Matrices and Determinants

Let $A=\left[\begin{array}{ccc}-1 & x & 3 \\ -4 & -5 & -6 \\ -7 & y & 9\end{array}\right]$

Cofactor of $-6=22$

$ \begin{array}{ll} \Rightarrow & (-1)[(-1)(y)-(-7)(x)]=22 \\ \Rightarrow & y-7 x=22 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Cofactor of $\quad-7=27$

$ \begin{array}{lrl} \Rightarrow &(x)(-6)-(-5)(3)=27 \\ \Rightarrow & -6 x+15 & =27 \\ \Rightarrow & -6 x & =12 \\ \Rightarrow & & x =-2 \end{array} $

$ \begin{aligned} &\text { On putting } x=-2 \text { in Eq. (i), }\\ &\begin{aligned} y & =7(-2)+22=-14+22=8 \\ \therefore 5 x+y & =-10+8=-2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \beta x+\alpha y-z=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 3 x-\beta y+\alpha z=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \alpha x+\beta y+z=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & \frac{\Delta_1}{\Delta}=x=-1 \\ & \Rightarrow \quad \frac{\Delta_2}{\Delta}=y=1 \\ & \text { and } \quad \frac{\Delta_3}{\Delta}=z=2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { From Eqs. (i), (ii) and (iii), we have }\\ &\begin{array}{rlrl} & & -\beta+\alpha-2 & =-1 \\ & \Rightarrow & \alpha-\beta & =1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & & -3-\beta+2 \alpha & =0 \\ & \Rightarrow& 2 \alpha-\beta & =3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & & -\alpha+\beta+2 & =1 \\\Rightarrow &&&-\alpha+\beta =-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{array} \end{aligned} $

Solving Eqs. (iv) and (v), we get

$ \begin{aligned} & & \alpha-\beta-2 \alpha+\beta & =1-3 \\ \Rightarrow & & -\alpha & =-2 \\ \Rightarrow & & \alpha & =2 \end{aligned} $

On putting $\alpha=2$ in Eq. (iv),

$ \begin{array}{rlrl} & & 2-\beta & =1 \\ & \Rightarrow & & \beta =1\\ & \therefore & & (\alpha, \beta) =(2,1) \end{array} $

$ \begin{array}{ll} \text { } & \left|\begin{array}{cc} 2+3 i & i \\ 1-2 i & -i \end{array}\right|=x+i y \\ \Rightarrow & -i(2+3 i)-i(1-2 i)=x+i y \\ \Rightarrow & -2 i-3 i^2-i+2 i^2=x+i y \\ \Rightarrow & -3 i-i^2=x+i y \\ \Rightarrow & -3 i+1=x+i y \\ \Rightarrow & 1-3 i=x+i y \end{array} $

On comparing real and imaginary parts, we get

$ \begin{array}{ll} & x=1 \text { and } y=-3 \\ \therefore & x+y=1+(-3)=-2 \end{array} $

             $ \begin{aligned} &\text { Given, } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5 \end{array}\right] \text { then, }\\ &A^T=\left[\begin{array}{lll} 1 & 4 & 3 \\ 2 & 3 & 4 \\ 3 & 2 & 5 \end{array}\right] \end{aligned} $

$ \begin{aligned} \left(A+A^T\right) & \left(A-A^T\right) \\ & =\left[\begin{array}{ccc} 2 & 6 & 6 \\ 6 & 6 & 6 \\ 6 & 6 & 10 \end{array}\right]\left[\begin{array}{ccc} 0 & -2 & 0 \\ 2 & 0 & -2 \\ 0 & 2 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} 12 & 8 & -12 \\ 12 & 0 & -12 \\ 12 & 8 & -12 \end{array}\right] \\ & =4\left[\begin{array}{ccc} 3 & 2 & -3 \\ 3 & 0 & -3 \\ 3 & 2 & -3 \end{array}\right] \end{aligned} $

Given,

$ f(x)=\left|\begin{array}{ccc} x & x+1 & x+3 \\ x+2 & x+4 & x+7 \\ x+6 & x+9 & x+13 \end{array}\right| $

Applying $R_3 \rightarrow R_3-R_1, R_2 \rightarrow R_2-R_1$,

$ f(x)=\left|\begin{array}{ccc} x & x+1 & x+3 \\ 2 & 3 & 4 \\ 6 & 8 & 10 \end{array}\right| $

$ \begin{aligned} & \mathrm{At} x=5, f(5)=\left|\begin{array}{llr} 5 & 6 & 8 \\ 2 & 3 & 4 \\ 6 & 8 & 10 \end{array}\right| \\ & =5(30-32)-6(20-24)+8(16-18) \\ & =-10+24-16=-2 \end{aligned} $

Given, $A=\left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2\end{array}\right]$

$ \because \quad A^{-1}=\alpha A^2+\beta A+\gamma I $

Multiply by $A$ both sides,

$ \begin{aligned} A^{-1} \cdot A & =\alpha A^3+\beta A^2+\gamma I \cdot A \\ I & =\alpha A^3+\beta A^2+\gamma A \end{aligned} $

For matrix $A$

$\operatorname{det} .(A-\lambda I)=0$

$ \left|\begin{array}{ccc} 2-\lambda & 1 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda \end{array}\right|=0 $

$ \text { Expanding along } R_1 $

$ \begin{aligned} & (2-\lambda)(1-\lambda)(2-\lambda)-1(0)+1(\lambda-1)=0 \\ & \Rightarrow \quad(1-\lambda)\left(\lambda^2-4 \lambda+4\right)+\lambda-1=0 \\ & \Rightarrow \lambda^2-4 \lambda+4-\lambda^3+4 \lambda^2-4 \lambda+\lambda-1=0 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & -\lambda^3+5 \lambda^2-7 \lambda+3=0 \\ \Rightarrow & \frac{1}{3} \lambda^3-\frac{5}{3} \lambda^2+\frac{7}{3} \lambda=1 \end{array} $

