Q: Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$.

$\Delta=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ $=x\left(-x^{2}-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)$ $=-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta$ $=-x^{3}-x+x\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$ $=-x^{3}-x+x$ $=-x^{3}$ Hence, $\Delta$ is independent of $\theta$.

Q: Without expanding the determinant, prove that $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$.

$$ \begin{array}{rlr} \text { LHS } & =\left|\begin{array}{lll} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{array}\right| & \\ & =\frac{1}{a b c}\left|\begin{array}{lll} a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c \end{array}\right| & {\left[R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, R_{3} \rightarrow c R_{3}\right]} \\ & =\frac{1}{a b c} \cdot a b c\left|\begin{array}{lll} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{array}\right| & {\left[\text { Taking out factor } a b c \text { from } C_{3}\right]} \\ & =\left|\begin{array}{lll} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{array}\right| & \\ & =\left|\begin{array}{lll} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{array}\right| & \\ & =R H S & {\left[C_{1} \leftrightarrow C_{3} \text { and } C_{2} \leftrightarrow C_{3}\right]} \end{array} $$ Hence, proved.

Q: Solve the equations $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0, a \neq 0$.

$\Rightarrow\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$ $\Rightarrow\left|\begin{array}{ccc}3 x+a & 3 x+a & 3 x+a \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0 \quad\left[R_{1} \rightarrow R_{1}+R_{2}+R_{3}\right]$ $\Rightarrow(3 x+a)\left|\begin{array}{ccc}1 & 1 & 1 \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$ $\Rightarrow(3 x+a)\left|\begin{array}{lll}1 & 0 & 0 \\ x & a & 0 \\ x & 0 & a\end{array}\right|=0 \quad\left[C_{2} \rightarrow C_{2}-C_{1}\right.$ and $\left.C_{3} \rightarrow C_{3}-C_{1}\right]$ Expanding along $R_{1}$, $\Rightarrow(3 x+a)\left[1 \times a^{2}\right]=0$ $\Rightarrow a^{2}(3 x+a)=0$ Since $a \neq 0$ Therefore, $$ \begin{aligned} & \Rightarrow 3 x+a=0 \\ & \Rightarrow x=-\frac{a}{3} \end{aligned} $$

Q: Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$.

$\Delta=\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|$ $=a b c\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ b & b+c & c\end{array}\right|$ [Taking out common factors $a, b$ and $c$ from $C_{1}, C_{2}$ and $C_{3}$ ] $=a b c\left|\begin{array}{ccc}a & c & a+c \\ b & b-c & -c \\ b-a & b & -a\end{array}\right| \left[R_{2} \rightarrow R_{2}-R_{1}\right.$ and $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$ $=a b c\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ b-a & b & -a\end{array}\right|$ $$ \left[R_{2} \rightarrow R_{2}+R_{1}\right] $$ $=a b c\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ 2 b & 2 b & 0\end{array}\right|$ $$ \left[R_{3} \rightarrow R_{3}+R_{2}\right] $$ $=2 a b^{2} c\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ 1 & 1 & 0\end{array}\right|$ $\Delta=2 a b^{2} c\left|\begin{array}{ccc}a & c-a & a+c \\ a+b & -a & a \\ 1 & 0 & 0\end{array}\right|$ $$ \left[C_{2} \rightarrow C_{2}-C_{1}\right] $$ Expanding along $R_{3}$, $$ \begin{aligned} \Delta & =2 a b^{2} c[a(c-a)+a(a+c)] \\ & =2 a b^{2} c\left[a c-a^{2}+a^{2}+a c\right] \\ & =2 a b^{2} c(2 a c) \\ & =4 a^{2} b^{2} c^{2} \end{aligned} $$ Hence, proved.

Q: Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$.

$\Delta=\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$ $=\left|\begin{array}{ccc}2(x+y) & 2(x+y) & 2(x+y) \\ y & x+y & x \\ x+y & x & y\end{array}\right| \quad\left[R_{1} \rightarrow R_{1}+R_{2}+R_{3}\right]$ $=2(x+y)\left|\begin{array}{ccc}1 & 1 & 1 \\ y & x+y & x \\ x+y & x & y\end{array}\right|$ $=2(x+y)\left|\begin{array}{ccc}1 & 0 & 0 \\ y & x & x-y \\ x+y & -y & -x\end{array}\right| \quad\left[C_{2} \rightarrow C_{2}-C_{1}\right.$ and $\left.C_{3} \rightarrow C_{3}-C_{1}\right]$ $=2(x+y)\left[-x^{2}+y(x-y)\right] \quad\left[\right.$ Expanding along $\left.R_{1}\right]$ $=-2(x+y)\left(x^{2}+y^{2}-y x\right)$ $=-2\left(x^{3}+y^{3}\right)$

Question 1

Prove that the determinant $\left|\begin{array}{ccc}x & \sin 	heta & \cos 	heta \ -\sin 	heta & -x & 1 \ \cos 	heta & 1 & x\end{array}ight|$ is independent of $	heta$.

