0.3em] -\cos \alpha & \sin \alpha \end{bmatrix}$$\\ \therefore A'=\begin{bmatrix} \sin \alpha & -\cos \alpha \\[0.3em] \cos \alpha & \sin \alpha \end{bmatrix}$$\\$ $A'A=\begin{bmatrix} \sin \alpha & -\cos \alpha \\[0.3em] \cos \alpha & \sin \alpha \end{bmatrix}\begin{bmatrix} \sin \alpha & \cos \alpha \\[0.3em] -\cos \alpha & \sin \alpha \end{bmatrix}$$\\ \begin{bmatrix} \sin \alpha & -\cos \alpha \\[0.3em] \cos \alpha & \sin \alpha \end{bmatrix}\begin{bmatrix} \sin \alpha & \cos \alpha \\[0.3em] -\cos \alpha & \sin \alpha \end{bmatrix}$$\\$ $=\begin{bmatrix} (\sin \alpha)(\sin \alpha )+(-\cos \alpha )(-\cos \alpha ) & (\sin \alpha )(\cos \alpha )+(-\cos \alpha )(\sin \alpha ) \\[0.3em] (\cos \alpha )(\sin \alpha )+(\sin \alpha )(-\cos \alpha )& (\cos \alpha )(\cos \alpha )+(\sin \alpha)(\sin \alpha ) \end{bmatrix}$$\\ =\begin{bmatrix} \sin^2 \alpha + \cos ^2 \alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\[0.3em] \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & \cos ^2 \alpha+\sin^2 \alpha \end{bmatrix}$$\\$ $=\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}=I$$\\ Hence, we have verified that A'A=1. 38 (i) Show that the matrix A=\begin{bmatrix} 1 & -1 & 5 \\[0.3em] -1 & 2 & 1 \\[0.3em] 5 & 1 & 3 \end{bmatrix}is a symmetric matrix\\ (ii) Show that the matrix A=\begin{bmatrix} 0 & 1 & -1 \\[0.3em] -1 & 0 & 1\\[0.3em] 1 & -1 & 0 \end{bmatrix}is a skew symmetric matrix ##### Solution : 39 (i) Show that the matrix A=\begin{bmatrix} 1 & -1 & 5 \\[0.3em] -1 & 2 & 1 \\[0.3em] 5 & 1 & 3 \end{bmatrix}is a symmetric matrix\\ (ii) Show that the matrix A=\begin{bmatrix} 0 & 1 & -1 \\[0.3em] -1 & 0 & 1\\[0.3em] 1 & -1 & 0 \end{bmatrix}is a skew symmetric matrix ##### Solution : (i) We have : \\ A'=\begin{bmatrix} 1 & -1 & 5 \\[0.3em] -1 & 2 & 1 \\[0.3em] 5 & 1 & 3 \end{bmatrix} = A\\ \therefore A'=A$$\\$ Hence, $A$ is a symmetric matrix.$\\$ (ii) We have: $\\$ $A'=\begin{bmatrix} 0 & -1 & 1 \\[0.3em] 1 & 0 & -1\\[0.3em] -1 & 1 & 0 \end{bmatrix}=- A=\begin{bmatrix} 0 & 1 & -1 \\[0.3em] -1 & 0 & 1\\[0.3em] 1 & -1 & 0 \end{bmatrix}=-A $$\\ \therefore A'=-A$$\\$ Hence, $A$ is a skew -symmetric matrix.

40   For the matrix $A =\begin{bmatrix} 1 & 5 \\[0.3em] 6 & 7 \end{bmatrix},$ verify that $\\$ (i)$(A+A')$ is a symmetric matrix $\\$ (ii) $(A-A')$ is a skew symmetric matrix

##### Solution :

41   For the matrix $A =\begin{bmatrix} 1 & 5 \\[0.3em] 6 & 7 \end{bmatrix},$ verify that $\\$ (i)$(A+A')$ is a symmetric matrix $\\$ (ii) $(A-A')$ is a skew symmetric matrix