$ \begin{aligned} &\text { Compare with } \alpha A^3+\beta A^2+\gamma A=I \text {, }\\ &\begin{aligned} & \text { So, } \alpha=\frac{1}{3}, \beta=\frac{-5}{3}, \gamma=\frac{7}{3} \\ & \therefore 17 \alpha+5 \beta+\gamma=\frac{17}{3}-\frac{25}{3}+\frac{7}{3}=-\frac{1}{3} \end{aligned} \end{aligned} $

Here,

$A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right], \operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and also $|A|>0$.

Now, using $|\operatorname{adj} A|=|A|^{n-1} ; n$ is order of square matrix $A$.

$ \begin{aligned} & \quad\left|\begin{array}{ccc} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{array}\right|=|A|^{3-1} \\ & \Rightarrow \quad|A|^2=1(2)+1(2)-1(0)=4 \\ & \Rightarrow \quad|A|=2 \quad[\because|A|>0] \end{aligned} $

$ \text { } \begin{aligned} Now, X & =A^{-1}\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]=\frac{\operatorname{Adj}(A)}{|A|}\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] \\ & =\frac{1}{2}\left[\begin{array}{ccc} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] \\ \Rightarrow \quad X & =\frac{1}{2}\left[\begin{array}{c} -2 \\ 0 \\ 4 \end{array}\right]=\left[\begin{array}{c} -1 \\ 0 \\ 2 \end{array}\right] \end{aligned} $

$ \text { } \begin{aligned}Given, A & =\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right] \\ A^2 & =\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right]=\left[\begin{array}{cc} 1 & k \\ k & 1+k^2 \end{array}\right] \end{aligned} $

$ \begin{aligned} A^3=A^2 \times A & =\left[\begin{array}{cc} 1 & k \\ k & 1+k^2 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right] \\ & =\left[\begin{array}{cc} k & 1+k^2 \\ 1+k^2 & 2 k+k^3 \end{array}\right] \end{aligned} $

After comparing with given value of $A^3$, we get

$ \begin{aligned} & a=k, b=1+k^2, c=1+k^2 \text { and } \\ & d=2 k+k^3 \end{aligned} $

Given, $d=228$

$ \Rightarrow \quad 2 k+k^3=228 \Rightarrow k^3+2 k-228=0 $

By hit and trial method, $k=6$

$ \begin{aligned} b+c & =1+k^2+1+k^2 \\ & =2\left(1+k^2\right) \\ & =2(1+36)=74 \end{aligned} $

$A$ and $B$ are squae matrices of order 3.

$C$ is a unit matrix of order 3.

$ D=(A B-C)^{-1} $

$\Rightarrow D$ and $A B-C$ are inverse of each other

$ \begin{aligned} & & D \times(A B-C) & =I \\ \Rightarrow & & |D| \times|A B-C| & =1 \end{aligned} $

$ \begin{aligned} \Rightarrow|A B-C|=\frac{1}{|D|} \text { or }|D| & =\frac{1}{|A B-C|} \\ \text { Now, }|B C-C| \times|B D A| & =\frac{1}{|D|} \times|B D A| \\ & =\frac{|B| \times|D| \times|A|}{|D|} \\ & =|B| \times|A|=|B A| \end{aligned} $

If $D$ and $A B-C$ are inverse of each other,

$ \text { } \begin{aligned}then \,\,\, (A B-C) \times D & =D \times(A B-C)=I \\ A B D-D & =D A B-D=I \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& A B D=D A B=D+I \\ \therefore \,\,\,\,\,\,\,& A B D=D A B \end{array}\\ &\text { Hence, both statements are true. } \end{aligned} $

$ \text { } \begin{aligned} \text { } A & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right], B=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ |A| & =\left[\left.\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array} \right\rvert\,=-1(0-1)=1\right. \\ \operatorname{adj} A & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]^T \\ A^{-1} & =\frac{1}{|A|} \operatorname{adj} A=A \\ |B| & =-1(1)=-1 \\ \operatorname{adj} B & =\left[\begin{array}{ccc} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right] \end{aligned} $

$ \begin{aligned} \therefore B^{-1}= & \frac{1}{|B|} \operatorname{adj} B=B \\ & \left(A^{-1} B\right)^{-1}+\left(A B^{-1}\right)^{-1} \\ & (A B)^{-1}+(A B)^{-1} \\ & {\left[\because A^{-1}=A \text { and } B^{-1}=B\right] } \\ & =2(A B)^{-1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

$ \begin{aligned} &\begin{aligned} A B & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{array}\right] \\ |A B| & =-1(1)=-1 \\ \operatorname{adj} A B & =\left[\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]^T=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right] \end{aligned}\\ &\text { From Eq. (i), }\\ &\begin{aligned} 2 A B^{-1} & =2 \frac{1}{|A B|} \operatorname{adj}(A B) \\ & =-2\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -2 & 0 \\ 0 & 0 & -2 \\ -2 & 0 & 0 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & {\left[\begin{array}{ccc} 7 & 5 & \alpha \\ \beta & 2 & 11 \\ 3 & \gamma & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 3 \\ 2 \end{array}\right]=\left[\begin{array}{c} \alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma \end{array}\right]} \\ & \Rightarrow \quad\left[\begin{array}{c} 22+2 \alpha \\ \beta+28 \\ 5+3 \gamma \end{array}\right]=\left[\begin{array}{c} \alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma \end{array}\right] \end{aligned} \end{aligned} $

