Matrices and Determinants

We have,

$A$ and $B$ are two square matrices

$ \begin{aligned} (A B+B A)^T+(A B-B A)^T & =2 B A \\ (A B)^T+(B A)^T+(A B)^T-(B A)^T & =2 B A \\ 2(A B)^T & =2 B A \\ (A B)^T & =B A \\ A^T B^T & =B A \end{aligned} $

If $A$ is symmetric matrix and $B$ is symmetric matrix.

$ \begin{aligned} \because \quad A^T & =A \text { and } B^T=B \\ A(B) & =B A \\ A B & =B A \quad \text { (which is not possible) } \end{aligned} $

Now, if $A$ and $B$ are skew-symmetric matrix.

$ \begin{aligned} \because \quad A^T & =-A \text { and } B^T=-B \\ -A-B & =B A \\ A B & =B A \quad \text { (which is not possible) } \end{aligned} $

Thus, $A$ and $B$ are neither symmetric nor skew-symmetric matrix.

Given,

$ \operatorname{adj}\left[\begin{array}{ccc} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{ccc} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{array}\right] $

Adjoint of matrix $=$ transpose of cofactor matrix

$ \begin{aligned} & C_{11}=\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|=5 \\ & C_{12}=-\left|\begin{array}{cc} -1 & -2 \\ 0 & 1 \end{array}\right|=1 \\ & C_{13}=\left|\begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array}\right|=-2 \end{aligned} $

$ \begin{aligned} C_{21} & =-\left|\begin{array}{ll} 0 & 2 \\ 2 & 1 \end{array}\right|=4 \\ C_{22} & =\left|\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right|=1 \\ C_{23} & =-\left|\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right|=-2 \\ C_{31} & =\left|\begin{array}{cc} 0 & 2 \\ 1 & -2 \end{array}\right|=-2 \\ C_{32} & =-\left|\begin{array}{cc} 1 & 2 \\ -1 & -2 \end{array}\right|=0 \\ C_{33}=\left|\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array}\right| & =1 \\ \text { So, Adj }\left[\begin{array}{ccc} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{array}\right] & =\left[\begin{array}{ccc} 5 & 1 & -2 \\ 4 & 1 & -2 \\ -2 & 0 & 1 \end{array}\right] \\ {\left[\begin{array}{ccc} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{array}\right] } & =\left[\begin{array}{ccc} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{array}\right] \end{aligned} $

Hence, $m=4, n=1$

$ \therefore m+n=4+1=5 $

We have,

$ \begin{gathered} A=\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right] \text { and } \\ f(x)=x+x^2+x^3+\ldots+x^{2023} \\ \text { Then, } A^2=A \cdot A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ A^3=A A^2=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ A^4=A A^3=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ A^n=A A^{n-1}=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{gathered} $

So, $f(A)=A+A^2+A^3+\ldots+A^{2023}$

$ \begin{aligned} & =\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\ldots+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & =\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right] \end{aligned} $

Hence, $f(A)+I=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$

$ \text { } \begin{aligned} A & =\left[\begin{array}{lll} b & a & 0 \\ c & 0 & b \\ a & a & b \end{array}\right], B=\left[\begin{array}{lll} 0 & a & b \\ b & 0 & c \\ b & a & a \end{array}\right] \\ A B & =\left[\begin{array}{lll} b & a & 0 \\ c & 0 & b \\ a & a & b \end{array}\right]\left[\begin{array}{lll} 0 & a & b \\ b & 0 & c \\ b & a & a \end{array}\right] \\ & =\left[\begin{array}{ccc} 0+a b+0 & a b+0+0 & b^2+c a+0 \\ 0+0+b^2 & a c+0+a b & b c+0+a b \\ 0+a b+b^2 & a^2+0+a b & a b+a c+a b \end{array}\right] \\ & =\left[\begin{array}{ccc} a b & a b & b^2+a c \\ a b+b^2 & a c+a b & b c+a b \\ a & a b+a c+a b \end{array}\right] \\ & =\left[\begin{array}{lll} 2 & 2 & 7 \\ 1 & 8 & 5 \\ 3 & 6 & 10 \end{array}\right] \end{aligned} $

$ b^2=1, \quad a b=2 $

$ \begin{array}{ll} \therefore & a^2+a b=6 \\ \Rightarrow & a^2+2=6 \Rightarrow a^2=4 \end{array} $

