Solution

$ \begin{aligned} \Delta & = \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} \\ & =x(x^{2}-1)-\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(\sin ^{2} \theta+\cos ^{2} \theta) \\ & =x^{3}-x+x \\ & =x^{3} \text{ (Independent of } \theta \text{ ) } \end{aligned} $

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

Solution

$ \Delta= \begin{vmatrix} \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{vmatrix} $

Expanding along $C_3$, we have:

$ \begin{aligned} \Delta & =-\sin \alpha(-\sin \alpha \sin ^{2} \beta-\cos ^{2} \beta \sin \alpha)+\cos \alpha(\cos \alpha \cos ^{2} \beta+\cos \alpha \sin ^{2} \beta) \\ & =\sin ^{2} \alpha(\sin ^{2} \beta+\cos ^{2} \beta)+\cos ^{2} \alpha(\cos ^{2} \beta+\sin ^{2} \beta) \\ & =\sin ^{2} \alpha(1)+\cos ^{2} \alpha(1) \\ & =1 \end{aligned} $

Solution

Given: $A^{-1}=\begin{vmatrix}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{vmatrix}$ and $B=\begin{vmatrix}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{vmatrix}$

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

$\begin{aligned} & |B|=\begin{vmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{vmatrix} \\ & \Rightarrow|B|=1(3-0)-2(-1-0)-2(2-0) \\ & \Rightarrow|B|=1(3)-2(-1)-2(2) \\ & \Rightarrow|B|=3+2-4=1 \end{aligned}$

Since, $|B| \neq 0$

Thus,$B^{-1}$ exists.

$\begin{aligned} & B=\begin{vmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{vmatrix} \\ & M_{11}=\begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix}=3(1)-(-2) 0=3 \\ & M_{12}=\begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix}=1(1)-(0) 0=-1 \\ & M_{13}=\begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix}=(-1)(-2)-O(3)=2 \\ & M_{21}=\begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix}=2(1)-(-2)(-2)=-2 \end{aligned}$

$\begin{aligned} & M_{22}=\begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix}=1(1) O(-2)=1 \\ & M_{23}=\begin{vmatrix} 1 & 2 \\ 0 & -2 \end{vmatrix}=1(-2)-0(2)=-2 \\ & M_{31}=\begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix}=2(0)-3(-2)=6 \\ & M_{32}=\begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix}=1(0)-(-1)(-2)=-2 \\ & M_{33}=\begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix}=1(3)-(-1) 2=5 \end{aligned}$

$\begin{aligned} & \therefore \operatorname{adj}(B)=\begin{vmatrix} M_{11} & -M_{21} & M_{31} \\ -M_{12} & M_{22} & -M_{31} \\ M_{13} & -M_{23} & M_{33} \end{vmatrix} \\ & \begin{vmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{vmatrix} \end{aligned}$

Now,

$B^{-1}=\frac{1}{|B|} \operatorname{adj}(B)$

$\Rightarrow B^{-1}=\frac{1}{1}\begin{vmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{vmatrix}$

$\begin{aligned} &\Rightarrow B^{-1}=\begin{vmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{vmatrix}\\ &B^{-1} A^{-1}=\begin{vmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{vmatrix}\begin{vmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{vmatrix}\\ &\Rightarrow \mathrm{B}^{-1} \mathrm{~A}^{-1}\\ &=\begin{vmatrix} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{vmatrix}\\ &B^{-1} A^{-1}=\begin{vmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \\ \end{vmatrix} \end{aligned}$

So, $(A B)^{-1}=B^{-1} A^{-1}=\begin{vmatrix}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{vmatrix}$

Therefore, $(A B)^{-1}=\begin{vmatrix}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{vmatrix}$

Solution

$ \begin{aligned} & A= \begin{vmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{vmatrix} \\ & \therefore|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13 \end{aligned} $

$ \begin{aligned} & \text{ Now, } 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 _{13}=-1 \\ & \therefore \text{ adj } A= \begin{vmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{vmatrix} \\ & \therefore A^{-1}=\frac{1}{|A|}(\text{ adj } A) \\ & \quad=-\frac{1}{13} \begin{vmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{vmatrix} =\frac{1}{13} \begin{vmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{vmatrix} \end{aligned} $

(i)

$ \begin{aligned} \text{ 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,

$ \begin{aligned} & adj(adj A)= \begin{vmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{vmatrix} \\ & \therefore[adj A]^{-1}=\frac{1}{|adj A|}(adj(adj A)) \\ & =\frac{1}{169} \begin{vmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{vmatrix} \\ & =\frac{1}{13} \begin{vmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{vmatrix} \\ & \text{ Now, } A^{-1}=\frac{1}{13} \begin{vmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{vmatrix} = \begin{vmatrix} -\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{vmatrix} \\ & \therefore adj(A^{-1})= \begin{vmatrix} -\frac{4}{169}-\frac{9}{169} & -(-\frac{11}{169}-\frac{15}{169}) & -\frac{33}{169}+\frac{20}{169} \\ -(-\frac{11}{169}-\frac{15}{169}) & -\frac{14}{169}-\frac{25}{169} & -(-\frac{42}{169}+\frac{55}{169}) \\ -\frac{33}{169}+\frac{20}{169} & -(-\frac{42}{169}+\frac{55}{169}) & \frac{56}{169}-\frac{121}{169} \end{vmatrix} \\ & =\frac{1}{169} \begin{vmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{vmatrix} =\frac{1}{13} \begin{vmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{vmatrix} \end{aligned} $