Accepted Answer

$\Delta=\left|\begin{array}{ccc}x & \sin 	heta & \cos 	heta \ -\sin 	heta & -x & 1 \ \cos 	heta & 1 & x\end{array}ight|$

$=x\left(-x^{2}-1ight)-\sin 	heta(-x \sin 	heta-\cos 	heta)+\cos 	heta(-\sin 	heta+x \cos 	heta)$

$=-x^{3}-x+x \sin ^{2} 	heta+\sin 	heta \cos 	heta-\sin 	heta \cos 	heta+x \cos ^{2} 	heta$

$=-x^{3}-x+x\left(\sin ^{2} 	heta+\cos ^{2} 	hetaight)$

$=-x^{3}-x+x$

$=-x^{3}$

Hence, $\Delta$ is independent of $	heta$.

Question 2

Without expanding the determinant, prove that $\left|\begin{array}{lll}a & a^{2} & b c \ b & b^{2} & c a \ c & c^{2} & a b\end{array}ight|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \ 1 & b^{2} & b^{3} \ 1 & c^{2} & c^{3}\end{array}ight|$.

Accepted Answer

$$

\begin{array}{rlr}

ext { LHS } & =\left|\begin{array}{lll}

a & a^{2} & b c \

b & b^{2} & c a \

c & c^{2} & a b

\end{array}ight| & \

& =\frac{1}{a b c}\left|\begin{array}{lll}

a^{2} & a^{3} & a b c \

b^{2} & b^{3} & a b c \

c^{2} & c^{3} & a b c

\end{array}ight| & {\left[R_{1} ightarrow a R_{1}, R_{2} ightarrow b R_{2}, R_{3} ightarrow c R_{3}ight]} \

& =\frac{1}{a b c} \cdot a b c\left|\begin{array}{lll}

a^{2} & a^{3} & 1 \

b^{2} & b^{3} & 1 \

c^{2} & c^{3} & 1

\end{array}ight| & {\left[	ext { Taking out factor } a b c 	ext { from } C_{3}ight]} \

& =\left|\begin{array}{lll}

a^{2} & a^{3} & 1 \

b^{2} & b^{3} & 1 \

c^{2} & c^{3} & 1

\end{array}ight| & \

& =\left|\begin{array}{lll}

1 & a^{2} & a^{3} \

1 & b^{2} & b^{3} \

1 & c^{2} & c^{3}

\end{array}ight| & \

& =R H S & {\left[C_{1} \leftrightarrow C_{3} 	ext { and } C_{2} \leftrightarrow C_{3}ight]}

\end{array}

$$

Hence, proved.

Question 3

Evaluate $\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \ -\sin \beta & \cos \beta & 0 \ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}ight|$.

Accepted Answer

Let $\Delta=\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \ -\sin \beta & \cos \beta & 0 \ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}ight|$

Expanding along $C_{3}$,

$$

\begin{aligned}

\Delta & =-\sin \alpha\left(-\sin \alpha \sin ^{2} \beta-\cos ^{2} \beta \sin \alphaight)+\cos \alpha\left(\cos \alpha \cos ^{2} \beta+\cos \alpha \sin ^{2} \betaight) \

& =\sin ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \betaight)+\cos ^{2} \alpha\left(\cos ^{2} \beta+\sin ^{2} \betaight) \

& =\sin ^{2} \alpha(1)+\cos ^{2} \alpha(1) \

& =1

\end{aligned}

$$

Question 4

If $a, b, c$ are real numbers and $\Delta=\left|\begin{array}{lll}b+c & c+a & a+b \ c+a & a+b & b+c \ a+b & b+c & c+a\end{array}ight|=0$, show that either $a+b+c=0$ or $a=b=c$.

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

b+c & c+a & a+b \

c+a & a+b & b+c \

a+b & b+c & c+a

\end{array}ight| & \

& =\left|\begin{array}{ccc}

2(a+b+c) & 2(a+b+c) & 2(a+b+c) \

c+a & a+b & b+c \

a+b & b+c & c+a

\end{array}ight| & {\left[R_{1} ightarrow R_{1}+R_{2}+R_{3}ight] } \

& =2(a+b+c)\left|\begin{array}{ccc}

1 & 1 & 1 \

c+a & a+b & b+c \

a+b & b+c & c+a

\end{array}ight| & {\left[C_{2} ightarrow C_{2}-C_{1} 	ext { and } C_{3} ightarrow C_{3}-C_{1}ight] } \

& =2(a+b+c)\left|\begin{array}{ccc}

1 & 0 & 0 \

c+a & b-c & b-a \

a+b & c-a & c-b

\end{array}ight| &

\end{aligned}

$$

Expanding $R_{1}$,

$$

\begin{aligned}

\Delta & =2(a+b+c)(1)[(b-c)(c-b)-(b-a)(c-a)] \

& =2(a+b+c)\left[-b^{2}-c^{2}+2 b c-b c+b a+a c-a^{2}ight] \

& =2(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}ight]

\end{aligned}

$$

It is given that $\Delta=0$.