$ \begin{aligned} \Rightarrow & & a & = \pm 2 \\ & \therefore & b^2+a c & =7 \\ \Rightarrow & & 1+a c & =7 \Rightarrow a c=6 \end{aligned} $

$ \begin{aligned} c & =\frac{6}{ \pm 2}= \pm 3 \Rightarrow c^2=9 \\ \therefore \quad a^2+b^2+c^2 & =9+1+4=14 \end{aligned} $

$A=\left[\begin{array}{lll}1 & a & 3 \\ b & 2 & c \\ 3 & d & 4\end{array}\right]$

Since, $A^T=A$

$ \begin{gathered} {\left[\begin{array}{lll} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{array}\right]=\left[\begin{array}{lll} 1 & b & 3 \\ a & 2 & d \\ 3 & c & 4 \end{array}\right]} \\ a=b \text { and } c=d \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ B=\left[\begin{array}{ccc} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{array}\right] \end{gathered} $

$ \begin{aligned} &\begin{aligned} & \text { Since, } B^T=-B \\ & \Rightarrow \quad\left[\begin{array}{ccc} 0 & -5 & 6 \\ 5 & 0 & c \\ b & -7 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -5 & -b \\ 5 & 0 & 7 \\ -6 & -c & 0 \end{array}\right] \end{aligned}\\ &\text { So, } b=-6, a=-6, c=7 \text { and } d=7\\ &\begin{aligned} A B & =\left[\begin{array}{ccc} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{array}\right]\left[\begin{array}{ccc} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0+30+18 & 5+0+21 & -6+42 \\ 0-10+42 & -30+49 & 22 \\ -11 & 43 & -67 \end{array}\right] \\ & =\left[\begin{array}{ccc} 48 & 26 & 36 \\ 32 & 19 & 22 \\ -11 & 43 & -67 \end{array}\right] \end{aligned} \end{aligned} $

$A=\left[\begin{array}{ccc}-1 & -3 & -2 \\ 0 & 1 & 2 \\ 3 & 4 & 5\end{array}\right]$

Co-factor matrix of $A$

$ \begin{aligned} & =\left[\begin{array}{ccc} -3 & +6 & -3 \\ +7 & 1 & -5 \\ -4 & 2 & -1 \end{array}\right] \\ \operatorname{adj}(A) & =\left[\begin{array}{ccc} -3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1 \end{array}\right] \\ |A| & =-1(5-8)+3(0-6)-2(0-3) \end{aligned} $

$ \begin{gathered} =3-18+6=-9 \\ \therefore A^{-1}=\frac{\operatorname{Adj} A}{|A|}=\left|\begin{array}{ccc} 1 / 3 & -7 / 9 & 4 / 9 \\ -2 / 3 & -1 / 9 & -2 / 9 \\ 1 / 3 & 5 / 9 & 1 / 9 \end{array}\right| \\ \therefore a_1+c_2+b_3=\frac{1}{3}+\frac{5}{9}+\left(-\frac{2}{9}\right) \\ \quad=\frac{3+5-2}{9}=\frac{6}{9}=\frac{2}{3} \end{gathered} $

Given, the system of linear equations

$ \begin{aligned} & 2 x-3 y+5 z=12 \\ & 5 x+2 y+3 z=11 \end{aligned} $

and $x+2 y-3 z=-3$

has the unique solution $(\alpha, \beta, \gamma)$.

Now, by using the Cramer's rule, we get

$ \begin{aligned} \alpha & =\frac{\left|\begin{array}{ccc} 12 & -3 & 5 \\ 11 & 2 & 3 \\ -3 & 2 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{array}\right|}=\frac{-76}{-38}=2 \\ \beta & =\frac{\left|\begin{array}{ccc} 2 & 12 & 5 \\ 5 & 11 & 3 \\ 1 & -3 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{array}\right|}=\frac{38}{-38}=-1 \\ \text { and } \gamma & =\frac{\left|\begin{array}{ccc} 2 & -3 & 12 \\ 5 & 2 & 11 \\ 1 & 2 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{array}\right|}=\frac{-38}{-38}=1 \end{aligned} $

$ \begin{array}{r} \therefore 2 \alpha+5 \beta+3 \gamma=2(2)+5(-1)+3(1) \\ =4-5+3=7-5=2 \end{array} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} A & =\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{array}\right], B=\left[\begin{array}{ccc} -3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3 \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} -2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3 \end{array}\right]=A+B \end{aligned} \end{aligned} $

