# Determinants

## Class 12 NCERT

### NCERT

1   Evaluate the determinants in Exercise $1$ and $2.$ $\begin{vmatrix} 2 & 4 \\[0.3em] -5 & -1 \end{vmatrix}$

##### Solution :

$\begin{vmatrix} 2 & 4 \\[0.3em] -5 & -1 \end{vmatrix}$ =$2(-1)-4(-5) =-2+20=18$

2   Evaluate the determinants in Exercise $1$ and $2.$$\\ (i) \begin{vmatrix} \cos \theta & -\sin \theta\\[0.3em] \sin \theta & \cos \theta \end{vmatrix} \ \ \ \ (ii) \begin{vmatrix} x^2-x+1 & x-1\\[0.3em] x+1 & x+1 \end{vmatrix} ##### Solution : (ii) \begin{vmatrix} x^2-x+1 & x-1\\[0.3em] x+1 & x+1 \end{vmatrix}$$\\$ $=(x^2-x+1)(x+1)-(x-1)(x+1)\\ =x^3-x^2+x+x^2-x+1-(x^2-1)\\ =x^3+1-x^2+1\\ =x^3-x^2+2$

$(i) \begin{vmatrix} \cos \theta & -\sin \theta\\[0.3em] \sin \theta & \cos \theta \end{vmatrix}$ = $(\cos \theta )(\cos \theta ) -(-\sin \theta)(\sin \theta ) =\cos^2\theta + \sin ^\theta =1$

3   If $A=\begin{bmatrix} 1 &2 \\[0.3em] 4 &2 \end{bmatrix}$ ,then show that $|2A|=4|A|$

##### Solution :

The given matrix is $A=\begin{bmatrix} 1 &2 \\[0.3em] 4 &2 \end{bmatrix}$$\\ \therefore 2A =2\begin{bmatrix} 1 &2 \\[0.3em] 4 &2 \end{bmatrix}=\begin{bmatrix} 2&4 \\[0.3em] 8&4 \end{bmatrix}\\ L.H.S:|2A|=\begin{vmatrix} 2&4 \\[0.3em] 8& 4 \end {vmatrix} \\ =2*4-4*8\\ =8-32=-24$$\\$ Now , $|A|=\begin{bmatrix} 1& 2 \\[0.3em] 4 &2 \end {bmatrix} =1*2-2*4=2*8=-6\\ R>H>S:4|A|=4*(-6) =-24\\ \therefore L.H.S.=\therefore R.H.S$

4   $A=\begin{bmatrix} 1 &0 &1 \\[0.3em] 0& 1& 2 \\[0.3em] 0 & 0 & 4 \end{bmatrix}$ , then show that $|3A|=27|A|.$

##### Solution :

The given matrix is $A=\begin {bmatrix} 1&0 &1 \\[0.3em] 0& 1& 2 \\[0.3em] 0 & 0 & 4 \end{bmatrix}$.$\\$ It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $(C 1 )$ for easier calculation.$\\$ $|A|=1\begin{vmatrix} 1 & 2\\[0.3em] 0 & 4 \end{vmatrix}$ $-0 \begin{vmatrix} 0 &1 \\[0.3em] 0 &4 \end{vmatrix}$ $+0 \begin{vmatrix} 0 & 1 \\[0.3em] 1 &2 \end{vmatrix}$ $=1(4-0)-0+0=4\\ \therefore 27|A|=27(4)=108 .....(i)$ $\\$ Now , $3A=3\begin{bmatrix} 1&0 & 1 \\[0.3em] 0 & 1 & 2 \\[0.3em] 0 & 0 & 4 \end {bmatrix}$ $\\$ $\therefore |3A|=3 \begin{vmatrix} 3 & 6\\[0.3em] 0 & 12 \end{vmatrix}$ $-0 \begin{vmatrix} 0 & 3 \\[0.3em] 0& 12 \end{vmatrix}$ $+0 \begin{vmatrix} 0 & 3 \\[0.3em] 3 & 6 \end{vmatrix}$$\\$$ =3(36-0)=3(36)=108 ....(ii)$$\\ From equations (i) and (ii), we have: |3 A = 27 A|$$\\$ Hence, the given result is proved.

5   Evaluate the determinants $\\$ $(i) \begin{vmatrix} 3 & -1 & -2 \\[0.3em] 0 & 0 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$ $\ \ \$ $(ii) \begin{vmatrix} 3 & -4 & 5 \\[0.3em] 1 & 1 & -2 \\[0.3em] 2 & 3 & 1 \end{vmatrix}$ $\ \ \$ $(iii) \begin{vmatrix} 0 & 1 & 2 \\[0.3em] -1 & 0 & -3 \\[0.3em] -2 & 3 & 0\end{vmatrix}$ $(iv) \begin{vmatrix} 2 & -1 & -2 \\[0.3em] 0 & 2 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$