Hence,

$$

2(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}ight]=0

$$

Either $(a+b+c)=0$ or $\left[a b+b c+c a-a^{2}-b^{2}-c^{2}ight]=0$

Now,

$\Rightarrow a b+b c+c a-a^{2}-b^{2}-c^{2}=0$

$\Rightarrow-2 a b-2 a c-2 c a+2 a^{2}+2 b^{2}+2 c^{2}=0$

$\Rightarrow(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$

$\Rightarrow(a-b)^{2}=(b-c)^{2}=(c-a)^{2}=0 \quad\left[(a-b)^{2},(b-c)^{2},(c-a)^{2}ight.$ are non-negative $]$

$\Rightarrow(a-b)=(b-c)=(c-a)=0$

$\Rightarrow a=b=c$

Hence, if $\Delta=0$, then either $(a+b+c)=0$ or $a=b=c$.

Question 5

Solve the equations $\left|\begin{array}{ccc}x+a & x & x \ x & x+a & x \ x & x & x+a\end{array}ight|=0, a 
eq 0$.

Accepted Answer

$\Rightarrow\left|\begin{array}{ccc}x+a & x & x \ x & x+a & x \ x & x & x+a\end{array}ight|=0$

$\Rightarrow\left|\begin{array}{ccc}3 x+a & 3 x+a & 3 x+a \ x & x+a & x \ x & x & x+a\end{array}ight|=0 \quad\left[R_{1} ightarrow R_{1}+R_{2}+R_{3}ight]$

$\Rightarrow(3 x+a)\left|\begin{array}{ccc}1 & 1 & 1 \ x & x+a & x \ x & x & x+a\end{array}ight|=0$

$\Rightarrow(3 x+a)\left|\begin{array}{lll}1 & 0 & 0 \ x & a & 0 \ x & 0 & a\end{array}ight|=0 \quad\left[C_{2} ightarrow C_{2}-C_{1}ight.$ and $\left.C_{3} ightarrow C_{3}-C_{1}ight]$

Expanding along $R_{1}$,

$\Rightarrow(3 x+a)\left[1 	imes a^{2}ight]=0$

$\Rightarrow a^{2}(3 x+a)=0$

Since $a 
eq 0$

Therefore,

$$

\begin{aligned}

& \Rightarrow 3 x+a=0 \

& \Rightarrow x=-\frac{a}{3}

\end{aligned}

$$

Question 6

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \ a^{2}+a b & b^{2} & a c \ a b & b^{2}+b c & c^{2}\end{array}ight|=4 a^{2} b^{2} c^{2}$.

Accepted Answer

$\Delta=\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \ a^{2}+a b & b^{2} & a c \ a b & b^{2}+b c & c^{2}\end{array}ight|$

$=a b c\left|\begin{array}{ccc}a & c & a+c \ a+b & b & a \ b & b+c & c\end{array}ight|$

[Taking out common factors $a, b$ and $c$ from $C_{1}, C_{2}$ and $C_{3}$ ]

$=a b c\left|\begin{array}{ccc}a & c & a+c \ b & b-c & -c \ b-a & b & -a\end{array}ight| \left[R_{2} ightarrow R_{2}-R_{1}ight.$ and $\left.R_{3} ightarrow R_{3}-R_{1}ight]$

$=a b c\left|\begin{array}{ccc}a & c & a+c \ a+b & b & a \ b-a & b & -a\end{array}ight|$

$$

\left[R_{2} ightarrow R_{2}+R_{1}ight]

$$

$=a b c\left|\begin{array}{ccc}a & c & a+c \ a+b & b & a \ 2 b & 2 b & 0\end{array}ight|$

$$

\left[R_{3} ightarrow R_{3}+R_{2}ight]

$$

$=2 a b^{2} c\left|\begin{array}{ccc}a & c & a+c \ a+b & b & a \ 1 & 1 & 0\end{array}ight|$

$\Delta=2 a b^{2} c\left|\begin{array}{ccc}a & c-a & a+c \ a+b & -a & a \ 1 & 0 & 0\end{array}ight|$

$$

\left[C_{2} ightarrow C_{2}-C_{1}ight]

$$

Expanding along $R_{3}$,

$$

\begin{aligned}

\Delta & =2 a b^{2} c[a(c-a)+a(a+c)] \

& =2 a b^{2} c\left[a c-a^{2}+a^{2}+a cight] \

& =2 a b^{2} c(2 a c) \

& =4 a^{2} b^{2} c^{2}

\end{aligned}

$$

Hence, proved.