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

(ii)

We have shown that:

$A^{-1}=\frac{1}{13} \begin{vmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{vmatrix} $

And, $adj A^{-1}=\frac{1}{13} \begin{vmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{vmatrix} $

Now,

$|A^{-1}|=(\frac{1}{13})^{3}[-14 \times(-13)+11 \times(-26)+5 \times(-13)]=(\frac{1}{13})^{3} \times(-169)=-\frac{1}{13}$

$\therefore(A^{-1})^{-1}=\frac{\text{ adj } A^{-1}}{|A^{-1}|}=\frac{1}{(-\frac{1}{13})} \times \frac{1}{13} \begin{vmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{vmatrix} = \begin{vmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{vmatrix} =A$

$\therefore(A^{-1})^{-1}=A$

Solution

$\Delta= \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{vmatrix} $

Applying $R_1 \to R_1+R_2+R_3$, we have:

$\Delta= \begin{vmatrix} 2(x+y) & 2(x+y) & 2(x+y) \\ y & x+y & x \\ x+y & x & y\end{vmatrix} $

$=2(x+y) \begin{vmatrix} 1 & 1 & 1 \\ y & x+y & x \\ x+y & x & y\end{vmatrix} $

Applying $C_2 \to C_2-C_1$ and $C_3 \to C_3-C_1$, we have:

$\Delta=2(x+y) \begin{vmatrix} 1 & 0 & 0 \\ y & x & x-y \\ x+y & -y & -x\end{vmatrix} $

Expanding along $R_1$, we have:

$ \begin{aligned} \Delta & =2(x+y)[-x^{2}+y(x-y)] \\ & =-2(x+y)(x^{2}+y^{2}-y x) \\ & =-2(x^{3}+y^{3}) \end{aligned} $

Solution

Applying $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$, we have:

$\Delta= \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x\end{vmatrix} $

Expanding along $C_1$, we have:

$\Delta=1(x y-0)=x y$

Using properties of determinants in Exercises 11 to 15, prove that:

Solution

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

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= \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{bmatrix} , X= \begin{bmatrix} p \\ q \\ r\end{bmatrix} $ and $B= \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} $.

Now,

$ \begin{aligned} |A| & =2(120-45)-3(-80-30)+10(36+36) \\ & =150+330+720 \\ & =1200 \end{aligned} $

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now,

$A _{11}=75, A _{12}=110, A _{13}=72$

$A _{21}=150, A _{22}=-100, A _{23}=0$

$A _{31}=75, A _{32}=30, A _{33}=-24$

$\therefore A^{-1}=\frac{1}{|A|}$ adj $A$

$ =\frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} $

Now,

$ \begin{aligned} & X=A^{-1} B \\ & \begin{aligned} \Rightarrow \begin{bmatrix} p \\ q \\ r \end{bmatrix} & =\frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} \\ & =\frac{1}{1200} \begin{bmatrix} 300+150+150 \\ 440-100+60 \\ 288+0-48 \end{bmatrix} \\ & =\frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix} \end{aligned} \\ & \therefore p=\frac{1}{2}, q=\frac{1}{3} \text{, and } r=\frac{1}{5} \end{aligned} $

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

Solution

$A= \begin{vmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{vmatrix} $

$\therefore|A|=x(y z-0)=x y z \neq 0$

Now, $A _{11}=y z, A _{12}=0, A _{13}=0$

$A _{21}=0, A _{22}=x z, A _{23}=0$

$A _{31}=0, A _{32}=0, A _{33}=x y$

$\therefore adj A= \begin{vmatrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|}$ adj $A$

$ \begin{aligned} &=\frac{1}{x y z} \begin{vmatrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{vmatrix} \\ &= \begin{vmatrix} \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{vmatrix} \\ &= \begin{vmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{vmatrix} = \begin{vmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{vmatrix} \end{aligned} $

The correct answer is A.

Solution

$ \begin{aligned} & A= \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix} \\ & \therefore|A|=1(1+\sin ^{2} \theta)-\sin \theta(-\sin \theta+\sin \theta)+1(\sin ^{2} \theta+1) \\ & \quad=1+\sin ^{2} \theta+\sin ^{2} \theta+1 \\ & \quad=2+2 \sin ^{2} \theta \\ & \quad=2(1+\sin ^{2} \theta) \end{aligned} $

Now, $0 \leq \theta \leq 2 \pi$

$\Rightarrow 0 \leq \sin \theta \leq 1$

$\Rightarrow 0 \leq \sin ^{2} \theta \leq 1$

$\Rightarrow 1 \leq 1+\sin ^{2} \theta \leq 2$

$\Rightarrow 2 \leq 2(1+\sin ^{2} \theta) \leq 4$

$\therefore Det(A) \in[2,4]$

The correct answer is D.