##### Solution :

Given : $A=\begin{bmatrix} 1 & 5 \\[0.3em] 6 & 7 \end{bmatrix},$ then $A'=\begin{bmatrix} 1 & 6 \\[0.3em] 5 & 7 \end{bmatrix}$$\\ (i)A+A'=\begin{bmatrix} 1 & 5 \\[0.3em] 6 & 7 \end{bmatrix}+\begin{bmatrix} 1 & 6 \\[0.3em] 5 & 7 \end{bmatrix}$$\\$ $=\begin{bmatrix} 2 & 11\\[0.3em] 11 & 14 \end{bmatrix}$$\\ \therefore (A+A')'=\begin{bmatrix} 2 & 11 \\[0.3em] 11 & 14 \end{bmatrix}=A+A'$$\\$ Hence, $(A + A ' )$ is a symmetric matrix.$\\$ (ii)$A-A'=\begin{bmatrix} 1 & 5 \\[0.3em] 6 & 7 \end{bmatrix}-\begin{bmatrix} 1 & 6 \\[0.3em] 5 & 7 \end{bmatrix}$$\\ =\begin{bmatrix} 0& -1 \\[0.3em] 1 & 0 \end{bmatrix}$$\\$ $(A-A')'= \begin{bmatrix} 0& -1 \\[0.3em] 1 & 0 \end{bmatrix}$$\\ =-\begin{bmatrix} 0& 1 \\[0.3em] -1 & 0 \end{bmatrix}=-(A-A)'$$\\$ Hence, $(A - A ' )$ is a skew-symmetric matrix.

42   Find $\dfrac{1}{2}(A-A')$ and $\dfrac{1}{2}(A-A'),$ when $A =\begin{bmatrix} 0 & a & b\\[0.3em] -a & 0 & c\\[0.3em] -b & -c & 0 \end{bmatrix}$

##### Solution :

The given matrix is $A =\begin{bmatrix} 0 & a & b\\[0.3em] -a & 0 & c\\[0.3em] -b & -c & 0 \end{bmatrix},$ then$\\$ $A'=\begin{bmatrix} 0 & -a & -b\\[0.3em] a & 0 & -c\\[0.3em] b & c & 0 \end{bmatrix}$$\\ A+A'=\begin{bmatrix} 0 & a & b\\[0.3em] -a & 0 & c\\[0.3em] -b & -c & 0 \end{bmatrix}+\begin{bmatrix} 0 & -a & -b\\[0.3em] a & 0 & -c\\[0.3em] b & c & 0 \end{bmatrix}$$\\$ $=\begin{bmatrix} 0 & 0 & 0\\[0.3em] 0 & 0 & 0\\[0.3em] 0 & 0 & 0 \end{bmatrix}$ $\\$ $\therefore \dfrac{1}{2}(A+A')=\begin{bmatrix} 0 & 0 & 0\\[0.3em] 0 & 0 & 0\\[0.3em] 0 & 0 & 0 \end{bmatrix}$$\\ Now , A-A'=\begin{bmatrix} 0 & a & b\\[0.3em] -a & 0 & c\\[0.3em] -b & -c & 0 \end{bmatrix}-\begin{bmatrix} 0 & -a & -b\\[0.3em] a & 0 & -c\\[0.3em] b & c & 0 \end{bmatrix}$$\\$ $=\begin{bmatrix} 0& 2a & 2b\\[0.3em] -2a & 0 & 2c\\[0.3em] -2b & -2c & 0 \end{bmatrix}$$\\ \therefore \dfrac{1}{2}(A-A')=\begin{bmatrix} 0 & a & b\\[0.3em] -a & 0 & c\\[0.3em] -b & -c & 0 \end{bmatrix} 43 Express the following matrices as the sum of a symmetric and a skew symmetric matrix:\\ (i) \begin{bmatrix} 3 & 5 \\[0.3em] 1 & -1 \end{bmatrix}$$\\$ (ii)$\begin{bmatrix} 6 & -2 & 2 \\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}$$\\ (iii) \begin{bmatrix} 3 & 3 & -1 \\[0.3em] -2 & -2 & 1 \\[0.3em] -4 & -5 & 2 \end{bmatrix}$$\\$ (iv)$\begin{bmatrix} 1 & 5 \\[0.3em] -1 & 2 \end{bmatrix}$