Question 7

If $A^{-1}=\left|\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}ight|$ and $B=\left|\begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1\end{array}ight|$, find $(A B)^{-1}$.

Accepted Answer

We know that $(A B)^{-1}=B^{-1} A^{-1}$.

It is given that

$$

B=\left|\begin{array}{ccc}

1 & 2 & -2 \

-1 & 3 & 0 \

0 & -2 & 1

\end{array}ight|

$$

Therefore,

$$

\begin{aligned}

|B| & =1(3)-2(-1)-2(-2) \

& =3+2-4 \

& =5-4 \

& =1

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

B_{11}=3 & B_{12}=1 & B_{13}=2 \

B_{21}=2 & B_{22}=1 & B_{23}=2 \

B_{31}=6 & B_{32}=2 & B_{33}=5

\end{array}

$$

Hence,

$$

\operatorname{adj} B=\left(\begin{array}{lll}

3 & 2 & 6 \

1 & 1 & 2 \

2 & 2 & 5

\end{array}ight)

$$

Now,

$$

\begin{aligned}

B^{-1} & =\frac{1}{|B|} \operatorname{adj} B \

& =\left(\begin{array}{lll}

3 & 2 & 6 \

1 & 1 & 2 \

2 & 2 & 5

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

(A B)^{-1} & =B^{-1} A^{-1} \

& =\left(\begin{array}{lll}

3 & 2 & 6 \

1 & 1 & 2 \

2 & 2 & 5

\end{array}ight)\left(\begin{array}{ccc}

3 & -1 & 1 \

-15 & 6 & -5 \

5 & -2 & 2

\end{array}ight) \

& =\left(\begin{array}{ccc}

9-30+30 & -3+12-12 & 3-10+12 \

3-15+10 & -1+6-4 & 1-5+4 \

6-30+25 & -2+12-10 & 2-10+10

\end{array}ight) \

& =\left(\begin{array}{ccc}

9 & -3 & 5 \

-2 & 1 & 0 \

1 & 0 & 2

\end{array}ight) \

(A B)^{-1}=\left(\begin{array}{ccc}

9 & -3 & 5 \

-2 & 1 & 0 \

1 & 0 & 2

\end{array}ight) &

\end{aligned}

$$

Thus,

Question 8

Let $A=\left(\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}ight)$ verify that

(i) $[\operatorname{adj} A]^{-1}=\operatorname{adj}(A)^{-1}$

(ii) $\left(A^{-1}ight)^{-1}=A$

Accepted Answer

It is given that

$$

A=\left(\begin{array}{ccc}

1 & -2 & 1 \

-2 & 3 & 1 \

1 & 1 & 5

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

|A| & =1(15-1)+2(-10-1)+1(-2-3) \

& =14-22-5 \

& =-13

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

A_{11}=14 & A_{12}=11 & A_{13}=-5 \

A_{21}=11 & A_{22}=4 & A_{23}=-3 \

A_{31}=-5 & A_{32}=-3 & A_{33}=-1

\end{array}

$$

Hence,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

14 & 11 & -5 \

11 & 4 & -3 \

-5 & -3 & -1

\end{array}ight)

$$

Now,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|}(\operatorname{adj} A) \

& =-\frac{1}{13}\left(\begin{array}{ccc}

14 & 11 & -5 \

11 & 4 & -3 \

-5 & -3 & -1

\end{array}ight) \

& =\frac{1}{13}\left(\begin{array}{ccc}

-14 & -11 & 5 \

-11 & -4 & 3 \

5 & 3 & 1

\end{array}ight)

\end{aligned}

$$

(i)

$$

\begin{aligned}

|\operatorname{adj} A| & =14(-4-9)-11(-11-15)-5(-33+20) \

& =14(-13)-11(-26)-5(-13) \

& =-182+286+65 \

& =169

\end{aligned}

$$

We have,

$$

\operatorname{adj}(\operatorname{adj} A)=\left(\begin{array}{ccc}

-13 & 26 & -13 \

26 & -39 & -13 \

-13 & -13 & -65

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

{[\operatorname{adj} A]^{-1} } & =\frac{1}{|\operatorname{adj} A|}(\operatorname{adj}(\operatorname{adj} A)) \

& =\frac{1}{169}\left(\begin{array}{ccc}

-13 & 26 & -13 \

26 & -39 & -13 \

-13 & -13 & -65

\end{array}ight) \

& =\left(\begin{array}{ccc}

\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \

\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \

\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}

\end{array}ight)