##### Solution :

(ii)Let $\begin{vmatrix} 3 & -4 & 5 \\[0.3em] 1 & 1 & -2 \\[0.3em] 2 & 3 & 1 \end{vmatrix}$$\\ By expanding along the first row, we have:\\ |A| =3\begin{vmatrix} 1 & -2 \\[0.3em] 3 & 1\end{vmatrix} +4\begin{vmatrix} 1 & -2 \\[0.3em] 2 & 1 \end{vmatrix} +5\begin{vmatrix} 1 & 1 \\[0.3em] 2 & 3 \end{vmatrix}$$\\$ $=3(1+6)+4(1+4)+5(3-2) \\ =3(7)+4(5)+5(1)\\ 21+20+5=46$

(iv) Let $A=\begin{vmatrix} 2 & -1 & -2 \\[0.3em] 0 & 2 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$$\\ By expanding along the first column, we have:\\ |A|=2\begin{vmatrix} 2 & -1\\[0.3em] -5 & 0\end{vmatrix} -0\begin{vmatrix} -1 & -2 \\[0.3em] -5 &0 \end{vmatrix}+3\begin{vmatrix} -1 & -2 \\[0.3em] 2 & -1 \end{vmatrix}$$\\$ $=2(0-5) -0+3(1+4)\\ =-10+15=5$

$(i) \begin{vmatrix} 3 & -1 & -2 \\[0.3em] 0 & 0 & -1 \\[0.3em] 3 & -5 & 0 \end{vmatrix}$$\\ It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.\\ |A|=-0\begin{vmatrix} -1 & -2 \\[0.3em] -5 & 0\end{vmatrix} +0\begin{vmatrix} 3 & -2 \\[0.3em] 3 & 0\end{vmatrix} -(-1)\begin{vmatrix} 3 & -1 \\[0.3em] 3 & -5\end{vmatrix} =(-15+3) =-12 (iii) Let A=\begin{vmatrix} 0 & 1 & 2 \\[0.3em] -1 & 0 & -3 \\[0.3em] -2 & 3 & 0\end{vmatrix}$$\\$ By expanding along the first row, we have:$\\$ $|A|=0\begin{vmatrix} 0 & -3 \\[0.3em] 3 & 0\end{vmatrix} -1\begin{vmatrix} -1 & -3 \\[0.3em] -2 & 0\end{vmatrix} +2\begin{vmatrix} -1 & 0\\[0.3em] -2 & 3 \end{vmatrix}$$\\ =0-1(0-6)+2(-3-0)\\ =-1(-6)+2(-3)\\ =6-6=0 6 If A=\begin{vmatrix} 1&1&-2\\[0.3em] 2&1&-3\\[0.3em] 5&4&-9 \end{vmatrix},find |A| ##### Solution : Let A=\begin{bmatrix} 1&1&-2\\[0.3em] 2&1&-3\\[0.3em] 5&4&-9 \end{bmatrix}$$\\$ By expanding along the first row, we have:$\\$ $A=\begin{bmatrix} 1&1&-2\\[0.3em] 2&1&-3\\[0.3em] 5&4&-9 \end{bmatrix}$$\\ |A|=1\begin{vmatrix} 1 & -3\\[0.3em] 4 & -9 \end{vmatrix}-1\begin{vmatrix} 2 & -3\\[0.3em] 5 & -9 \end{vmatrix}-2\begin{vmatrix} 2 & 1 \\[0.3em] 5 & 4 \end{vmatrix}$$\\$ $=1(-9+12)-1(-18+15)-2(8-5)\\ =1(3)-1(-3)-2(3)\\ =3+3-6\\ =6-6\\ =0$$\\$$\\$$\\ 7 Find values of x,if\\ (i)\begin{vmatrix} 2& 4\\[0.3em] 5 & 1\end{vmatrix}=\begin{vmatrix} 2x& 4\\[0.3em] 6 & x\end{vmatrix} \ \ \ \ (ii) \begin{vmatrix} 2& 3\\[0.3em] 4 & 5\end{vmatrix}=\begin{vmatrix} x& 3\\[0.3em] 2x & 5\end{vmatrix} ##### Solution : (i)\begin{vmatrix} 2& 4\\[0.3em] 5 & 1\end{vmatrix}=\begin{vmatrix} 2x& 4\\[0.3em] 6 & x\end{vmatrix} \ \ \ \$$\\$ $\Rightarrow 2*1-5*4=2x*x-6*4\\ \Rightarrow 2-20=2x^2-24\\ \Rightarrow 2x^2=6\\ \Rightarrow x^2=3\\ \Rightarrow x=\pm3$$\\ (ii) \begin{vmatrix} 2& 3\\[0.3em] 4 & 5\end{vmatrix}=\begin{vmatrix} x& 3\\[0.3em] 2x & 5\end{vmatrix} \\ \Rightarrow 2*5-3*4=x*5-3*2x\\ \Rightarrow10-12=5x-6x\\ \Rightarrow-2=-x\\ \Rightarrow x=2$$\\$

8   If$\begin{vmatrix} x & 2\\[0.3em] 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2\\[0.3em] 18 & 6 \end{vmatrix}$,then x is equal to $\\$