$ \Rightarrow \quad \begin{aligned} 22+2 \alpha & =\alpha+\beta \\ -\alpha+\beta & =22 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, $-2 \alpha+\beta-2 \gamma=\beta+28$

$ \Rightarrow \quad-2(\alpha+\gamma)=28 $

or $\quad \alpha+\gamma=-14\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Also, $5+3 \gamma=\alpha+2 \beta+3 \gamma$

$ \alpha+2 \beta=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On adding Eqs. (i) and (iii),

$ 3 \beta=27 \Rightarrow \beta=9 $

From Eq. (i), $-\alpha+9=22$

$ \alpha=-13 $

From Eq. (ii), $-13+\gamma=-14$

$ \begin{aligned} & \gamma=-1 \\ & \begin{aligned} \therefore 100+\frac{2 \alpha+11 \beta}{\gamma} & =100+\left[\frac{(-26)+99}{-1}\right] \\ & =100-73=27 \end{aligned} \end{aligned} $

Let $A=\left[\begin{array}{ccc}0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0\end{array}\right]$

If $A$ is a skew-symmetric, then $A^T=-A$

$ \begin{aligned} {\left[\begin{array}{ccc} 0 & b & -3 \\ 2 & 0 & c \\ a & 4 & 0 \end{array}\right] } & =\left[\begin{array}{ccc} 0 & -2 & -a \\ -b & 0 & -4 \\ 3 & -c & 0 \end{array}\right] \\ \Rightarrow \quad b=-2, a & =3 \text { and } c=-4 \\ {\left[\begin{array}{ll} a & b \\ b & a \end{array}\right]\left[\begin{array}{ll} b & c \\ c & b \end{array}\right] } & =\left[\begin{array}{cc} 3 & -2 \\ -2 & 3 \end{array}\right]\left[\begin{array}{ll} -2 & -4 \\ -4 & -2 \end{array}\right] \\ & =\left[\begin{array}{cc} 2 & -8 \\ -8 & 2 \end{array}\right] \end{aligned} $

$ \text {Let } A=\left[\begin{array}{ccc} -1 & 2 & b \\ a & 5 & 6 \\ 3 & c & 7 \end{array}\right] $

If $A$ is a symmetric matrix, then $A^T=A$

$ \left[\begin{array}{ccc} -1 & a & 3 \\ 2 & 5 & c \\ b & 6 & 7 \end{array}\right]=\left[\begin{array}{ccc} -1 & 2 & b \\ a & 5 & 6 \\ 3 & c & 7 \end{array}\right] $

On comparing both the matrices, we get $a=2, b=3$ and $c=6$

$ \begin{aligned} \therefore & \left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=\left|\begin{array}{lll} 2 & 3 & 6 \\ 3 & 6 & 2 \\ 6 & 2 & 3 \end{array}\right| \\ & =2(18-4)-3(9-12)+6(6-36) \\ & =28+9-180=-143 \end{aligned} $

$A^2-4 A-5 I=0$

Post multiplying by $A^{-1}$ on both the sides, we get

$ \begin{aligned} & A-4 I-5 A^{-1}=0 \\ \Rightarrow & 5 A^{-1}=A-4 I \\ \Rightarrow & A^{-1}=\frac{1}{5}\{A-4 I\} \\ \therefore & A^{-1}=\frac{1}{5}\left\{\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right]\right\} \\ \because & A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \end{aligned} $

$\because B$ is the unique solution of $A Y=Q$ and $A$ is an invertible matrix

$ \begin{aligned} \Rightarrow & & A B & =Q \\ \Rightarrow & & B & =A^{-1} Q \end{aligned} $

Now, $\quad A X=B$

$ \begin{aligned} \Rightarrow \quad X & =A^{-1} B \\ & =A^{-1}\left(A^{-1} Q\right)=\left(A^{-1}\right)^2 Q \end{aligned} $

$ \begin{aligned} &\text { } f(x)=\left|\begin{array}{ccc} -\sin x & 2 \sin 2 x & 4 \cos ^2 x \\ \cos x & 4 \sin ^2 x & 2 \sin 2 x \\ 0 & -\cos x & \sin x \end{array}\right|\\ &\begin{gathered} =-\sin x\left|\begin{array}{cc} 4 \sin ^2 x & 2 \sin 2 x \\ -\cos x & \sin x \end{array}\right| \\ -\cos x\left|\begin{array}{cc} 2 \sin 2 x & 4 \cos ^2 x \\ -\cos x & \sin x \end{array}\right| \\ =-\sin x\left[4 \sin ^3 x+4 \cos ^2 x \sin x\right] \\ -\cos x\left[4 \sin ^2 x \cos x+4 \cos ^3 x\right] \\ {[\because \sin 2 x=2 \sin x \cos x]} \\ =4\left[-\sin ^4 x-\cos ^2 x \sin ^2 x\right. \\ \left.-\sin ^2 x \cos ^2 x-\cos ^4 x\right] \\ =-4\left[\sin ^4 x+\cos ^4 x+2 \sin ^2 x \cos ^2 x\right] \\ =-4\left(\sin ^2 x+\cos ^2 x\right)^2=-4 \\ \therefore\,\,\,\,\,\,\,\,\,\, f^{\prime}(x)=0 \\ f\left(\frac{5 \pi}{4}\right)+f^{\prime}\left(\frac{5 \pi}{4}\right)=-4+0=-4 \end{gathered} \end{aligned} $

