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?

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 $18$ 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 :

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, 4$4\\$ 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}$

$ 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 $A=\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\[0.3em] \frac{5}{2} & 2 & \dfrac{3}{2} & 1 \\[0.3em] 4 & \frac{7}{2} & 3 &\frac{5}{2} \end{bmatrix}$

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

Solution :

(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 $z = 3$

$(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

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

Solution :

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 :

$(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}$$\\$

$(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}$$\\$

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 :

$(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}$$\\$ $=\begin{bmatrix} 6-1+9 & -9-0+3 \\[0.3em] -2+0+6 & 3+ 0+2 \end{bmatrix}=\begin{bmatrix} 14 & -6 \\[0.3em] 4 & 5 \end{bmatrix}$

$(iii) \begin{bmatrix} 1 & -2 \\[0.3em] 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 2 & 3 & 1 \end{bmatrix}$$\\$ $=\begin{bmatrix} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\[0.3em] 2(1)+3(2) & 2(2)+ 3(3) & 2(3)+3(1) \end{bmatrix}$$\\$ $=\begin{bmatrix} 1-4 & 2-6 & 3-2 \\[0.3em] 2+6 & 4+9 & 6+3 \end{bmatrix}$ $\\$ $= \begin{bmatrix} -3 & -4 & 1 \\[0.3em] 8 & 13 & 9 \end{bmatrix}$

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   If$A=\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 :

$(i)X+Y=\begin{bmatrix} 7 & 0 \\[0.3em] 2 & 5 \end{bmatrix}$ ......(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}$

$(ii) 2X+3Y =\begin{bmatrix} 2 & 3\\[0.3em] 4 & 0\end{bmatrix}$.......(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 ,$\\$

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 $2 \begin{bmatrix} 1 & 3\\[0.3em] 0 & x \end{bmatrix}+\begin{bmatrix} y & 0\\[0.3em] 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6\\[0.3em] 1 & 8 \end{bmatrix}$

Solution :

$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.

Solution :

$ 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},$ $\\$ $\implies \begin{bmatrix} 3x & 3y\\[0.3em] 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & 6+x+y\\[0.3em] -1+z+w & 2w +3 \end{bmatrix}$$\\$ Comparing the corresponding elements of these two matrices, we get:$\\$ $3 x = x + 4\\ \implies 2 x = 4\\ \implies x = 2 $$\\$ $3 x = 6 + x + y\\ \implies 2 y = 6 + x =6 + 2 = 8\\ \implies y = 4$$\\$ $3 w = 2 w + 3 \implies w = 3$$\\$ $3 z =- 1 + z + w\\ \implies 2 z =- 1 +w =- 1 +3 = 2 \implies z = 1$$\\$ $\therefore x =2, y = 4, z = 1, \text{and } \ w = 3$

22   If$F(x)= \begin{bmatrix} \cos x& -\sin x& 0 \\[0.3em] \sin x& \cos x & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$, show that $ F(x)F(y)=F(x+y.)$

Solution :

F(x)= \begin{bmatrix} \cos x& -\sin x& 0 \\[0.3em] \sin x& \cos x & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$,F(x)= \begin{bmatrix} \cos y& -\sin y& 0 \\[0.3em] \sin y& \cos y& 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$$\\$ $R.H.S:F(x+y)= \begin{bmatrix} \cos (x+y)& -\sin (x+y)& 0 \\[0.3em] \sin (x+y)& \cos (x+y) & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$$\\$ $L.H.S:F(x)F(y)= \begin{bmatrix} \cos x& -\sin x& 0 \\[0.3em] \sin x& \cos x & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos y& -\sin y& 0 \\[0.3em] \sin y& \cos y & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$$\\$ $= \begin{bmatrix} \cos x \cos y -\sin x \sin y +0& -\cos x\sin y-\sin x \cos y +0 & 0 \\[0.3em] \sin x \cos y + \cos x \sin y +0 & -\sin x \sin y + \cos x \cos y +0 & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$ $\\$ $= \begin{bmatrix} \cos (x+y)& -\sin (x+y)& 0 \\[0.3em] \sin (x+y)& \cos (x+y) & 0\\[0.3em] 0 & 0 & 1 \end{bmatrix}$$\\$ $=F(x+y) \\ \therefore F(x)F(y) =F(x+y)$