Let $\Delta_1=\left|\begin{array}{lll}2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3\end{array}\right|=-70$

$ \begin{aligned}and\,\, \Delta_2 & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 7 & 11 & 13 \\ 49 & 121 & 169 \end{array}\right| \\ & C_2 \rightarrow C_2-C_1 \\ & C_3 \rightarrow C_3-C_1 \\ & =\left|\begin{array}{ccc} 1 & 0 & 0 \\ 7 & 4 & 6 \\ 49 & 72 & 120 \end{array}\right|=48 \end{aligned} $

Now, $\Delta_1+\Delta_2=-70+48=-22$

$\operatorname{det} A=190$

$ \begin{aligned} & {\left[\begin{array}{ccc} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{array}\right]=190 } \\ \Rightarrow & (k-3)\left(k^2+3 k+19\right)=0 \Rightarrow k=3 \end{aligned} $

$\operatorname{adj} A=$ Transpose of co-factor matrix of $A$

$ =\left[\begin{array}{ccc} -1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19 \end{array}\right] $

System of equations

$ \begin{aligned} 3 x-2 y+z & =5 k \\ 2 x+3 y-2 z & =-5 k \\ x+4 y+3 z & =k \end{aligned} $

It has unique solution.

If $\Delta \neq 0$ and $x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}$

$ \begin{array}{ll} \text { Now, } & \Delta=\left|\begin{array}{ccc} 3 & -2 & 1 \\ 2 & 3 & -2 \\ 1 & 4 & 3 \end{array}\right|=72 \neq 0 \\ \text { and } & \Delta_3=\left|\begin{array}{ccc} 3 & -2 & 5 k \\ 2 & 3 & -5 k \\ 1 & 4 & k \end{array}\right| \\ & =k[69+14+25]=108 k \\ \therefore & z=3\,\,\,\, (given) \\ \Rightarrow & \frac{\Delta_3}{\Delta}=3 \Rightarrow \frac{108 k}{72}=3 \\ \Rightarrow & k=2 \end{array} $

$\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|$

Taking $\sqrt{3}, \sqrt{5}$ and $\sqrt{5}$ from $C_1, C_2$ and $C_3$, we get

$ \begin{aligned} & \sqrt{3} \times \sqrt{5} \times \sqrt{5}\left|\begin{array}{ccc} 1 & 2 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5} \end{array}\right| \\ & =5 \sqrt{3}\left|\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{5} & 0 & \sqrt{2} \\ \sqrt{3} & 0 & \sqrt{5} \end{array}\right| \end{aligned} $

$ \begin{aligned} & \text { (Applying } C_2 \rightarrow C_2-C_1, \\ & \quad-5 \sqrt{3}(5-\sqrt{6}) \\ & =-25 \sqrt{3}+15 \sqrt{2}=15 \sqrt{2}-25 \sqrt{3} \end{aligned} $

Given that $ A $ is a non-singular matrix, and it satisfies the equation:

$ (A-2I)(A-3I) = 0 $

Since $ A $ is non-singular, it implies that $ |A| \neq 0 $, which means $ A^{-1} $ exists.

Expanding the given equation:

$ A^2 - 2A - 3A + 6I = 0 $

Simplifies to:

$ A^2 - 5A + 6I = 0 $

Multiplying both sides of this equation by $ A^{-1} $ yields:

$ A - 5I + 6A^{-1} = 0 $

Dividing through by 5 gives:

$ \frac{A}{5} + \frac{6}{5} A^{-1} = I $

Given that $A B=$ Scalar matrix

$ =\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] $

where $\lambda=$ scalar

Also, given that $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ and

$ \begin{aligned} & \operatorname{det}(3 A)=27 \\ & \Rightarrow \quad \operatorname{det} A=\frac{27}{9}=3 \end{aligned} $

Now, let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$ \therefore \quad A B=\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] $