$ \begin{aligned} & \text { } A+B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{array}\right] . A B=\left[\begin{array}{lll} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{array}\right] \\ & (A+B)^2=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{array}\right]=\left[\begin{array}{lll} 5 & 8 & 8 \\ 4 & 5 & 2 \\ 2 & 8 & 4 \end{array}\right] \\ & (A+B)^2=A^2+B^2+A B+B A \\ & \therefore \quad A^2+B^2+B A=(A+B)^2-A B \\ & \text { or } A^2+B(A+B)=\left[\begin{array}{lll} 5 & 8 & 8 \\ 4 & 5 & 2 \\ 2 & 8 & 4 \end{array}\right] -\left[\begin{array}{lll} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{array}\right] =\left[\begin{array}{lll} 4 & 6 & 6 \\ 3 & 4 & 2 \\ 1 & 6 & 3 \end{array}\right]\end{aligned} $

$ \begin{aligned} & \text { } \quad|-B|=5 \\ & \Rightarrow \quad(-1)^3|B|=5 \Rightarrow|B|=-5 \\ & \left|B A^T\right|=15 \Rightarrow|B|\left|A^T\right|=15 \\ & \Rightarrow \quad(-5)|A|=15 \quad\left[\because\left|A^T\right|=|A|\right] \\ & \Rightarrow \quad|A|=-3 \end{aligned} $

$ \begin{array}{ll} \text { Again, } & \left|P^T A P\right|=-27 \\ \Rightarrow & \left|P^T\|A\| P\right|=-27 \\ \Rightarrow & |P||A||P|=-27\,\,\,\,\,\, \left[\because\left|P^T\right|=|P|\right]\\ \Rightarrow & |P|^2(-3)=-27 \Rightarrow|P|^2=9 \end{array} $

$ \therefore \,\,\,\,\,\,\,\,\,\,\,\quad|P|= \pm 3 $

$ \begin{aligned} & \text { Given, }|A|=\frac{1}{2} \\ & \qquad \begin{array}{l} \left|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))\right|^{-1}=\frac{1}{\left|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))\right|} \\ =\frac{1}{\left|A^{-1}\right| \cdot|\operatorname{Adj}(\operatorname{Adj} A)|}=\frac{1}{\left|A^{-1}\right| \cdot|\operatorname{Adj} A|^{3-1}} \\ =\frac{1}{\left|A^{-1}\right| \cdot\left(|A|^{3-1}\right)^2}=\frac{1}{\frac{1}{|A|} \cdot|A|^4} \\ =\frac{1}{|A|^3}=\frac{1}{\left(\frac{1}{2}\right)^3}=8 \end{array} \end{aligned} $

$ \begin{aligned} & \text { } \begin{array}{c} \text { } 2 x+3 y-2 z=-4 \\ 3 x-4 y+3 z=-5 \\ \Rightarrow \quad k x-2 y+z=-3 \\ D=\left|\begin{array}{ccc} 2 & 3 & -2 \\ 3 & -4 & 3 \\ k & -2 & 1 \end{array}\right|=k+7 \\ D_x=\left|\begin{array}{ccc} -4 & 3 & -2 \\ -5 & -4 & 3 \\ -3 & -2 & 1 \end{array}\right| \\ =-4(2)-3(4)-2(-2)=-16 \\ \therefore \quad \alpha=D_x / D=-2 \\ \Rightarrow \quad \frac{-16}{k+7}=-2 \Rightarrow k+7=8 \\ \Rightarrow \quad k=1 \text { or }\left|\begin{array}{ll} 3 & 5 \\ 1 & 2 \end{array}\right| \end{array} 1 \end{aligned} $

$ \begin{aligned} & \text { } \frac{x^2+7}{\left(x^2+1\right)(x-2)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1} \\ & \Rightarrow x^2+7=A\left(x^2+1\right)+(B x+C)(x-2) \\ & \text { On putting } x=2, A=11 / 5 \\ & \text { On putting } x=0, C=\frac{-12}{5} \\ & \text { On putting } x=1, B=\frac{-6}{5} \end{aligned} $

$ \begin{aligned} &\text { ∴ The value of the required determinant }\\ &\begin{aligned} & =\left|\begin{array}{cc} \frac{11}{5} & \frac{-6}{5} \\ \frac{-12}{5} & \frac{2}{5} \end{array}\right| \\ & =\frac{22}{25}-\frac{72}{25}=-2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, trace of } A=-4\\ &\begin{aligned} & \therefore \quad a+(-1)+(-4)=-4 \\ & \Rightarrow \quad a=1 \\ & \text { Now, } A B=\left[\begin{array}{ccc} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} {\left[\begin{array}{ccc} 1 & 3 & 5 \\ 5 & -1 & 3 \\ 2 & 3 & -4 \end{array}\right]\left[\begin{array}{ccc} b & 1 & 4 \\ 4 & c & 1 \\ -3 & 1 & d \end{array}\right] } & =\left[\begin{array}{ccc} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} b-3 & 3 c+6 & 7+5 d \\ 5 b-13 & 8-c & 19+3 d \\ 2 b+24 & 3 c-2 & 11-4 d \end{array}\right] & =\left[\begin{array}{ccc} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{array}\right] \end{aligned}\\ &\text { On comparing, }\\ &\begin{aligned} b-3=-1 \Rightarrow b & =2 \\ 3 c+6=0 \Rightarrow c & =-2 \\ 7+5 d=17 \Rightarrow d & =2 \\ \therefore a+b+c+d=1+2-2+2 & =3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \text { Given, }\left|\begin{array}{ccc} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{array}\right| \\ & \begin{array}{r} =\left(b^2 c^3-b^3 c^2\right)-\left(a^2 c^3-a^3 c^2\right) +\left(a^2 b^3-a^3 b^2\right) \\ =\left(a^2 b^3-a^2 c^3\right)+\left(b^2 c^3-a^3 b^2\right) +\left(a^3 c^2-b^3 c^2\right) \\ =a^2\left(b^3-c^3\right)+b^2\left(c^3-a^3\right)+c^2\left(a^3-b^3\right) \end{array} \end{aligned} $