\end{aligned}

$$

Now,

$$

A^{-1}=-\frac{1}{13}\left(\begin{array}{ccc}

14 & 11 & -5 \

11 & 4 & -3 \

-5 & -3 & -1

\end{array}ight)=\left(\begin{array}{ccc}

\frac{-14}{13} & \frac{-11}{13} & \frac{5}{13} \

\frac{-11}{13} & \frac{-4}{13} & \frac{3}{13} \

\frac{5}{13} & \frac{3}{13} & \frac{1}{13}

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

\operatorname{adj}(A)^{-1} & =\left(\begin{array}{lll}

\frac{-13}{169} & \frac{26}{169} & \frac{-13}{169} \

\frac{26}{169} & \frac{-39}{169} & \frac{-13}{169} \

\frac{-13}{169} & \frac{-13}{169} & \frac{-65}{169}

\end{array}ight) \

& =\left(\begin{array}{ccc}

\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \

\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \

\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}

\end{array}ight)

\end{aligned}

$$

Hence, $[\operatorname{adj} A]^{-1}=\operatorname{adj}(A)^{-1}$ proved.

$A^{-1}=\frac{1}{13}\left(\begin{array}{ccc}-14 & -11 & 5 \ -11 & -4 & 3 \ 5 & 3 & 1\end{array}ight)$

(ii)

Hence,

$$

\operatorname{adj}(A)^{-1}=\left(\begin{array}{ccc}

\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \

\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \

\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}

\end{array}ight)

$$

Now,

$$

\begin{aligned}

\left|A^{-1}ight| & =\left(\frac{1}{13}ight)^{3}[-14(-4-9)+11(-11-26)+5(-33+20)] \

& =\left(\frac{1}{13}ight)^{3}[-169] \

& =-\frac{1}{13}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\left(A^{-1}ight)^{-1} & =\frac{\operatorname{adj} A^{-1}}{|A|}=\frac{1}{\left(-\frac{1}{13}ight)} 	imes\left(\begin{array}{ccc}

\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \

\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \

\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}

\end{array}ight) \

& =\left(\begin{array}{ccc}

1 & -2 & 1 \

-2 & 3 & 1 \

1 & 1 & 5

\end{array}ight)=A

\end{aligned}

$$

Hence, $\left(A^{-1}ight)^{-1}=A$ proved.

Question 9

Evaluate $\left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}ight|$.

Accepted Answer

$\Delta=\left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}ight|$

$=\left|\begin{array}{ccc}2(x+y) & 2(x+y) & 2(x+y) \ y & x+y & x \ x+y & x & y\end{array}ight| \quad\left[R_{1} ightarrow R_{1}+R_{2}+R_{3}ight]$

$=2(x+y)\left|\begin{array}{ccc}1 & 1 & 1 \ y & x+y & x \ x+y & x & y\end{array}ight|$

$=2(x+y)\left|\begin{array}{ccc}1 & 0 & 0 \ y & x & x-y \ x+y & -y & -x\end{array}ight| \quad\left[C_{2} ightarrow C_{2}-C_{1}ight.$ and $\left.C_{3} ightarrow C_{3}-C_{1}ight]$

$=2(x+y)\left[-x^{2}+y(x-y)ight] \quad\left[ight.$ Expanding along $\left.R_{1}ight]$

$=-2(x+y)\left(x^{2}+y^{2}-y xight)$

$=-2\left(x^{3}+y^{3}ight)$

Question 10

Evaluate $\left|\begin{array}{ccc}1 & x & y \ 1 & x+y & y \ 1 & x & x+y\end{array}ight|$.

Accepted Answer

$$

\begin{array}{rlr}

\Delta & =\left|\begin{array}{ccc}

1 & x & y \

1 & x+y & y \

1 & x & x+y

\end{array}ight| & \

& =\left|\begin{array}{ccc}

1 & x & y \

0 & y & 0 \

0 & 0 & x

\end{array}ight| & {\left[R_{2} ightarrow R_{2}-R_{1} 	ext { and } R_{3} ightarrow R_{3}-R_{1}ight]} \

& =1(x y-0) & {\left[	ext { Expanding along } C_{1}ight]} \

& =x y &

\end{array}

$$

Question 11

Using properties of determinants prove that:

$$

\left|\begin{array}{lll}

\alpha & \alpha^{2} & \beta+\gamma \

\beta & \beta^{2} & \gamma+\alpha \

\gamma & \gamma^{2} & \alpha+\beta

\end{array}ight|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)

$$

Accepted Answer

$$

\begin{array}{rlr}

\Delta & =\left|\begin{array}{ccc}

\alpha & \alpha^{2} & \beta+\gamma \

\beta & \beta^{2} & \gamma+\alpha \

\gamma & \gamma^{2} & \alpha+\beta

\end{array}ight| & \

& =\left|\begin{array}{ccc}

\alpha & \alpha^{2} & \beta+\gamma \

\beta-\alpha & \beta^{2}-\alpha^{2} & \alpha-\beta \

\gamma-\alpha & \gamma^{2}-\alpha^{2} & \alpha-\gamma

\end{array}ight| & \

& =(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}

\alpha & \alpha^{2} & \beta+\gamma \

1 & \beta+\alpha & -1 \

1 & \gamma+\alpha & -1

\end{array}ight| & \

& =(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}

\alpha & \alpha^{2} & \beta+\gamma \

1 & \beta+\alpha & -1 \

0 & \gamma-\beta & 0

\end{array}ight| & {\left[R_{3} ightarrow R_{3}-R_{1} 	ext { and } R_{3} ightarrow R_{3}-R_{1}ight]} \

& =(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)] & {\left[	ext { Expanding along } R_{3}ight]} \

& =(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma) & \

& =(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) &

\end{array}

$$

Hence, proved.