##### Solution :

(iii)Let $A= \begin{bmatrix} 3 & 3 & -1 \\[0.3em] -2 & -2 & 1 \\[0.3em] -4 & -5 & 2 \end{bmatrix},$$\\ then A'= \begin{bmatrix} 3 & -2 & -4 \\[0.3em] 3 & -2 & -5 \\[0.3em] -1 & 1 & 2 \end{bmatrix}$$\\$ Now, $A+A'= \begin{bmatrix} 3 & 3 & -1 \\[0.3em] -2 & -2 & 1 \\[0.3em] -4 & -5 & 2 \end{bmatrix}+ \begin{bmatrix} 3 & -2 & -4 \\[0.3em] 3 & -2 & -5 \\[0.3em] -1 & 1 & 2 \end{bmatrix}$$\\ = \begin{bmatrix} 6 & 1 & -5 \\[0.3em] 1 & -4 & -4 \\[0.3em] -5 & -4 & 4 \end{bmatrix}$$\\$ $P=\dfrac{1}{2}(A+A')=\dfrac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\[0.3em] 1 & -4 & -4 \\[0.3em] -5 & -4 & 4 \end{bmatrix}$$\\ = \begin{bmatrix} 3 & \dfrac{1}{2} & -\dfrac{5}{2} \\[0.3em] \dfrac{1}{2} & -2 & -2 \\[0.3em] -\dfrac{5}{2} 7 -2 & 2 \end{bmatrix}$$\\$ Now , $P'=\begin{bmatrix} 3 & \dfrac{1}{2} & -\dfrac{5}{2} \\[0.3em] \dfrac{1}{2} & -2 & -2 \\[0.3em] -\dfrac{5}{2} 7 -2 & 2 \end{bmatrix}=P$$\\ Thus P=\dfrac{1}{2}(A+A') is symmetric matrix.\\ Now,A-A'=\begin{bmatrix} 3 & 3 & -1 \\[0.3em] -2 & -2 & 1 \\[0.3em] -4 & -5 & 2 \end{bmatrix}- \begin{bmatrix} 3 & -2 & -4 \\[0.3em] 3 & -2 & -5 \\[0.3em] -1 & 1 & 2 \end{bmatrix}$$\\$ $= \begin{bmatrix} 0 & 5 & 3 \\[0.3em] -5 & 0 & 6 \\[0.3em] -3 & -6 & 0 \end{bmatrix}$$\\ Now , P'=\begin{bmatrix} 3 & 3 \\[0.3em] 3 & -1 \end{bmatrix} =p$$\\$ Thus , $P=\dfrac{1}{2}(A+A')$ is a symmetric matrix.$\\$ Now,$A-A'=\begin{bmatrix} 3 & 5\\[0.3em] 1 & -1 \end{bmatrix}-\begin{bmatrix} 3 & 1 \\[0.3em] 5 & -1 \end{bmatrix}$$\\ =\begin{bmatrix} 0 & 4 \\[0.3em] -4& 0 \end{bmatrix}$$\\$

now ,$P=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}=p$$\\ Thus , P =\dfrac{1}{2}(A+A')is a symmetric matrix.\\ Now, A-A'=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}+\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}$$\\$ $= \begin{bmatrix} 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \end{bmatrix}$$\\ Let Q =\dfrac{1}{2}(A-A')= \begin{bmatrix} 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \end{bmatrix}$$\\$ Now,$Q'= \begin{bmatrix} 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \end{bmatrix}=-Q$$\\ Thus Q =\dfrac{1}{2}(A-A')is a skew-symmetric matrix.\\ Representing A as the sum of P and Q$$\\$ $P+Q =\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}+\begin{bmatrix} 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \end{bmatrix}$$\\ =\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}=A$$\\$