$ \begin{aligned} &\text { Let } \alpha, \beta, \gamma \text { be real numbers, then }\\ &\begin{aligned} & A=\left[\begin{array}{ccc} 7 & 3 & \alpha \\ \beta & 1 & -11 \\ -5 & \gamma & 19 \end{array}\right] \\ & \Rightarrow A\left[\begin{array}{c} 5 \\ -13 \\ 11 \end{array}\right]=\left[\begin{array}{c} -290 \\ -119 \\ 210 \end{array}\right] \\ & \Rightarrow {\left[\begin{array}{ccc} 7 & 3 & \alpha \\ \beta & 1 & -11 \\ -5 & \gamma & 19 \end{array}\right]\left[\begin{array}{c} 5 \\ -13 \\ 11 \end{array}\right]=\left[\begin{array}{c} -290 \\ -119 \\ 210 \end{array}\right] } \\ & 11 \alpha-4=-290 \Rightarrow 11 \alpha=-286 \\ & \Rightarrow \quad \alpha=-265 \beta-134=-119 \\ & \Rightarrow 5 \beta=15 \Rightarrow \beta=3-13 \gamma+184=210 \\ & \Rightarrow-13 \gamma=26 \Rightarrow \gamma=-2 \\ & \quad A=\left[\begin{array}{ccc} 7 & 3 & -26 \\ 3 & 1 & -11 \\ -5 & -2 & 19 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} |A| & =7(-3)-3(2)-26(-1) \\ & =-27+26=-1 \\ A^{-1} & =\frac{\operatorname{adj}(A)}{|A|} \\ \Rightarrow \operatorname{adj}(A) & =A^{-1}|A| \end{aligned} $

$ \begin{aligned} (\operatorname{adj} A)^{-1}+\operatorname{adj} A^{-1} & =(\operatorname{adj} A)^{-1}+(\operatorname{adj} A)^{-1} \\ & =2(\operatorname{adj} A)^{-1} \\ =2\left(A^{-1}|A|\right)^{-1} & =2 A \cdot|A|^{-1} \\ =\frac{2 A}{|A|}=\frac{2 A}{-1} & =-2 A \end{aligned} $

Given, $[\alpha, \beta \gamma]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5\end{array}\right]=\left[\begin{array}{lll}3 & 5 & 2\end{array}\right]$

$ \begin{aligned} & \text { Let } A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{array}\right] \\ & {\left[\begin{array}{lll} \alpha & \beta & \gamma \end{array}\right] \cdot A=\left[\begin{array}{lll} 3 & 5 & 2 \end{array}\right]} \\ & {\left[\begin{array}{lll} \alpha & \beta & \gamma \end{array}\right]=\left[\begin{array}{lll} 3 & 5 & 2 \end{array}\right] A^{-1}} \\ & |A|=1(15+10)-2(10+5)+3(4-3) \\ & |A|=25-30+3=-2 \\ & \operatorname{adj}(A)=\left[\begin{array}{ccc} 25 & -4 & -19 \\ -15 & 2 & 11 \\ 1 & 0 & -1 \end{array}\right] \end{aligned} $

Now, $\left[\begin{array}{lll}\alpha & \beta & \gamma\end{array}\right]=\left[\begin{array}{lll}3 & 5 & 2\end{array}\right] \frac{\operatorname{adj} A}{|A|}$

$ \left[\begin{array}{lll} \alpha & \beta & \gamma \end{array}\right]=\frac{1}{-2}\left[\begin{array}{lll} 3 & 5 & 2 \end{array}\right]\left[\begin{array}{ccc} 25 & -4 & -19 \\ -15 & 2 & 11 \\ 1 & 0 & -1 \end{array}\right] $

$\left[\begin{array}{lll}\alpha & \beta & \gamma\end{array}\right]=-\frac{1}{2}\left[\begin{array}{lll}2 & -2 & -4\end{array}\right]=\left[\begin{array}{lll}-1 & 1 & 2\end{array}\right]$

On comparing $\alpha=-1, \beta=1$ and $\gamma=2$

$ \therefore \quad \alpha^3+\beta^3+\gamma^3=-1+1+8=8 $

Here,

$\begin{aligned} & A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right] \\ & A^T=\left[\begin{array}{ccc} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{array}\right] \\ & A A^T=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{array}\right] \\ & A A^T=\left[\begin{array}{ccc} 9+9+16 & 6+9+16 & 0+3+4 \\ 6+9+16 & 4+9+16 & 0+3+4 \\ 0+3+4 & 0+3+4 & 0+1+1 \end{array}\right] \\ & A A^T=\left[\begin{array}{ccc} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{array}\right] \\ & \left(A A^T\right)^T=\left[\begin{array}{ccc} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{array}\right] \\ \end{aligned}$

So, $A A^T$ is a symmetric matrix.

The given system of equation is $5 x-y+2 z=3, x+y+z=6$ and $2 x+y-z=-5$ In matrix form, if may be represented as where

$A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 5 & -1 & 2 \\ 2 & 1 & -1 \end{array}\right], x=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } D=\left[\begin{array}{c} 6 \\ 3 \\ -5 \end{array}\right]$

Thus, $A$ is a non-singular matrix, so $A^{-1}$ exist.