23   Show that $ (i)\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}$$\\$ $(ii) \begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\neq \begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}$

Solution :

$\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}$$\\$ $=\begin{bmatrix} 10+6 & -2+7\\[0.3em] 15+24 & -3+28 \end{bmatrix}=\begin{bmatrix} 16 & 5\\[0.3em] 39 & 25\end{bmatrix}$$\\$ $\therefore \begin{bmatrix} 5 & -1\\[0.3em] 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1\\[0.3em] 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1 \\[0.3em] 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\[0.3em] 6 & 7 \end{bmatrix}$

$(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}$$\\$ $\therefore \begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\neq \begin{bmatrix} -1 & 1 & 0 \\[0.3em] 0 & -1 & 1 \\[0.3em] 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 1 & 1 & 0 \end{bmatrix}$

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 :

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

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$

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 :

$=\begin{bmatrix} 1-2\sin^2 \dfrac{\alpha}{2} +2\sin^2\dfrac{\alpha}{2} & -2\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} +2 \sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} -\tan \dfrac {\alpha}{2} \\[0.3em] -2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} + \tan \dfrac{\alpha}{2} +2 \sin\dfrac{\alpha}{2}\cos \dfrac{\alpha}{2} & 2 \sin^2 \dfrac{\alpha}{2} +1-2\sin^2\dfrac{\alpha}{2} \end{bmatrix}$ $\\$ $=\begin{bmatrix} 1 & -\tan \dfrac{\alpha}{2}\\[0.3em] \tan \dfrac{\alpha}{2} & 1 \end{bmatrix}$ $\\$ 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 $L.H.S. = R.H.S.$

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

$L.H.S.$ $\\$ $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$\\$ $A. p * 2\\ B. 2 * n\\ C. n * 3\\ D. p * n$

Solution :

The correct answer is B.$\\$ Matrix $X$ is of the order $2 * n .$$\\$ Therefore, matrix $7 X$ is also of the same order.$\\$ Matrix $Z$ is of the order $2 * p ,$ i.e, $2 * n$$\\$ [ since $n * p $]$\\$ Therefore, matrix $5Z$ is also of the same order.$\\$ Now, both the matrices $7 X$ and $5Z$ are of the order $2 * n .$$\\$ Thus, matrix $7 X - 5 Z$ is well-defined and is of the order $2 * n .$

32   Find the transpose of each of the following matrices:$\\$ (i)$\begin{bmatrix} 5\\[0.3em] \frac{1}{2}\\[0.3em] -1 \end{bmatrix}$$\\$ (ii)$\begin{bmatrix} 1 & -1\\[0.3em] 2 & 3 \end{bmatrix}$$\\$ (iii)$\begin{bmatrix} -1 & 5 & 6 \\[0.3em] \sqrt{3} & 5 & 6 \\[0.3em] 2 & 3 & -1\end{bmatrix}$

Solution :

(i)Let A=$\begin{bmatrix} 5 \\[0.3em] \frac{1}{2} \\[0.3em] -1\end{bmatrix}$, then $\\$ $A^T =\begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}$$\\$ $\\$ (ii)Let A=$ \begin{bmatrix} 1 & -1 \\[0.3em] 2 & 3 \end{bmatrix},$ then $\\$ $A^T = \begin{bmatrix} 1 & 2 \\[0.3em] -1 & 3 \end{bmatrix}$$\\$ $\\$ (iii)Let A=$\begin{bmatrix} -1 & 5 & 6 \\[0.3em] \sqrt{3} & 5 & 6 \\[0.3em] 2 & 3 & -1 \end{bmatrix},$ then $\\$ $A^T=\begin{bmatrix} -1 & \sqrt{3} & 2 \\[0.3em] 5 & 5 & 3\\[0.3em] 6 & 6 & -1\end{bmatrix}$$\\$