Let $A =\begin{bmatrix} 3 & 5 \\[0.3em] 1 & -1 \end{bmatrix},$ then $A'=\begin{bmatrix} 3 & 1 \\[0.3em] 5 & -1 \end{bmatrix}$$\\ Now , A+A'=\begin{bmatrix} 3 & 5 \\[0.3em] 1 & -1 \end{bmatrix}+\begin{bmatrix} 3 & 1 \\[0.3em] 5 & -1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 6 & 6 \\[0.3em] 6 & -2 \end{bmatrix}$$\\ Let P=\dfrac{1}{2}(A+A')=$$\\$ $\dfrac{1}{2}\begin{bmatrix} 6 & 6 \\[0.3em] 6 & -2 \end{bmatrix}=\begin{bmatrix} 3 & 3 \\[0.3em] 3 & -1 \end{bmatrix}$$\\ Now , P'=\begin{bmatrix} 3 & 3 \\[0.3em] 3 & -1 \end{bmatrix} =p$$\\$ Thus , $P=\dfrac{1}{2}(A+A')$ is a symmetric matrix.$\\$ Now,$A-A'=\begin{bmatrix} 3 & 5\\[0.3em] 1 & -1 \end{bmatrix}-\begin{bmatrix} 3 & 1 \\[0.3em] 5 & -1 \end{bmatrix}$$\\ =\begin{bmatrix} 0 & 4 \\[0.3em] -4& 0 \end{bmatrix}$$\\$

Let $Q=\dfrac{1}{2}(A-A')=$$\\ \dfrac{1}{2}\begin{bmatrix} 0 & 4\\[0.3em] -4 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 2\\[0.3em] -2 & 0 \end{bmatrix}$$\\$ Now , $Q'=\begin{bmatrix} 0 & 2\\[0.3em] -2 & 0 \end{bmatrix}=-Q $$\\ Thus, Q =\dfrac{1}{2}(A-A') is a skew-symmetric matrix.\\ Representing A as the sum of P and Q :\\ P+Q=\begin{bmatrix} 3 & 3\\[0.3em] 3 & -1\end{bmatrix}+\begin{bmatrix} 0 & 2\\[0.3em] -2 & 0\end{bmatrix}$$\\$ $=\begin{bmatrix} 3 & 5\\[0.3em] 1& -1\end{bmatrix}=A$$\\ (ii) Let A=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}, then \\ A'=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}$$\\$ Now $A+A'=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}+\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}$$\\ =\begin{bmatrix} 12 & -4 & 4\\[0.3em] -4 & 6 & -2 \\[0.3em] 4 & -2 & 6 \end{bmatrix}$$\\$ Let $P =\dfrac{1}{2}(A+A')$$\\ =\dfrac{1}{2} \begin{bmatrix} 12 & -4 & 4\\[0.3em] -4 & 6 & -2 \\[0.3em] 4 & -2 & 6 \end{bmatrix}=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}$$\\$

44   Find the inverse of each of the matrices, if it exists. $\\$ $\begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_2 \rightarrow R_2 - 2R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_2 \rightarrow \dfrac{1}{5}R_2$ $\\$ $\implies$ $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] \dfrac{-2}{5} & \dfrac{1}{5} \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_2 \rightarrow R_1 + R_2$ $\\$ $\implies$ $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{3}{5} & \dfrac{1}{5} \\[0.3em] \dfrac{-2}{5} & \dfrac{1}{5} \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Post - multiply both sides by, we get$ $\\$ $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $\mathrm{A}^{-1}$= $\begin{bmatrix} \dfrac{3}{5} & \dfrac{1}{5} \\[0.3em] \dfrac{-2}{5} & \dfrac{1}{5} \\[0.3em] \end{bmatrix}$ $A \cdot \mathrm{A}^{-1}$ $\\$ $\implies$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} \dfrac{3}{5} & \dfrac{1}{5} \\[0.3em] \dfrac{-2}{5} & \dfrac{1}{5} \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Thus, \mathrm{A}^{-1}$ = $\begin{bmatrix} \dfrac{3}{5} & \dfrac{1}{5} \\[0.3em] \dfrac{-2}{5} & \dfrac{1}{5} \\[0.3em] \end{bmatrix}$

45   Find the inverse of each of the matrices, if it exists. $\\$ $\begin{bmatrix} 2 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 2 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}$ $\\$ $We$ $know$ $that$ $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1 \rightarrow R_1 - R_2$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_2 \rightarrow R_2 - R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ $\begin{bmatrix} 1 & -1 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$