$\begin{vmatrix} x & 2\\[0.3em] 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2\\[0.3em] 18 & 6 \end{vmatrix}$ $\Rightarrow x^2-36=36-36\\ \Rightarrow x^2-36=0\\ \Rightarrow x^2=36\\ \Rightarrow x=\pm6$$\\ Hence, the correct answer is B. 9 Using the property of determinants and without expanding, prove that:\\ \begin{vmatrix} x&a&x+a \\[0.3em] y & b& y+b \\[0.3em] z & c & Z+c \end{vmatrix}=0 ##### Solution : \begin{vmatrix} x&a&x+a \\[0.3em] y & b& y+b \\[0.3em] z & c & z+c \end{vmatrix}=\begin{vmatrix} x&a&x \\[0.3em] y & b& y \\[0.3em] z & c & z \end{vmatrix}+\begin{vmatrix} x&a&a \\[0.3em] y & b& b \\[0.3em] z & c & c \end{vmatrix}$$\\$ Clearly, the two determinants have two identical columns. Thus,$\\$ $=0+0=0$

10   Using the property of determinants and without expanding, prove that:$\\$ $\begin{vmatrix} a-b &b-c & c-a \\[0.3em] b-c &c-a & a-b \\[0.3em] c-a & a-b & b-c\end{vmatrix}=0$

##### Solution :

$\Delta =$ $\begin{vmatrix} a-b &b-c & c-a \\[0.3em] b-c &c-a & a-b \\[0.3em] c-a & a-b & b-c\end{vmatrix}=0$ $\\$ Applying $R _1 \to R _1 + R_ 2$ , we have:$\\$ $\Delta =\begin{vmatrix} a-c & b-a & c-b\\[0.3em] b-c & c-a & a-b\\[0.3em] -(a-c)& -(b-a) & -(c-b)\end{vmatrix}$$\\ =-\begin{vmatrix} a-c & b-a & c-b \\[0.3em] b-c & c-a & a-b \\[0.3em] a-c & b-a & c-b\end{vmatrix}$$\\$ Here, the two rows $R _1$ and $R _3$ are identical.$\\$ $\therefore \Delta =0$$\\$

11   Using the property of determinants and without expanding, prove that: $\\$ $\begin{vmatrix} 2 & 7 & 65 \\[0.3em] 3 & 8 & 75 \\[0.3em] 5 & 9 & 86 \\[0.3em] \end{vmatrix} =0$

##### Solution :

$\begin{vmatrix} 2 & 7 & 65 \\[0.3em] 3 & 8 & 75 \\[0.3em] 5 & 9 & 86 \\[0.3em] \end{vmatrix} = \begin{vmatrix} 2 & 7 & 63+2 \\[0.3em] 3 & 8 & 72+3 \\[0.3em] 5 & 9 & 81+5 \\[0.3em] \end{vmatrix}$ $\\$ = $\begin{vmatrix} 2 & 7 & 63 \\[0.3em] 3 & 8 & 72 \\[0.3em] 5 & 9 & 81 \\[0.3em] \end{vmatrix} = \begin{vmatrix} 2 & 7 & 2 \\[0.3em] 3 & 8 & 3 \\[0.3em] 5 & 9 & 5 \\[0.3em] \end{vmatrix}$ $\\$ $\begin{vmatrix} 2 & 7 & 9(7) \\[0.3em] 3 & 8 & 9(8) \\[0.3em] 5 & 9 & 9(9) \\[0.3em] \end{vmatrix} +0$ $\\$ = $\begin{vmatrix} 2 & 7 & 7 \\[0.3em] 3 & 8 & 8 \\[0.3em] 5 & 9 & 9 \\[0.3em] \end{vmatrix}$ $\\$ $=0$

12   Using the property of determinants and without expanding, prove that: $\\$ $\begin{vmatrix} 1 & bc & a(b+c)\\[0.3em] 1 & ca & b(c+a)\\[0.3em] 1 & ca & c(a+b) \end{vmatrix}=0$ $\\$

##### Solution :

$\bigtriangleup = \begin{vmatrix} 1 & bc & a\left( b+c\right) \\[0.3em] 1 & ca & b\left( c+a\right) \\[0.3em] 1 & ca & c\left( a+b\right) \end{vmatrix}$ $\\$ By applying$C 3 \to C 3 + C 2 .$ We have: $\\$ $\bigtriangleup = \begin{vmatrix} 1 & bc & ab+bc+ca \\[0.3em] 1 & ca & ab+bc+ca \\[0.3em] 1 & ca & ab+bc+ca \end{vmatrix}$ $\\$ Here. Two columns $C1$ and $C3$ are proportional. $\\$ $\therefore \bigtriangleup =0.$

13   Using the property of determinants and without expanding, prove that: $\\$ $\begin{vmatrix} b+c & q+r & y+z \\[0.3em] c+a & r+p & z+x \\[0.3em] a+b & p+q & x+y \end{vmatrix}$ = 2 $\begin{vmatrix} a & p & x \\[0.3em] b & q & y \\[0.3em] c & r& z \end{vmatrix}$