$\begin{aligned} & C_{11}=(-1)^{1+1}\left|\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right|=-1 \\ & C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 5 & 2 \\ 2 & -1 \end{array}\right|=9 \\ & C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 5 & -1 \\ 2 & 1 \end{array}\right|=7 \\ & C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right|=2 \end{aligned}$

$\begin{aligned} & C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right|=-3 \\ & C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=1 \\ & C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array}\right|=3 \\ & C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 5 & 2 \end{array}\right|=3 \\ & C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 5 & -1 \end{array}\right|=-6 \end{aligned}$

$\begin{aligned} \text { Then, } \operatorname{adj}(A) & =\left[\begin{array}{ccc} -1 & 9 & 7 \\ 2 & -3 & 1 \\ 3 & 3 & -6 \end{array}\right]^T=\left[\begin{array}{ccc} -1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6 \end{array}\right] \\ (\operatorname{adj} A) \cdot D & =\left[\begin{array}{ccc} -1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6 \end{array}\right]\left[\begin{array}{c} 6 \\ 3 \\ -5 \end{array}\right] \\ & =\left[\begin{array}{c} -6+6-15 \\ 54-9-15 \\ 42+3+30 \end{array}\right]=\left[\begin{array}{c} -15 \\ 30 \\ 75 \end{array}\right] \end{aligned}$

$\begin{aligned} & A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right], B=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow A^2=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \\ & \Rightarrow A^2=\left[\begin{array}{ll} 1+0 & 0+0 \\ 2+2 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 4 & 1 \end{array}\right] \\ & \Rightarrow A^3=\left[\begin{array}{ll} 1 & 0 \\ 4 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \quad\left[A^2 \cdot A=A^3\right]\\ & \Rightarrow A^3=\left[\begin{array}{ll} 1+0 & 0+0 \\ 4+2 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 6 & 1 \end{array}\right] \\ & \text { Similarly, } A^6=\left[\begin{array}{ll} 1 & 0 \\ 12 & 1 \end{array}\right] \quad \left[A^n=A_{21}=n \times 2\right]\\ & \text { Now, } B=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow B^2=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow B^2=\left[\begin{array}{ll} 1+0 & 3+3 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow B^3=\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 9 \\ 0 & 1 \end{array}\right] \end{aligned}$

Similarly, $B^6=\left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]\quad \left[B^n=B_{12}=n \times 3\right]$

$\begin{aligned} &\begin{aligned} & \therefore \quad\left(A^6+B^6\right)=\left[\begin{array}{cc} 1 & 0 \\ 12 & 1 \end{array}\right]+\left[\begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array}\right] \\ & A^6+B^6=\left[\begin{array}{cc} 2 & 18 \\ 12 & 2 \end{array}\right] \\ & \end{aligned}\\ &\operatorname{Det}\left(A^6+B^6\right)=4-216=-212 \end{aligned}$

$\text { Here, } G(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$

$G(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right]$

So, $x+y=0 \Rightarrow y=(-x)$

Now, $G(-x)=\left[\begin{array}{ccc}\cos (-x) & -\sin (-x) & 0 \\ \sin (-x) & \cos (-x) & 0 \\ 0 & 0 & 1\end{array}\right]$

$\begin{aligned} G(x) \cdot G(y) & =\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}$

Given, $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 1\end{array}\right]$

$\begin{aligned} \left(A^T\right)^2 & +(12 A)^T \\ & =\left[\begin{array}{cc} 2 & -4 \\ -3 & 1 \end{array}\right]^2+\left[\begin{array}{cc} 24 & -36 \\ -48 & 12 \end{array}\right]^T \\ & =\left[\begin{array}{cc} 16 & -12 \\ -9 & 13 \end{array}\right]+\left[\begin{array}{cc} 24 & -48 \\ -36 & 12 \end{array}\right] \\ & =\left[\begin{array}{cc} 16+24 & -12-48 \\ -9-36 & 13+12 \end{array}\right]=\left[\begin{array}{cc} 40 & -60 \\ -45 & 25 \end{array}\right] \end{aligned}$

Given, $a, b$ and $c$ are respectively 5th, 8th, and 13th terms of arithmetic progression.

$\begin{aligned} & \therefore a=P+(5-1) d \\ & \Rightarrow a=P+4 d \\ & b=P+(8-1) d \\ & \Rightarrow \quad b=P+7 d \\ & c=P+(13-1) d \\ & \Rightarrow c=P+12 d \\ & \therefore \quad\left|\begin{array}{ccc} a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1 \end{array}\right|=\left|\begin{array}{ccc} P+4 d & 5 & 1 \\ P+7 d & 8 & 1 \\ P+12 d & 13 & 1 \end{array}\right| \\ & =\left|\begin{array}{ccc} P & 5 & 1 \\ P & 8 & 1 \\ P & 13 & 1 \end{array}\right|+\left|\begin{array}{ccc} 4 d & 5 & 1 \\ 7 d & 8 & 1 \\ 12 d & 13 & 1 \end{array}\right| \\ \end{aligned}$

$\begin{aligned} & =P\left|\begin{array}{ccc} 1 & 5 & 1 \\ 1 & 8 & 1 \\ 1 & 13 & 1 \end{array}\right|+d\left|\begin{array}{ccc} 4+1 & 5 & 1 \\ 7+1 & 8 & 1 \\ 12+1 & 13 & 1 \end{array}\right| \\ & =P \cdot 0+d\left|\begin{array}{ccc} 5 & 5 & 1 \\ 8 & 8 & 1 \\ 13 & 13 & 1 \end{array}\right|=P \cdot 0+d \cdot 0=0 \end{aligned}$