46   Find the inverse of each of the matrices, if it exists. $\\$ $\begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 1 & 3 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 1 & 3 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_2$ $\rightarrow$ $R_2 - R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & 3 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1$ $\rightarrow$ $R_1 - 3R_2$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 7 & -3 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 7 & -3 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$

47   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 2 & 3 \\[0.3em] 5 & 7 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 2 & 3 \\[0.3em] 5 & 7 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & 3 \\[0.3em] 5 & 7 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1 \rightarrow \dfrac{1}{2} R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & \dfrac{3}{2} \\[0.3em] 5 & 7 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_2 \rightarrow R_1 + 3R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & \dfrac{3}{2} \\[0.3em] 0 & \dfrac{-1}{2} \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 \\[0.3em] \dfrac{-5}{2} & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_2 \rightarrow - 2R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} -7 & 3 \\[0.3em] 5 & -2 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} -7 & 3 \\[0.3em] 5 & -2 \\[0.3em] \end{bmatrix}$

48   Find the inverse of each of the matrices, if it exists. $\\$ $\begin{bmatrix} 2 & 1 \\[0.3em] 7 & 4 \\[0.3em] \end{bmatrix}$ $\\$

##### Solution :

Let $A = \begin{bmatrix} 2 & 1 \\[0.3em] 7 & 4 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & 1 \\[0.3em] 7 & 4 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1 \rightarrow \dfrac{1}{2}R_1$ $\\$ $\therefore$ $\begin{bmatrix} 1 & \dfrac{1}{2} \\[0.3em] 7 & 4 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_2 \rightarrow R_2 - 7R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & \dfrac{1}{2} \\[0.3em] 0 & \dfrac{1}{2} \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 \\[0.3em] \dfrac{-7}{2} & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1\rightarrow R_1 - R_2$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & \dfrac{1}{2} \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 4 & -1 \\[0.3em] -7& 2 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 4 & -1 \\[0.3em] -7 & 2 \\[0.3em] \end{bmatrix}$

49   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$ $\\$

##### Solution :

Let $A = \begin{bmatrix} 2 & 5 \\[0.3em] 1 & 3 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & 5 \\[0.3em] 1 & 3 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1 \rightarrow \dfrac{1}{2} R_1$ $\\$ $\implies$ $\begin{bmatrix} 1 & \dfrac{5}{2} \\[0.3em] 0 & \dfrac{1}{2} \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 \\[0.3em] \dfrac{-1}{2} & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_2 \rightarrow 2R_2$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 3 & -5 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 3 & -5 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$ $\\$

50   Find the inverse of each matrices, if exists. Let $A = \begin{bmatrix} 3 & 1 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}$ $\\$

##### Solution :

Let $A = \begin{bmatrix} 3 & 1 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 3 & 1 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 1 \\[0.3em] 1 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 1 \\[0.3em] 1 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \\[0.3em] -2 & 3 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 2 & -1 \\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 2 & -1 \\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}$

51   Find the inverse of each of the matrices, if it exists. $\\$ $\begin{bmatrix} 3 & 10 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 3 & 10 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 3 & 10 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 3 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 3 \\[0.3em] 2 & 7 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 3 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \\[0.3em] -2 & 3 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 7 & -10 \\[0.3em] -2 & 3 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 7& -10 \\[0.3em] -2 & 3 \\[0.3em] \end{bmatrix}$

52   Find The inverse of each of the matrices, if it exists. $\begin{bmatrix} 3 & -1 \\[0.3em] -4 & 2 \\[0.3em] \end{bmatrix}$ $\\$

##### Solution :

Let $A = \begin{bmatrix} 3 & -1 \\[0.3em] -4 & 2 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 3 & -1 \\[0.3em] -4 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 2 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 \\[0.3em] 2 & 3 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & \dfrac{1}{2} \\[0.3em] 2 & \dfrac{3}{2} \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Thus, \mathrm{A}^{-1}$ = $\begin{bmatrix} 1 & \dfrac{1}{2} \\[0.3em] 2 & \dfrac{3}{2} \\[0.3em] \end{bmatrix}$