##### Solution :

$\bigtriangleup = \begin{vmatrix} b+c & q+r & y+z \\[0.3em] c+a & r+p & z+x \\[0.3em] a+b & p+q & z+y \end{vmatrix}$ $\\$ = $\bigtriangleup = \begin{vmatrix} b+c & q+r & y+z \\[0.3em] c+a & r+p & z+x \\[0.3em] a & p & x \end{vmatrix}$ + $\begin{vmatrix} b+c & q+r & y+z \\[0.3em] c+a & r+p & z+x \\[0.3em] b & q & y \end{vmatrix}$ $\\$ $=\bigtriangleup_1+ \bigtriangleup_2$ (say) $\\$ Now, $\bigtriangleup_1 = \begin{vmatrix} b+c & q+r & y+z \\[0.3em] c+a & r+p & z+x \\[0.3em] a & p & x \end{vmatrix}$ $\\$ Applying $R_1 \to R_1 -R_2$, we have: $\\$ $\bigtriangleup_1 = \begin{vmatrix} b & q & y \\[0.3em] c & r & z \\[0.3em] a & p & x \end{vmatrix}$ $\\$ Applying $R_1 \leftrightarrow R_3$ and $R_2 \leftrightarrow R_3$, we have: $\\$

$\bigtriangleup_1 = \left( -1\right)^2 \begin{vmatrix} a & p & x \\[0.3em] b & q & y \\[0.3em] c & r & z \end{vmatrix} = \begin{vmatrix} a & p & x \\[0.3em] b & q & y \\[0.3em] c & r & z \end{vmatrix}$ $\\$ $\bigtriangleup_2 = \begin{vmatrix} b+c & q+r & y+z \\[0.3em] c+a & r+p & z+x \\[0.3em] b & q & y \end{vmatrix}$ $\\$ Applying $R_1 \to R_1 - R_1,$ we have: $\\$ $\bigtriangleup_2 = \begin{vmatrix} c & r & z \\[0.3em] c+a & r+p & z+x \\[0.3em] b & q & y \end{vmatrix}$ $\\$ Applying $R_2 \to R_2 - R_3,$ we have: $\\$ $\bigtriangleup_2 = \begin{vmatrix} c & r & z \\[0.3em] a & p & x \\[0.3em] b & q & y \end{vmatrix}$ $\\$ Applying $R_1 \to R_2$ and $R_2 \to R_3$, we have: $\\$

$\bigtriangleup_2 = \left( -1\right)^2 \begin{vmatrix} a & p & x \\[0.3em] b & q & y \\[0.3em] c & r & z \end{vmatrix} = \begin{vmatrix} a & p & x \\[0.3em] b & q & y \\[0.3em] c & r & z \end{vmatrix}$ $\\$ From (1),(2), and(3), we have: $\bigtriangleup =2 \begin{vmatrix} a & p & x \\[0.3em] b & q & y \\[0.3em] c & r & z \end{vmatrix}$ $\\$ Hence, the given result is proved.

14   By using properties of determinants, show that: $\\$ $\begin{vmatrix} 0 & a & -b \\[0.3em] -a & 0 & -c \\[0.3em] b & c& 0 \end{vmatrix} =0$

##### Solution :

We have, $\Delta =$ $\begin{vmatrix} 0 & a & -b \\[0.3em] -a & 0 & -c \\[0.3em] b & c & 0 \end{vmatrix}=0$ $\\$ Applying $R_1 \to cR_1$ , we have : $\\$ $\Delta =$ $\dfrac{1}{c} \begin{vmatrix} 0 & ac & -bc \\[0.3em] -a & 0 & -c \\[0.3em] b & c & 0 \end{vmatrix}=0$ $\\$ Applying $R_1 \to R_1-bR_1$ , we have : $\\$

$\Delta =$ $\dfrac{1}{c} \begin{vmatrix} ab & ac & 0 \\[0.3em] -a & 0 & -c \\[0.3em] b & c & 0 \end{vmatrix}=0$ $\\$ = $\dfrac{a}{c} \begin{vmatrix} b & c & 0 \\[0.3em] -a & 0 & -c \\[0.3em] b & c & 0 \end{vmatrix}=0$ $\\$ Here, the two rows $R_1$ and $R_3$ are identical. $\\$ $\therefore \bigtriangleup =0.$

15   By using properties of determinants, show that: $\\$ $\Delta =$ $\begin{vmatrix} a^2& ab & ac \\[0.3em] ba & -b2 & bc \\[0.3em] ca & cb & -c^2\end{vmatrix}= 4a^2b^2c^2$ $\\$

##### Solution :

Applying $R_2 \to R_2 +R_1$ and $R_3 \to R_3 + R_1,$ we have: $\\$ $\Delta = a^2b^2c^2$ $\begin{vmatrix} -1 & 1 & 1 \\[0.3em] 0 & 0 & 2 \\[0.3em] 0 & 2 & 0\end{vmatrix}=0$ $\\$ $= a^2b^2c^2 \left( -1\right)$ $\begin{vmatrix} 0 & 2 \\[0.3em] 2 & 0\end{vmatrix}$ $\\$ $=- a^2b^2c^2 \left( 0-4\right) = 4a^2b^2c^2$