Question 12

Using properties of determinants prove that:

$$

\left|\begin{array}{lll}

x & x^{2} & 1+p x^{3} \

y & y^{2} & 1+p y^{3} \

z & z^{2} & 1+p z^{3}

\end{array}ight|=(1+p x y z)(x-y)(y-z)(z-x)

$$

Accepted Answer

$\Delta=\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \ y & y^{2} & 1+p y^{3} \ z & z^{2} & 1+p z^{3}\end{array}ight|$

$\Delta=\left|\begin{array}{ccc}x & x^{2} & 1+p x^{3} \ y-x & y^{2}-x^{2} & p\left(y^{3}-x^{3}ight) \ z-x & z^{2}-x^{2} & p\left(z^{3}-x^{3}ight)\end{array}ight| \quad\left[R_{2} ightarrow R_{2}-R_{1}ight.$ and $\left.R_{3} ightarrow R_{3}-R_{1}ight]$

$\Delta=(y-x)(z-x)\left|\begin{array}{ccc}x & x^{2} & 1+p x^{3} \ 1 & y+x & p\left(y^{2}+x^{2}+x yight) \ 1 & z+x & p\left(z^{2}+x^{2}+x zight)\end{array}ight|$

$\Delta=(y-x)(z-x)\left|\begin{array}{ccc}x & x^{2} & 1+p x^{3} \ 1 & y+x & p\left(y^{2}+x^{2}+x yight) \ 0 & z-y & p(z-y)(x+y+z)\end{array}ight| \quad\left[R_{3} ightarrow R_{3}-R_{2}ight]$

$\Delta=(y-x)(z-x)(z-y)\left|\begin{array}{ccc}x & x^{2} & 1+p x^{3} \ 1 & y+x & p\left(y^{2}+x^{2}+x yight) \ 0 & 1 & p(x+y+z)\end{array}ight|$

$\Delta=(x-y)(z-y)(z-x)\left[(-1)(p)\left(x y^{2}+x^{3}+x^{2} yight)+1+p x^{3}+p(x+y+z)(x y)ight]$

[Expanding along $R_{3}$ ]

$=(x-y)(y-z)(z-x)\left[-p x y^{2}-p x^{3}-p x^{2} y+1+p x^{3}+p x^{2} y+p x y^{2}+p x y zight]$

$=(x-y)(y-z)(z-x)(1+p x y z)$

Hence, proved.

Question 13:

Using properties of determinants prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \ -b+a & 3 b & -b+c \ -c+a & -c+b & 3 c\end{array}ight|=3(a+b+c)(a b+b c+c a)$

\section*{Solution:}

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

3 a & -a+b & -a+c \

-b+a & 3 b & -b+c \

-c+a & -c+b & 3 c

\end{array}ight| \

& =\left|\begin{array}{ccc}

a+b+c & -a+b & -a+c \

a+b+c & 3 b & -b+c \

a+b+c & -c+b & 3 c

\end{array}ight| \

& =(a+b+c)\left|\begin{array}{ccc}

1 & -a+b & -a+c \

1 & 3 b & -b+c \

1 & -c+b & 3 c

\end{array}ight| \

& =(a+b+c)\left|\begin{array}{ccc}

1 & -a+b & -a+c \

0 & 2 b+a & a-b \

0 & a-c & 2 c+a

\end{array}ight| \

& =(a+b+c)[(2 b+a)(2 c+a)-(a-b)(a-c)] \quad\left[R_{2} ightarrow R_{2}-R_{1} 	ext { and } R_{3} ightarrow R_{3}-R_{1}ight] \

& =(a+b+c)\left[4 b c+2 a b+2 a c+a^{2}-a^{2}+a c+b a-b cight] \quad\left[	ext { Expanding along } C_{1}ight] \

& =(a+b+c)(3 a b+3 b c+3 a c) \

& =3(a+b+c)(a b+b c+c a)

\end{aligned}

$$

Hence, proved.