53   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 2 & -6 \\[0.3em] 1 & -2 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 2 & -6 \\[0.3em] 1 & -2 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & -6 \\[0.3em] 1 & -2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 2 & 0 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 3 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 2 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} -2 & 3 \\[0.3em] -1 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} -1 & 3 \\[0.3em] \dfrac{-1}{2} & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} -1 & 3 \\[0.3em] \dfrac{-1}{2} & 1 \\[0.3em] \end{bmatrix}$

54   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 6 & -3 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 6 & -3 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 6 & -3 \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & \dfrac{-1}{2} \\[0.3em] -2 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{6} & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & \dfrac{-1}{2} \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{6} & 0 \\[0.3em] \dfrac{1}{3} & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Now, in the above equation, we can see all the zeros in the second row of the matrix on the$ $\\$ $L.H.S,$ $\\$ Therefore, $\mathrm{A}^{-1}$ does not exist.

55   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 2 & -3 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 2 & -3 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & -3 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & -1 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & -1 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 \\[0.3em] 1 & 2 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\implies$ $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 2 & 3 \\[0.3em] 1 & 2 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 2 & 1 \\[0.3em] 4 & 2 \\[0.3em] \end{bmatrix}$

56   Find the inverse of each of the matrices, if it exist. $\\$ $\begin{bmatrix} 2 & 1 \\[0.3em] 4 & 2 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 2 & 1 \\[0.3em] 4 & 2 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & 1 \\[0.3em] 4 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying$ $R_1 \rightarrow R_1 - \dfrac{1}{2}R_1$ , we have: $\\$ $\implies$ $\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 2 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & \dfrac{-1}{2} \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ Now, in the above equation, we can see all the zeros in the first row of the matrix on the L.H.S. $\\$ Therefore, $\mathrm{A}^{-1}$ does not exist.

57   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 1 & 3 & -2 \\[0.3em] -3 & 0 & -5 \\[0.3em] 2 & 5 & 0 \\[0.3em] \end{bmatrix}$

##### Solution :

Let $A = \begin{bmatrix} 1 & 3 & -2 \\[0.3em] -3 & 0 & -5 \\[0.3em] 2 & 5 & 0 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 1 & 3 & -2 \\[0.3em] -3 & 0 & -5 \\[0.3em] 2 & 5 & 0 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_2 \rightarrow R_2 +3R_1$ and $R_3 \rightarrow R_3 - 2R_1 ,$ we have: $\\$ $\implies$ $\begin{bmatrix} 1 & 3 & -2 \\[0.3em] -3 & 0 & -5 \\[0.3em] 2 & 5 & 0 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$

$Applying R_1 \rightarrow R_1 +3R_3$ and $R_2 \rightarrow R_2 - 2R_3 ,$ we have: $\\$ $\begin{bmatrix} 1 & 0 & 10 \\[0.3em] 0 & 1 & 21 \\[0.3em] 0 & -1 & 4 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} -5 & 0 & 3 \\[0.3em] -13 & 1 & 8 \\[0.3em] -2 & 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_3 \rightarrow R_3 + R_2 ,$ we have: $\\$

$\begin{bmatrix} 1 & 0 & 10 \\[0.3em] 0 & 1 & 21 \\[0.3em] 0 & -1 & 25 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} -5 & 0 & 3 \\[0.3em] -13 & 1 & 8 \\[0.3em] -15 & 1 & 9 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_3 \rightarrow \dfrac{1}{25}R_3 ,$ we have: $\\$ $\begin{bmatrix} 1 & 0 & 10 \\[0.3em] 0 & 1 & 21 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} -5 & 0 & 3 \\[0.3em] -13 & 1 & 8 \\[0.3em] \dfrac{-3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \\[0.3em] \end{bmatrix}$ $A$ $\\$ Applying $R_1 \rightarrow R_1-10R_3 ,$ and $R_2 \rightarrow R_2-21R_3,$ we have: $\\$