$\Delta =$ $\begin{vmatrix} -a^2 & ab & ac \\[0.3em] ba & -b^2 & bc \\[0.3em] ca & cb & -c^2\end{vmatrix}=0$ $\\$ = abc $\begin{vmatrix} -a & b & c \\[0.3em] a & -b & c \\[0.3em] a & b & -c\end{vmatrix}=0$ $\\$ == $a^2b^2c^2$ $\begin{vmatrix} -1 & 1 & 1 \\[0.3em] 1 & -1 & 1 \\[0.3em] 1 & 1 & -1 \end{vmatrix}$ $\\$

16   By using properties of determinants, show that: $\\$ $(i) \begin{vmatrix} 1 & a & a^2 \\[0.3em] 1 & b & b^2 \\[0.3em] 1 & c& c^2 \end{vmatrix}= \left( a+b\right) \left( b-c\right) \left( c-a\right)$ $\\$ $(ii) \begin{vmatrix} 1 & 1 & 1 \\[0.3em] 1 & b & c \\[0.3em] a^3 & b^3 & c^3 \end{vmatrix}= \left( a-b\right) \left( b-c\right) \left( c-a\right) \left( a+b+c\right)$ $\\$

##### Solution :

$\bigtriangleup = \left( a-b\right) \left( b-c\right) \left( c-a\right) \left( a+b+c\right) (-1) \begin{vmatrix} 0 & 1\\[0.3em] 1 & c\end{vmatrix}$ $\\$ $= \left( a-b\right) \left( b-c\right) \left( c-a\right) \left( a+b+c\right)$ $\\$ Hence, the given result is proved.

$= \left( c-a \right) \left( b-c \right) \left( c-a \right) \begin{vmatrix} 0 & 0 & -a+b \\[0.3em] 0 & 1 & b+c \\[0.3em] 1 & c & c^2 \end{vmatrix}$ $\\$ Expanding along $C_1$, we have: $\\$ $\bigtriangleup = \left( c-a \right) \left( b-c \right) \left( c-a \right) \begin{vmatrix} 0 & -1 \\[0.3em] 1 & b+c \\[0.3em] \end{vmatrix} = \left( c-a \right) \left( b-c \right) \left( c-a \right)$ $\\$ Hence, the given result is proved. $\\$

$\bigtriangleup = \left( a-c\right) \left( b-c\right) \begin{vmatrix} 0 & 0 & 1 \\[0.3em] -1 & 1 & c \\[0.3em] -\left( a^2 + ac+c^2\right) & \left( b^2 + bc+c^2\right) & c^3 \end{vmatrix}$ $\\$ $= \left( a-c\right) \left( b-c\right) \left( a-b\right) \begin{vmatrix} 0 & 0 & 1 \\[0.3em] -1 & 1 & c \\[0.3em] -\left( a + b+c\right) & \left( b^2 + bc+c^2\right) & c^3 \end{vmatrix}$ $\\$ $= \left( a-b\right) \left( b-c\right) \left( c-a\right) \left( a+b+c\right) \begin{vmatrix} 0 & 0 & 1 \\[0.3em] -1 & 1 & c \\[0.3em] -1 & \left( b^2 + bc+c^2\right) & c^3 \end{vmatrix}$ $\\$ Expanding along C 1 , we have: $\\$

(i) Let $\bigtriangleup = \begin{vmatrix} 1 & a & a^2 \\[0.3em] 1 & b & b^2 \\[0.3em] 1 & c& c^2 \end{vmatrix}$ $\\$ Applying $R_1 \to R_1 -R_3$ and $R_2 \to R_2-R_3$, we have: $\\$ $\bigtriangleup = \begin{vmatrix} 0 & a-c & a^2-c^2 \\[0.3em] 0 & b-c & b^2-c^2 \\[0.3em] 1 & c & c^2 \end{vmatrix}$ $\\$ =$\left( c-a \right) \left( b-c \right)$ $\begin{vmatrix} 0 & -1 & -a-c \\[0.3em] 0 & 1 & b+c \\[0.3em] 1 & c & c^2\end{vmatrix}=0$ $\\$ Applying $R_1 \to R_1 +R_2$, we have: $\\$ $\bigtriangleup = \left( b-c \right) \left( c-a \right) \begin{vmatrix} 0 & 0 & -a+b \\[0.3em] 0 & 1 & b+c \\[0.3em] 1 & c & c^2 \end{vmatrix}$ $\\$

(ii) Let $\bigtriangleup = \begin{vmatrix} 1 & 1 & 1 \\[0.3em] a & b & c \\[0.3em] a^3 & b^3& c^3 \end{vmatrix}$ $\\$ Applying $C_1 \to C_1 -C_3$ and $C_2 \to C_2-C_3$, we have: $\\$ $\bigtriangleup = \begin{vmatrix} 0 & 0 & 1 \\[0.3em] a-c & b-c & c \\[0.3em] a^3-c^3 & b^3- c^3 & c^3 \end{vmatrix}$ $\\$ =$\begin{vmatrix} 0 & 0 & 1 \\[0.3em] a-c & b-c & c \\[0.3em] \left( a-c\right) \left( a^2 + ac+c^2\right) & \left( b-c\right) \left( b^2 + bc+c^2\right) & c^3 \end{vmatrix}$ $\\$ =$\left( a-c\right) \left( b-c\right) \begin{vmatrix} 0 & 0 & 1 \\[0.3em] -1 & 1 & c \\[0.3em] -\left( a^2 + ac+c^2\right) & \left( b^2 + bc+c^2\right) & c^3 \end{vmatrix}$ $\\$ Applying $C_1 \to C_1 +C_2,$ we have: $\\$