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 :

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$(A+B)'=A'+B'$

(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 $(A-B)'=A'-B'.$

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 :

(i) It is known that $ A=(A')'$$\\$ therefore , we have: $\\$ $A=\begin{bmatrix} 3 & -1 & 0 \\[0.3em] 4 & 2 & 1 \end{bmatrix}$$\\$ $B'=\begin{bmatrix} -1 & 1 \\[0.3em] 2 & 2 \\[0.3em] 1 & 3 \end{bmatrix}$$\\$ $ A+B=\begin{bmatrix} 3 & -1 & 0 \\[0.3em] 4 & 2 & 1 \end{bmatrix}+\begin{bmatrix} -1 & 2 & 1 \\[0.3em] 1 & 2 & 3 \end{bmatrix}$$\\$ $=\begin{bmatrix} 2 & 1 & 1\\[0.3em] 5 & 4 & 4 \end{bmatrix}$$\\$ $\therefore (A+B)'=\begin{bmatrix} 2 & 5 \\[0.3em] 1 & 4 \\[0.3em] 1 & 4 \end{bmatrix}$$\\$ $A'+B'=\begin{bmatrix} 3 & 4 \\[0.3em] -1 & 2 \\[0.3em] 0 & 1 \end{bmatrix}+\begin{bmatrix} -1 & 1 \\[0.3em] 2 & 2 \\[0.3em] 1 & 3 \end{bmatrix}$$\\$ $=\begin{bmatrix} 2 & 5 \\[0.3em] 1 & 4 \\[0.3em] 1 & 4 \end{bmatrix}$$\\$ Thus, we verified that$(A+B)'=A'+B'$

(ii)$A-B=\begin{bmatrix} 3 & -1 & 0 \\[0.3em] 4 & 2 & 1 \end{bmatrix}-\begin{bmatrix} -1 & 2 & 1 \\[0.3em] 1 & 2 & 3 \end{bmatrix}$$\\$ $=\begin{bmatrix} 4 & -3 &-1 \\[0.3em] 3 & 0 & -2 \end{bmatrix}$$\\$ $ \therefore (A-B)'=\begin{bmatrix} 4 & 3 \\[0.3em] -3 & 0 \\[0.3em] -1 & -2 \end{bmatrix}$$\\$ $A'-B'=\begin{bmatrix} 3 & 4 \\[0.3em] -1 & 2 \\[0.3em] 0 & 1 \end{bmatrix}-\begin{bmatrix} -1 & 1 \\[0.3em] 2 & 2 \\[0.3em] 1 & 3 \end{bmatrix}$$\\$ $=\begin{bmatrix} 4 & 3 \\[0.3em] -3 & 0 \\[0.3em] -1 & -2 \end{bmatrix}$$\\$ Thus, we have verified that$(A-B)'=A'-B'.$

35   If $ A'=\begin{bmatrix} -2 & 3 \\[0.3em] 1 & 2\end{bmatrix}$ and $B=\begin{bmatrix} -1 & 0 \\[0.3em] 1 & 2\end{bmatrix}$,then find $(A+2B)$

Solution :

We know that $ A =(A')'$$\\$ $\therefore A=\begin{bmatrix} -2 & 1 \\[0.3em] 3 & 2 \end{bmatrix}$$\\$ $\therefore A+2B=\begin{bmatrix} -2 & 1 \\[0.3em] 3 & 2 \end{bmatrix}+2\begin{bmatrix} -1 & 0 \\[0.3em] 1 & 2 \end{bmatrix}$$\\$ $=\begin{bmatrix} -2 & 1 \\[0.3em] 3 & 2 \end{bmatrix}+\begin{bmatrix} -2 & 0 \\[0.3em] 2 & 4 \end{bmatrix}$$\\$ $=\begin{bmatrix} -4 & 1 \\[0.3em] 5 & 6 \end{bmatrix}$$\\$ $\therefore (A+2B)'=\begin{bmatrix} -4& 5 \\[0.3em] 1 & 6 \end{bmatrix}$