Question 13

Using properties of determinants prove that:

$$

\left|\begin{array}{ccc}

1 & 1+p & 1+p+q \

2 & 3+2 p & 4+3 p+2 q \

3 & 6+3 p & 10+6 p+3 q

\end{array}ight|=1

$$

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

1 & 1+p & 1+p+q \

2 & 3+2 p & 4+3 p+2 q \

3 & 6+3 p & 10+6 p+3 q

\end{array}ight| & & \

& =\left|\begin{array}{ccc}

1 & 1+p & 1+p+q \

0 & 1 & 2+p \

0 & 3 & 7+3 p

\end{array}ight| & & {\left[R_{2} ightarrow R_{2}-2 R_{1} 	ext { and } R_{3} ightarrow R_{3}-3 R_{1}ight] } \

& =\left|\begin{array}{ccc}

1 & 1+p & 1+p+q \

0 & 1 & 2+p \

0 & 0 & 1

\end{array}ight| & & {\left[R_{3} ightarrow R_{3}-3 R_{2}ight] } \

& =1\left|\begin{array}{cc}

1 & 2+p \

0 & 1

\end{array}ight| & & {\left[	ext { Expanding along } C_{1}ight] } \

& =1(1-0)=1 & &

\end{aligned}

$$

Hence, proved.

Question 14

Using properties of determinants prove that:

$$

\left|\begin{array}{ccc}

\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \

\sin \beta & \cos \beta & \cos (\beta+\delta) \

\sin \gamma & \cos \gamma & \cos (\gamma+\delta)

\end{array}ight|=0

$$

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{lll}

\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \

\sin \beta & \cos \beta & \cos (\beta+\delta) \

\sin \gamma & \cos \gamma & \cos (\gamma+\delta)

\end{array}ight| \

& =\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{lll}

\sin \alpha \sin \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \

\sin \beta \sin \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \

\sin \gamma \sin \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta

\end{array}ight| \

& =\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{lll}

\cos \alpha \cos \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \

\cos \beta \cos \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \

\cos \gamma \cos \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta

\end{array}ight| \quad\left[C_{1} ightarrow C_{1}+C_{3}ight]

\end{aligned}

$$

Here, two columns $C_{1}$ and $C_{2}$ are identical.

Therefore, $\Delta=0$

Hence, proved.

Question 15

Solve the system of the following equations:

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Accepted Answer

Let $\frac{1}{x}=p, \frac{1}{y}=q$ and $\frac{1}{z}=r$.

Then the given system of equations is as follows:

$2 p+3 q+10 r=4$

$4 p-6 q+5 r=1$

$6 p+9 q-20 r=2$

This system can be written in the form of $A X=B$, where

$$

A=\left(\begin{array}{ccc}

2 & 3 & 10 \

4 & -6 & 5 \

6 & 9 & -20

\end{array}ight), X=\left[\begin{array}{l}

p \

q \

r

\end{array}ight] 	ext { and } \quad B=\left[\begin{array}{l}

4 \

1 \

2

\end{array}ight]

$$

Therefore,

$$

\begin{aligned}

|A| & =2(120-45)-3(-80-30)+10(36+36) \

& =150+330+720 \

& =1200

\end{aligned}

$$

Thus, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Now,

$$

\begin{array}{lll}

A_{11}=75 & A_{12}=110 & A_{13}=72 \

A_{21}=150 & A_{22}=-100 & A_{23}=0 \

A_{31}=75 & A_{32}=30 & A_{33}=-24

\end{array}

$$

Hence,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|}(\operatorname{adj} A) \

& =\frac{1}{1200}\left(\begin{array}{ccc}

75 & 150 & 75 \

110 & -100 & 30 \

72 & 0 & -24

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \Rightarrow X=A^{-1} B \

& \Rightarrow\left[\begin{array}{c}

p \

q \

r

\end{array}ight]=\frac{1}{1200}\left(\begin{array}{ccc}

75 & 150 & 75 \

110 & -100 & 30 \

72 & 0 & -24

\end{array}ight)\left[\begin{array}{c}

4 \

1 \

2

\end{array}ight] \

& \Rightarrow\left[\begin{array}{c}

p \

q \

r

\end{array}ight]=\frac{1}{1200}\left[\begin{array}{c}

300+150+150 \

440-100+60 \

288+0-48

\end{array}ight] \

& \Rightarrow\left[\begin{array}{c}

p \

q \

r

\end{array}ight]=\frac{1}{1200}\left[\begin{array}{c}

600 \

400 \

240

\end{array}ight] \

& \Rightarrow\left[\begin{array}{c}

p \

q \

r

\end{array}ight]=\left[\begin{array}{c}

\frac{1}{2} \

\frac{1}{3} \

\frac{1}{5}

\end{array}ight]

\end{aligned}

$$

Therefore,

$p=\frac{1}{2}, q=\frac{1}{3}$ and $r=\frac{1}{5}$

Hence, $x=2, y=3$ and $z=5$.