17   By using properties of determinants, show that: $\\$ $\begin{vmatrix} x & x^2 & yz \\[0.3em] y & y^2 & zx \\[0.3em] z & z^2 & xy\end{vmatrix}= \left( x-y\right) \left( y-z\right) \left( z-x\right) \left( xy+yz+zx\right)$

##### Solution :

$\bigtriangleup = \left( x-y\right) \left( z-x\right) \begin{vmatrix} x &x^2 &yz \\[0.3em] -1 & -x-y & z \\[0.3em] 1 & z-y & z-y \end{vmatrix}$ $\\$ $\ = [\left( x-y\right) \left( z-x\right) \left( z-y\right)] \begin{vmatrix} x &x^2 &yz \\[0.3em] -1 & -x-y & z \\[0.3em] 0 & 1 & 1 \end{vmatrix}$ $\\$ Expanding along R_3, we have: $\\$ $\bigtriangleup = \left( x-y\right) \left( z-x\right) \begin{bmatrix} (-1) \begin{vmatrix} x & yz \\[0.3em] -1 & z \\[0.3em] \end{vmatrix} +1 \begin{vmatrix} x & x^2 \\[0.3em] -1 & -x-y \\[0.3em] \end{vmatrix} Let$\bigtriangleup = \begin{vmatrix} x &x^2 &yz \\[0.3em] y & y^2 & zx \\[0.3em] z & z^2 & xy \end{vmatrix} \\$Applying$ R_2 \to R_2-R_1$and$ R_3 \to R_3-R_1,$we have:$\\\bigtriangleup = \begin{vmatrix} x &x^2 &yz \\[0.3em] y-x & y^2-x^2 & zx-yz \\[0.3em] z-x & z^2-x^2 & xy-yz \end{vmatrix} \\$=$\begin{vmatrix} x &x^2 &yz \\[0.3em] -(x-y) & -(x-y)(x+y) & z(x-y) \\[0.3em] (z-x) & (z-x)(z+x) & -y(z-x) \end{vmatrix} \\$=$ \left( x-y\right) \left( z-x\right) \begin{vmatrix} x &x^2 &yz \\[0.3em] -1 & -x-y & z \\[0.3em] 1 & z-y & z-y \end{vmatrix} \\$Applying$ R_2 \to R_3+R_2, $we have:$\\$\end{bmatrix}$ $\\$ $= \left( x-y\right) \left( z-x\right) \left( z-y\right) \begin{bmatrix} \left( -xz-yz\right) + \left( -x^2-xy +x^2\right) \end{bmatrix}$ $\\$ $= -\left( x-y\right) \left( z-x\right) \left( z-y\right) \left( xy+yz +zx\right)$ $\\$ $= \left( x-y\right) \left( y-z\right) \left( z-x\right) \left( xy+yz +zx\right)$ $\\$ Hence, the given result is proved.

18   By using properties of determinants, show that: $\\$ (i)$\begin{vmatrix} x+4 & 2x & 2x \\[0.3em] 2x & x+4 & 2x \\[0.3em] 2x & 2x & x+4\end{vmatrix}= \left( 5x+4 \right) \left( 4-x \right)^2$ $\\$ (ii) $\begin{vmatrix} y+k & y & y \\[0.3em] y & y+k & y \\[0.3em] y & y & y+k\end{vmatrix}= k^2 \left( 3x+k \right)^2$ $\\$

##### Solution :

$= \left( 5x+4 \right) \left( 4 - x \right) \left( 4 - x \right)$ $\begin{vmatrix} 1 & 0 & 0 \\[0.3em] 2x & 1 & 0 \\[0.3em] 2x & 0 & 1 \end{vmatrix}$ $\\$ Expanding along $C_3,$ we have: $\\$ $\Delta = \left( 5x+4 \right)\left( 4 -x\right)^2$ $\begin{vmatrix} 1 & 0 \\[0.3em] 2x & 1 \\[0.3em] \end{vmatrix}=0$ $\\$ Hence, the given result is proved.

$\Delta = \left( 3y+k\right)$ $\begin{vmatrix} 1 & 0 & 0\\[0.3em] y & k & 0\\[0.3em] y & 0 & k \end{vmatrix}$ $\\$ $\Delta = k^2\left( 3x+k\right)$ $\begin{vmatrix} 1 & 0 & 0\\[0.3em] y & 1 & 0\\[0.3em] y & 0 & 1 \end{vmatrix}$ $\\$ Expanding alone $C_3,$ we have: $\\$ $\Delta = k^2\left( 3x+k\right)$ $\begin{vmatrix} 1 & 0 \\[0.3em] y & 1\\[0.3em] \end{vmatrix}$ = k^2 ( 3y+k\right) $\\$ Hence, the given result is proved.