Given that,

$\begin{aligned} A & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ b+c a+b & 0 & 1 \end{array}\right] \end{aligned}$

$\begin{aligned} & \text { Since, } A^2=I \text { (given) } \\ & \Rightarrow b+c a+b=0 \\ & \Rightarrow b=\frac{-c a}{2} \end{aligned}$

Given,

$A=\left[\begin{array}{ccc} -2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1 \end{array}\right]$

Given, $\operatorname{det} A=\left|\begin{array}{ccc}-2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1\end{array}\right|=0$

$\begin{array}{ll} \Rightarrow & -2(-1-3)+x(2+x)+1(3 x-2)=0 \\ \Rightarrow & 8+2 x+x^2+3 x-2=0 \\ \Rightarrow & x^2+5 x+6=0 \Rightarrow(x+2)(x+3)=0 \\ \Rightarrow & x=-2 \text { and }-3 \\ \therefore & l=-2 \text { and } m=-3 \\ & l^3-m^3=(-2)^3-(-3)^3=-8+27=19 \end{array}$

$\begin{aligned} & \text { Given, } a_i^2+b_i^2+c_i^2=1 \\ & \Rightarrow \quad a_1^2+b_1^2+c_1^2=1 \\ & \qquad a_2^2+b_2^2+c_2^2=1 \\ & \qquad a_3^2+b_3^2+c_3^2=1 \\ \end{aligned}$

Given, $a_i a_j+b_i b_j+c_i c_j=0\quad [i, j=1,2,3, i \neq j]$

$\begin{aligned} \text { Let } \mathbf{p} & =a_1 \mathbf{i}+b_1 \mathbf{j}+c_1 \mathbf{k} \\ \mathbf{q} & =a_2 \mathbf{i}+b_2 \mathbf{j}+c_2 \mathbf{k} \\ \mathbf{r} & =a_3 \mathbf{i}+b_3 \mathbf{j}+c_3 \mathbf{k} \\ |\mathbf{p}| & =|\mathbf{q}|=|\mathbf{r}|=1 \\ \mathbf{p} \cdot \mathbf{q} & =\left(a_1 \mathbf{i}+b_1 \mathbf{j}+c_1 \mathbf{k}\right) \cdot\left(a_2 \mathbf{i}+b_2 \mathbf{j}+c_3 \mathbf{k}\right) \\ & =a_1 a_2+b_1 b_2+c_1 c_2=0 \\ \Rightarrow \mathbf{p} & \perp \mathbf{q} \end{aligned}$

Similarly, $\mathbf{q} \perp \mathbf{r}, \mathbf{r} \perp \mathbf{p}$

$\begin{aligned} & \text { Now, } A=\left|\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right|=\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=\left[\begin{array}{lll} \mathbf{p} & \mathbf{q} & \mathbf{r} \end{array}\right]=1 \\ & \Rightarrow \operatorname{det}\left(A A^\tau\right)=1 \end{aligned}$

$\begin{aligned} & \text { } \begin{aligned} & A=\frac{1}{7}\left[\begin{array}{ccc} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{array}\right] \\ &|A|=\frac{1}{7^3}\left[\begin{array}{ccc} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{array}\right] \quad\left[\because \operatorname{det}(k A)=k^n \operatorname{det} A\right] \\ &=\frac{1}{7^3}[3(-9-12)+2(-18+4)+6(-36-6)] \\ &=\frac{1}{7^3}[-63-28-252] \\ &=\frac{-343}{7^3}=-1 \neq-0 \end{aligned} \end{aligned}$

$\operatorname{Adj}(A)=\frac{1}{7^2}\left[\begin{array}{ccc}-21 & 14 & -42 \\ 42 & 21 & -14 \\ 14 & -42 & -21\end{array}\right]$

$\left[\because \operatorname{adj}(k A)=k^{n-1} \operatorname{adj} A\right]$

$=\frac{1}{7^2}\left[\begin{array}{ccc} -21 & 42 & 14 \\ 14 & 21 & -42 \\ -42 & -14 & -21 \end{array}\right]$

Now, $A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{-1}{7^2}\left[\begin{array}{ccc}-21 & 42 & 14 \\ 14 & 21 & -42 \\ -42 & -14 & -21\end{array}\right]$

$\begin{aligned} & =\frac{1}{7}\left[\begin{array}{ccc} 3 & -6 & -2 \\ -2 & -3 & 6 \\ 6 & 2 & 3 \end{array}\right] \\ A^{-1} & =A^{T} \end{aligned}$

$A=\left[\begin{array}{cc}\alpha^2 & 5 \\ 5 & -\alpha\end{array}\right]$ and $\left|A^{10}\right|=1024$

$\begin{aligned} & |A|=-\alpha^3-25 \\ & \because \quad\left|A^{10}\right|=1024 \\ & \Rightarrow \quad|A|^{10}=2^{10} \\ & \Rightarrow \quad|A|= \pm 2 \\ \end{aligned}$

Taking positive value, $-\alpha^3-25=2 \Rightarrow-\alpha^3=27$

$\begin{aligned} & \Rightarrow \alpha^3=-27 \\ & \Rightarrow \alpha=-3 \end{aligned}$

Taking negative value, $\alpha^3-25=-2 \Rightarrow-\alpha^3=23$

$\Rightarrow \alpha=(-23)^{\frac{1}{3}}$

$A=\left[\begin{array}{ccc} 5 & \sin ^2 \theta & \cos ^2 \theta \\ -\sin ^2 \theta & -5 & 1 \\ \cos ^2 \theta & 1 & 5 \end{array}\right]$