36   For the matrices A and B , verify that $(AB)'=B'A'$ where $\\$ (i)$A=\begin{bmatrix} 1\\[0.3em] -4\\[0.3em] 3\end{bmatrix},B=\begin{bmatrix} -1 & 2 & 1\end{bmatrix}$$\\$ (ii)$A=\begin{bmatrix} 0\\[0.3em] 1\\[0.3em] 2\end{bmatrix},B=\begin{bmatrix} 1&5&7\end{bmatrix}$

Solution :

(i)$A=\begin{bmatrix} 1\\[0.3em] -4\\[0.3em] 3\end{bmatrix}\begin{bmatrix} -1 & 2 & 1\end{bmatrix}$$\\$ $=\begin{bmatrix} -1 & 2 &1\\[0.3em] 4 & 8 &-4\\[0.3em] -3 & 6 &3\end{bmatrix}$$\\$ $\therefore (AB)'=\begin{bmatrix} -1 & 4 & -3\\[0.3em] 2 & -8 & 6\\[0.3em] 1 & -4 & 3\end{bmatrix}$$\\$ Now,$A=\begin{bmatrix} 1 & -4 &3\end{bmatrix},B=\begin{bmatrix} -1\\[0.3em] 2\\[0.3em] 1\end{bmatrix}$$\\$ $\therefore B'A'=\begin{bmatrix} -1\\[0.3em] 2\\[0.3em] 1\end{bmatrix}\begin{bmatrix} 1 & -4 &3\end{bmatrix}$$\\$ $=\begin{bmatrix} -1 & 4 & -3\\[0.3em] 2 & -8 & 6\\[0.3em] 1 & -4 & 3 \end{bmatrix}$$\\$ Hence, we have verified $(AB )= B ' A ' .$

(ii)$AB=\begin{bmatrix} 0 \\[0.3em] 1\\[0.3em] 2\end{bmatrix}\begin{bmatrix} 1 & 5 & 7\end{bmatrix}$$\\$ $=\begin{bmatrix} 0& 0 & 0 \\[0.3em] 1& 5 & 7 \\[0.3em] 2& 10 & 14 \end{bmatrix}$$\\$ $ \therefore (AB)'=\begin{bmatrix} 0& 1 & 2 \\[0.3em] 0 & 5 & 10\\[0.3em] 0 & 7 & 14\end{bmatrix}$$\\$ Now , $ A'=\begin{bmatrix} 0 & 1 & 2\end{bmatrix},B=\begin{bmatrix} 1 \\[0.3em] 5\\[0.3em] 7\end{bmatrix}$$\\$ $\therefore B'A'=\begin{bmatrix} 1 \\[0.3em] 5\\[0.3em] 7\end{bmatrix}\begin{bmatrix} 0 & 1 & 2\end{bmatrix}$$\\$ $=\begin{bmatrix} 0 & 1 & 2\\[0.3em] 0 & 5 & 10 \\[0.3em] 0 & 7 & 14\end{bmatrix}$$\\$ Hence, we have verified that$(AB)'=B'A'.$

37   If (i) $A=\begin{bmatrix} \cos \alpha & \sin \alpha \\[0.3em] -\sin \alpha & \cos \alpha \end{bmatrix},$ then verify that $ A'A=1$$\\$ (ii) $A=\begin{bmatrix} \sin \alpha & \cos \alpha \\[0.3em] -\cos \alpha & \sin \alpha \end{bmatrix},$ then verify that $ A'A=1 $

Solution :

$=\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.$

$=\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.$

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 :

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')$$\\$ $=\dfrac{1}{2} \begin{bmatrix} 12 & -4 & 4\\[0.3em] -4 & 6 & -2 \\[0.3em] 4 & -2 & 6 \end{bmatrix}=\begin{bmatrix} 6 & -2 & 2\\[0.3em] -2 & 3 & -1 \\[0.3em] 2 & -1 & 3 \end{bmatrix}$$\\$

(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}$$\\$