(i)$\Delta =$ $\begin{vmatrix} x+4 & 2x & 2x \\[0.3em] 2x & x+4 & 2x \\[0.3em] 2x & 2x & x+4\end{vmatrix}$ $\\$ Applying $R_1\to R_1+R_2+R_3,$ we have: $\Delta =$ $\begin{vmatrix} 5x+4 & 5x+4 & 5x+4 \\[0.3em] 2x & x+4 & 2x \\[0.3em] 2x & 2x & x+4\end{vmatrix}$ $\\$ $= \left( 5x+4 \right)$ $\begin{vmatrix} 1 & 0 & 0 \\[0.3em] 2x & x+4 & 0 \\[0.3em] 2x & 0 & -x+4\end{vmatrix}$ $\\$

(ii) $\Delta =$ $\begin{vmatrix} y+k & y & y \\[0.3em] y & y+k & y \\[0.3em] y & y & y+k \end{vmatrix}$ $\\$ Applying $R_1 \to R_1 + R_2R_3,$ we have: $\\$ $\Delta =$ $\begin{vmatrix} 3y+k & 3y+k & 3y+k \\[0.3em] y & y+k & y \\[0.3em] y & y & y+k \end{vmatrix}$ $\\$ = $\left( 3y+k\right)$ $\begin{vmatrix} 1 & 0 & 0 \\[0.3em] y & y+k & y \\[0.3em] y & y & y+k \end{vmatrix}$ $\\$ Applying $C_2 \to C_2-C_1$ and$C_3 \to C_3-C_1,$ we have: $\\$

19   By using properties of determinants, show that: $\\$ (i) $\begin{vmatrix} a-b-c & 2a & 2a \\[0.3em] 2b & b-c-a & 2b \\[0.3em] 2c & 2c & c-a-b \end{vmatrix}= \left( a+b+c \right)^3$ $\\$ (ii) $\begin{vmatrix} x+y+2z & x & y \\[0.3em] z & y+z+2x & y \\[0.3em] z & x & z+x+2y \end{vmatrix}= 2 \left( x+y+z \right)^3$ $\\$

##### Solution :

$= \left( a+b+c \right)^3$ $\begin{vmatrix} 1 & 0 & 0 \\[0.3em] 2b & -1 & 0 \\[0.3em] 2c & 0 & -1\end{vmatrix}=0$ $\\$ Expanding along $C_3$, we have: $\Delta = \left( a+b+c \right)^3 \left( -1\right)\left( -1 \right) = \left( a+b+c \right)^3$ $\\$ Hence, the given result is proved. $\\$

(i) $\begin{vmatrix} a-b-c & 2a & 2a \\[0.3em] 2b & b-c-a & 2b \\[0.3em] 2c & 2c & c-a-b \end{vmatrix}= \left( a+b+c \right)^3$ $\\$ Applying R_1\to R_1 + R_2 +R_3, we have: $\\$ $\Delta =$ $\begin{vmatrix} a+b+c & a+b+c & a+b+c \\[0.3em] 2b &b-c-a & 2b \\[0.3em] 2c & 2c & c-a-b\end{vmatrix}=0$ $\\$ Applying $C_2 \to C_2 - C_1 , C_3 \to C_3 - C_1$, we have: $\\$ $\Delta = \left( a+b+c \right)$ $\begin{vmatrix} 1 & 0 & 0 \\[0.3em] 2b &-(a+b+c) & 0 \\[0.3em] 2c & 0 & -(a+b+c)\end{vmatrix}=0$ $\\$

$\Delta =$ $\begin{vmatrix} 2(x+y+z) & x & y\\[0.3em] 2(x+y+z) & y+z+2x & y \\[0.3em] 2(x+y+z) & x & z+x+2y \end{vmatrix}=0$ $\\$ Appling $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1,$ we have: $\\$ $\Delta = 2\left( x+y+z \right)^3 \left( 1\right)\left( 1-0 \right) = 2\left( x+y+z \right)^3$ $\\$ Hence, the given result is proved.

20   By using properties of determinants, show that: $\\$ $\begin{vmatrix} 1 & x & x^2 \\[0.3em] x^2 & 1 & x \\[0.3em] x & x^2 & 1\end{vmatrix} = \left( 1-x^3 \right)^2$ $\\$

##### Solution :

$\Delta = \begin{vmatrix} 1 & x & x^2 \\[0.3em] x^2 & 1 & x \\[0.3em] x & x^2 & 1\end{vmatrix}$ $\\$ Applying $R_1 \to R_1 + R_2 + R_3,$ we have: $\\$ $\Delta = \begin{vmatrix} 1+x+x^2 & 1+x+x^2 & 1+x+x^2 \\[0.3em] x^2 & 1 & x \\[0.3em] x & x^2 & 1\end{vmatrix}$ $\\$ Applying $C_2 \to C_2 -C_1$ and $C_3 \to C_3 -C_1,$ we have : $\\$ $\Delta = (1+x+x^2) \begin{vmatrix} 1 & 0 & 0 \\[0.3em] x^2 & 1-x^2 & x-x^2 \\[0.3em] x & x^2-x & 1-x\end{vmatrix}$ $\\$ $= (1+x+x^2) (1-x )(1-x ) \begin{vmatrix} 1 & 0 & 0 \\[0.3em] x^2 & 1+ x & x \\[0.3em] x & x^2-x & 1-x\end{vmatrix}$ $\\$

=$(1-x^3) (1-x )\begin{vmatrix} 1 & 0 & 0 \\[0.3em] x^2 & 1+ x & x \\[0.3em] x & -x & 1\end{vmatrix}$ $\\$ Expanding along R_1, we have: $\\$ $\Delta = (1-x^3) (1-x) (1) \begin{vmatrix} 1+x & x \\[0.3em] -x & 1 \\[0.3em] \end{vmatrix}$ $\\$ $= (1-x^3) (1-x^3)$ $\\$ = $(1-x^3) ^2$ $\\$ Hence, the given result is proved.