Question 16

If $a, b, c$ are in A.P, then the determinant $\left|\begin{array}{lll}x+2 & x+3 & x+2 a \ x+3 & x+4 & x+2 b \ x+4 & x+5 & x+2 c\end{array}ight|$ is

(A) 0

(B) 1

(C) $x$

(D) $2 x$

Accepted Answer

$$

\begin{aligned}

\Delta & =\left|\begin{array}{ccc}

x+2 & x+3 & x+2 a \

x+3 & x+4 & x+2 b \

x+4 & x+5 & x+2 c

\end{array}ight| & & \

& =\left|\begin{array}{ccc}

x+2 & x+3 & x+2 a \

x+3 & x+4 & x+(a+c) \

x+4 & x+5 & x+2 c

\end{array}ight| & & (2 b=a+c 	ext { as } a, b, c 	ext { are in A.P }) \

& =\left|\begin{array}{ccc}

-1 & -1 & a-c \

x+3 & x+4 & x+(a+c) \

1 & 1 & c-a

\end{array}ight| & & {\left[R_{1} ightarrow R_{1}-R_{2} 	ext { and } R_{3} ightarrow R_{3}-R_{2}ight] } \

& =\left|\begin{array}{ccc}

0 & 0 & 0 \

x+3 & x+4 & x+a+c \

1 & 1 & c-a

\end{array}ight| & & {\left[R_{1} ightarrow R_{1}+R_{3}ight] }

\end{aligned}

$$

Here, all the elements of the first row are zero.

Hence, we have $\Delta=0$

Thus, the correct option is A.

Question 17

If $x, y, z$ are non-zero real numbers, then the inverse of matrix $A=\left(\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}ight)$ is

(A) $\left(\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}ight)$

(B) $x y z\left(\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}ight)$

(C) $\frac{1}{x y z}\left(\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}ight)$

(D) $\frac{1}{x y z}\left(\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight)$

Accepted Answer

It is given that $A=\left(\begin{array}{lll}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}ight)$

Hence,

$$

\begin{aligned}

|A| & =x(y z-0) \

& =x y z \

& 
eq 0

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

A_{11}=y z & A_{12}=0 & A_{13}=0 \

A_{21}=0 & A_{22}=x z & A_{23}=0 \

A_{31}=0 & A_{32}=0 & A_{33}=x y

\end{array}

$$

Therefore,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|}(\operatorname{adj} A) \

& =\frac{1}{x y z}\left(\begin{array}{ccc}

y z & 0 & 0 \

0 & x z & 0 \

0 & 0 & x y

\end{array}ight) \

& =\left(\begin{array}{ccc}

\frac{y z}{x y z} & 0 & 0 \

0 & \frac{x z}{x y z} & 0 \

0 & 0 & \frac{x y}{x y z}

\end{array}ight) \

& =\left(\begin{array}{ccc}

\frac{1}{x} & 0 & 0 \

0 & \frac{1}{y} & 0 \

0 & 0 & \frac{1}{z}

\end{array}ight) \

& =\left(\begin{array}{ccc}

x^{-1} & 0 & 0 \

0 & y^{-1} & 0 \

0 & 0 & z^{-1}

\end{array}ight)

\end{aligned}

$$

Thus, the correct option is A.

Question 18

Let $A=\left(\begin{array}{ccc}1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1\end{array}ight)$, where $0 \leq 	heta \leq 2 \pi$, then:

(A) $\operatorname{Det}(A)=0$

(B) $\operatorname{Det}(A) \in(2, \infty)$

(C) $\operatorname{Det}(A) \in(2,4)$

(D) $\operatorname{Det}(A) \in[2,4]$

Accepted Answer

$$

A=\left(\begin{array}{ccc}

1 & \sin 	heta & 1 \

-\sin 	heta & 1 & \sin 	heta \

-1 & -\sin 	heta & 1

\end{array}ight)

$$

It is given that

Hence,

$$

\begin{aligned}

|A| & =1\left(1+\sin ^{2} 	hetaight)-\sin 	heta(-\sin 	heta+\sin 	heta)+1\left(\sin ^{2} 	heta+1ight) \

& =1+\sin ^{2} 	heta+\sin ^{2} 	heta+1 \

& =2+2 \sin ^{2} 	heta \

& =2\left(1+\sin ^{2} 	hetaight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \Rightarrow 0 \leq 	heta \leq 2 \pi \

& \Rightarrow-1 \leq \sin 	heta \leq 1 \

& \Rightarrow 0 \leq \sin ^{2} 	heta \leq 1 \

& \Rightarrow 1 \leq 1+\sin ^{2} 	heta \leq 2 \

& \Rightarrow 2 \leq 2\left(1+\sin ^{2} 	hetaight) \leq 4

\end{aligned}

$$

Therefore,

$$

\operatorname{Det}(A) \in[2,4]

$$

Thus, the correct option is D.