$\begin{aligned} & |A|=5(-25-1)-\sin ^2 \theta\left(-5 \sin ^2 \theta-\cos ^2 \theta\right) \\ & +\cos ^2 \theta\left(-\sin ^2 \theta+5 \cos ^2 \theta\right) \\ & =-130-\sin ^2 \theta\left[-5\left(1-\cos ^2 \theta\right)-\cos ^2 \theta\right] \\ & +\cos ^2 \theta\left[-\sin ^2 \theta+5\left(1-\sin ^2 \theta\right)\right] \\ & =-130-\sin ^2 \theta\left(-5+4 \cos ^2 \theta\right)+\cos ^2 \theta\left(-6 \sin ^2 \theta+5\right) \\ & =-130+5 \sin ^2 \theta-4 \sin ^2 \theta \cos ^2 \theta \\ & -6 \sin ^2 \theta \cos ^2 \theta+5 \cos ^2 \theta \\ & =-130+5-10 \sin ^2 \theta \cos ^2 \theta \\ & |A|=-125-10 \sin ^2 \theta \cos ^2 \theta \\ & \therefore \quad|A|_{\max }=-125 \\ & \end{aligned}$

$\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$

$\begin{aligned} \text { Let } x^2+1 & =k \\ \Rightarrow \quad x^2 & =k-1 \end{aligned}$

Substituting $x^2=k-1$ in LHS, we get

$\begin{aligned} & \frac{(k-1)^2+24(k-1)+28}{k^3}=\frac{k^2+22 k+5}{k^3} \\ &=\frac{1}{k}+\frac{22}{k^2}+\frac{5}{k^3} \\ &=\frac{1}{\left(x^2+1\right)}+\frac{22}{\left(x^2+1\right)^2}+\frac{5}{\left(x^2+1\right)^3} \end{aligned}$

On comparing with RHS, we get

$\begin{aligned} & A x+B=1, C x+D=22 \text { and } E x+F=5 \\ & \text { At } x=1, \\ & A+B=1, C+D=22 \text { and } E+F=5 \\ & \therefore A+B+C+D+E+F=1+22+5=28 \end{aligned}$

Since, the system has no solution.

$\left| {\matrix{ 2 & { - 1} & 2 \cr 1 & { - 2} & { - 1} \cr 1 & 1 & \lambda \cr } } \right| = 0$

$ \Rightarrow 2( - 2\lambda + 1) + 1(\lambda + 1) + 2(3) = 0$

$ \Rightarrow - 4\lambda + 2 + \lambda + 1 + 6 = 0$

$ \Rightarrow 3\lambda = 9 \Rightarrow \lambda = 3$

$A = \left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]$

$\therefore$ $\left| A \right| = \left| {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right|$

$ = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 0$

$ \Rightarrow Adj\,A = {\left[ {\matrix{ 4 & { - 5} & 1 \cr 2 & 0 & { - 2} \cr 2 & 5 & 3 \cr } } \right]^T} = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & 5 \cr 1 & { - 2} & 3 \cr } } \right]$

$ \Rightarrow B = {A^{ - 1}} = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & 5 \cr 1 & { - 2} & 3 \cr } } \right]$ [$\because$ ${A^{ - 1}} = {1 \over {|A|}}Adj\,A$]

$ \Rightarrow 10B = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & 5 \cr 1 & { - 2} & 3 \cr } } \right]$

Hence, $\alpha = 5$

Given,

$\left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right]{\left[ {\matrix{ 1 & {\tan \theta } \cr { - \tan \theta } & 1 \cr } } \right]^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right]\,.\,{1 \over {1 + {{\tan }^2}\theta }}\left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow {1 \over {1 + {{\tan }^2}\theta }}\left[ {\matrix{ {1 - {{\tan }^2}\theta } & { - 2\tan \theta } \cr {2\tan \theta } & {1 - {{\tan }^2}\theta } \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {{{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }}} & {{{2\tan \theta } \over {1 + {{\tan }^2}\theta }}} \cr {{{2\tan \theta } \over {1 + {{\tan }^2}\theta }}} & {{{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }}} \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {\cos 2\theta } & { - \sin 2\theta } \cr {\sin 2\theta } & {\cos 2\theta } \cr } } \right] = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$

$ \Rightarrow a = \cos 2\theta ,b = \sin 2\theta $

As the given system of equations has a non-trivial solution.

$\Delta = \left| {\matrix{ p & a & a \cr b & q & b \cr c & c & r \cr } } \right| = 0$

Applying ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$

$\Delta = \left| {\matrix{ p & {a - p} & {a - p} \cr b & {q - b} & 0 \cr c & 0 & {r - c} \cr } } \right| = 0$

Expanding along C₃, we get

$(a - p)\left| {\matrix{ b & {q - b} \cr c & 0 \cr } } \right| + (r - c)\left| {\matrix{ p & {a - p} \cr b & {q - b} \cr } } \right| = 0$

$ \Rightarrow (a - p)( - c)(q - b) + (r - c)\{ p(q - b) - b(a - p)\} = 0$

$ \Rightarrow (p - a)(q - b)c + p(r - c)(q - b) + b(r - c)(p - a) = 0$

Dividing $(p - a)(q - b)(r - c)$, we get

${c \over {r - c}} + {p \over {p - a}} + {b \over {q - b}} = 0$

$ \Rightarrow {p \over {p - a}} + {q \over {q - b}} + {r \over {r - c}} = {{q - b} \over {q - b}} + {{r - c} \over {r - c}} = 2$