21   By using properties of determinants, show that: $\\$ $\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\[0.3em] 2ab & 1-a^2+b^2 & 2a \\[0.3em] 2a & -2a & 1-a^2-b^2\end{vmatrix}=(1+a^2 +b^2)^3$ $\\$

##### Solution :

=$\ (1+a^2+b^2)$ $\begin{vmatrix} 1 & 0 & -b \\[0.3em] 0 & 1 & a \\[0.3em] 2b & -2a & 1-a^2-b^2\end{vmatrix}$ $\\$ Expanding along $R_1,$ we have: $\\$ $\Delta = (1+a^2+b^2)^2$ $\begin{bmatrix} (1) \begin{vmatrix} 1 & a \\[0.3em] -2a & 1-a^2-b^2 \\[0.3em] \end{vmatrix}-b \begin{vmatrix} 0 & 1 \\[0.3em] 2b & -2a\\[0.3em] \end{vmatrix} \end{bmatrix}$ $\\$ $= (1+a^2+b^2)^2 \begin{bmatrix} 1-a^2-b^2+2a^2-b(-2b) \end{bmatrix}$ $\\$ $= (1+a^2+b^2)^2 (1+a^2+b^2)$ $\\$ $= (1+a^2+b^2)^3$

$\Delta \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\[0.3em] 2ab & 1-a^2+b^2 & 2a \\[0.3em] 2a & -2a & 1-a^2-b^2\end{vmatrix}$ $\\$ Applying $R_1 \to R_1 +bR_3$ and $R_2 \to R_2 -aR_3$, we have: $\\$ $\Delta =$ $\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\[0.3em] 2ab & 1-a^2+b^2 & 2a \\[0.3em] 2a & -2a & 1-a^2-b^2\end{vmatrix}$ $\\$ Applying $R_1 \to R_1 +bR_3$ and $R_2 \to R_2 -aR_3$, we have: $\\$ $\Delta =$ $\begin{vmatrix} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\[0.3em] 0 & 1+a^2+b^2 & a(1+a^2+b^2) \\[0.3em] 2b & -2a & 1-a^2-b^2\end{vmatrix}$ $\\$

22   By using properties of determinants, show that: $\\$ $\begin{vmatrix} a^2+1 & ab & ac \\[0.3em] ab & b^2+1 & bc \\[0.3em] ca & cb & c^2+1\end{vmatrix}=1+a^2+b^2+c^2$ $\\$

##### Solution :

Applying $C_1 \to aC_1, C_2 \to bC_2$ and $C_3 \to cC_3$, we have: $\\$ $\Delta$ = abc x $\dfrac{1}{abc}$ $\begin{vmatrix} a^2 +1 & b^2 & c^2 \\[0.3em] -1 & 1 & 0 \\[0.3em] -1 & 0 & 1 \end{vmatrix}$ $\\$ Expanding along $R_3$, we have: $\\$ $\Delta$ =-1 $\begin{vmatrix} b^2 & c^2 \\[0.3em] 1 & 0 \\[0.3em] \end{vmatrix}$ +1 $\begin{vmatrix} a^2+1 & b^2 \\[0.3em] -1 & 1 \\[0.3em] \end{vmatrix}$ $\\$ $= -1(-c^2) + (a^2 +1+b^2) = 1+a^2+b^2+c^2$ $\\$ Hence, the given result is proved.

$\Delta =$ $\begin{vmatrix} a^2+1 & ab & ac \\[0.3em] ab & b^2+1 & bc \\[0.3em] ca & cb & c^2+1\end{vmatrix}$ $\\$ Taking out common factors a,b and c from $R_1, R_2$ and $R_3$ respectively, we have: $\\$ $\Delta = abc$ $\begin{vmatrix} a+\dfrac{1}{a} & b & c \\[0.3em] a & b+\dfrac{1}{b} & bc \\[0.3em] ca & b & c+\dfrac{1}{c} \end{vmatrix}$ $\\$ Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 -R_1$, we have: $\\$ $\Delta = abc$ $\begin{vmatrix} a+\dfrac{1}{a} & b & c \\[0.3em] -\dfrac{1}{a} & \dfrac{1}{b} & bc \\[0.3em] -\dfrac{1}{a} & 0 & \dfrac{1}{c} \end{vmatrix}$ $\\$

23   Choose the correct answer. $\\$ Let A be a square matrix of order $3 x 3$ , then $kA$ is equal to