$\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & \dfrac{-2}{5} & \dfrac{-3}{5} \\[0.3em] \dfrac{-2}{5} & \dfrac{4}{25} & \dfrac{11}{25} \\[0.3em] \dfrac{-3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 1 & \dfrac{-2}{5} & \dfrac{-3}{5} \\[0.3em] \dfrac{-2}{5} & \dfrac{4}{25} & \dfrac{11}{25} \\[0.3em] \dfrac{-3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \\[0.3em] \end{bmatrix}$

58   Find the inverse of each of the matrices, if it exists. $\begin{bmatrix} 2 & 0 & -1 \\[0.3em] 5 & 1 & 0 \\[0.3em] 0 & 1 & 3 \\[0.3em] \end{bmatrix}$ $\\$

##### Solution :

$\begin{bmatrix} 1 & 0 & \dfrac{-1}{2} \\[0.3em] 0 & 1 & \dfrac{5}{2} \\[0.3em] 0 & 0 & \dfrac{1}{2} \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 & 0 \\[0.3em] \dfrac{-5}{2} & 1 & 0 \\[0.3em] \\[0.3em] \dfrac{5}{2} & -1 & 1 \\[0.3em] \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_3 \rightarrow 2R_3,$ we have: $\\$

Let $A = \begin{bmatrix} 2 & 0 & -1 \\[0.3em] 5 & 1 & 0 \\[0.3em] 0 & 1 & 3 \\[0.3em] \end{bmatrix}$ $\\$ We know that $A = IA$ $\\$ $\therefore$ $\begin{bmatrix} 2 & 0 & -1 \\[0.3em] 5 & 1 & 0 \\[0.3em] 0 & 1 & 3 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_1 \rightarrow \dfrac{1}{2}R_1,$ we have: $\\$

$\begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} 3 & -1 & 1 \\[0.3em] -15 & 6 & -5 \\[0.3em] \\[0.3em] 5 & -2 & 2 \\[0.3em] \\[0.3em] \end{bmatrix}$ $A$ $\\$ $\therefore$ $\mathrm{A}^{-1}$ = $\begin{bmatrix} 3 & -1 & 1 \\[0.3em] -15 & 6 & -5 \\[0.3em] \\[0.3em] 5 & -2 & 2 \\[0.3em] \\[0.3em] \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & \dfrac{-1}{2} \\[0.3em] 0 & 1 & \dfrac{5}{2} \\[0.3em] 0 & 0 & \dfrac{1}{2} \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 & 0 \\[0.3em] \dfrac{-5}{2} & 1 & 0 \\[0.3em] \\[0.3em] 0 & 0 & 1 \\[0.3em] \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_3 \rightarrow R_3-R_2,$ we have: $\\$

$\begin{bmatrix} 1 & 0 & \dfrac{-1}{2} \\[0.3em] 0 & 1 & \dfrac{5}{2} \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ = $\begin{bmatrix} \dfrac{1}{2} & 0 & 0 \\[0.3em] \dfrac{-5}{2} & 1 & 0 \\[0.3em] \\[0.3em] 5 & -2 & 2 \\[0.3em] \\[0.3em] \end{bmatrix}$ $A$ $\\$ $Applying R_1 \rightarrow R_1 + \dfrac{1}{2} R_3 , and R_2 \rightarrow R_2 - \dfrac{5}{2}R_3$ we have: $\\$

$\mathrm{A}^{-1}$ = $\begin{bmatrix} 3 & -1 & 1 \\[0.3em] -15 & 6 & -5 \\[0.3em] 5 & -2 & 2 \\[0.3em] \end{bmatrix}$

59   Matrices $A$ and $B$ will be inverse of each other only if

##### Solution :

We know that if $A$ is a square matrix of order m , and if there exists another square matrix $B$ of the same order m , such that $AB = BA = I$ , then $B$ is said to be the inverse of $A$ . In this case, it is clear that $A$ is the inverse of $B$ . Thus, matrices $A$ and $B$ will be inverses of each other only if $AB = BA = I$.

60   Let $A = \begin{bmatrix} 0 & 1 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$, show that $\left(aI + b A\right)^n = a^nI + n \mathrm{a}^{n-1}\ bA,$ where I is the identity matrix of order 2 and $n \in N$