$ \begin{aligned} &\begin{aligned} & \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll} a & 2 a+3 b \\ c & 2 c+3 d \end{array}\right]=\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] \end{aligned}\\ &\text { Comparing on both sides, we get }\\ &\begin{aligned} & a=\lambda, c=0,2 a+3 b=0,2 c+3 d=\lambda \\ & \therefore \quad b=-\frac{2 \lambda}{3} \text { and } d=\frac{\lambda}{3} \\ & \text { Thus, } A=\left[\begin{array}{cc} \lambda & -\frac{2 \lambda}{3} \\ 0 & \frac{\lambda^3}{3} \end{array}\right] \\ & \qquad|A|=\left|\begin{array}{cc} \lambda & -2 \lambda \\ 0 & \frac{\lambda}{3} \end{array}\right|=\frac{\lambda^2}{3}=3 \quad \text { (given) } \\ & \Rightarrow \quad \lambda^2=9 \Rightarrow \lambda= \pm 3 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \quad A^2=\left[\begin{array}{cc} \lambda & -\frac{2 \lambda}{3} \\ 0 & \frac{\lambda^2}{3} \end{array}\right]\left[\begin{array}{cc} \lambda & -\frac{2 \lambda}{3} \\ 0 & \frac{\lambda^2}{3} \end{array}\right] \\ & \quad=\left[\begin{array}{cc} \lambda^2 & -\frac{2 \lambda^2}{3}-\frac{2 \lambda^2}{9} \\ 0 & \frac{\lambda^2}{9} \end{array}\right]=\left[\begin{array}{cr} \lambda^2 & -\frac{8 \lambda^2}{9} \\ 0 & \frac{\lambda^2}{9} \end{array}\right] \\ & \quad A^2=\left[\begin{array}{cc} 9 & -8 \\ 0 & 1 \end{array}\right] \\ & \text { Clearly, } A^{-1}=\frac{1}{3}\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] \\ & \Rightarrow \quad 3 A^{-1}=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] \end{aligned}\\ &\text { Thus, }\\ &3 A^{-1}+A^2=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]+\left[\begin{array}{cc} 9 & -8 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 10 & -6 \\ 0 & 4 \end{array}\right] \end{aligned} $

If $ A $ is a symmetric matrix with real entries, it means $ A^T = A $.

Now, suppose the inverse $ A^{-1} $ exists.

For symmetric matrices, the transpose of the inverse is equal to the inverse of the transpose, so:

$ (A^{-1})^T = (A^T)^{-1} = A^{-1} $

This equation shows that $ A^{-1} $ is also symmetric. Therefore, if the inverse of a symmetric matrix exists, it is necessarily symmetric as well.

$ \begin{aligned} &\text { }\left|\begin{array}{ccc} \omega+\omega^2 & \omega^2+\omega^9 & \omega^9+\omega \\ \omega^{27}+\omega^{31} & \omega^{31}+\omega^{17} & \omega^{17}+\omega^{27} \\ \omega^{30}+\omega^{41} & \omega^{41}+\omega^{19} & \omega^{19}+\omega^{30} \end{array}\right|\\ &\begin{aligned} = & \left|\begin{array}{ccc} \omega & \omega^2 & \omega^9 \\ \omega^{27} & \omega^{31} & \omega^{17} \\ \omega^{30} & \omega^{41} & \omega^{19} \end{array}\right|+\left|\begin{array}{ccc} \omega^2 & \omega^9 & \omega \\ \omega^{31} & \omega^{17} & \omega^{27} \\ \omega^{41} & \omega^{19} & \omega^{30} \end{array}\right| \\ = & \omega \times \omega^{17} \times \omega^{19}\left|\begin{array}{ccc} 1 & \omega & \omega^8 \\ \omega^{10} & \omega^{14} & 1 \\ \omega^{11} & \omega^{22} & 1 \end{array}\right| +\omega \times \omega^{17} \times \omega^{19}\left|\begin{array}{ccc} \omega & \omega^8 & 1 \\ \omega^{14} & 1 & \omega^{10} \\ \omega^{22} & 1 & \omega^{11} \end{array}\right| \\ = & \omega^{37}\left|\begin{array}{ccc} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & \omega & 1 \end{array}\right|+\omega^{37}\left|\begin{array}{ccc} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ \omega & 1 & \omega^2 \end{array}\right| \\ = & \omega^{37} \times 0+\omega^{37} \times 0 \\ = & 0+0=0 \end{aligned} \end{aligned} $

Given a non-singular matrix $ P $ such that:

$ I + P + P^2 + \ldots + P^n = 0 $

we aim to determine $ P^{-1} $.

First, recognize the expression as a series whose sum is zero:

$ I + P + P^2 + \ldots + P^n = 0 $

This expression can be seen as a finite geometric series. Now, apply the formula for the sum of a geometric series:

$ S = I \frac{P^{n+1} - I}{P - I} $

Given that the entire series equals zero:

$ \frac{P^{n+1} - I}{P - I} = 0 $

This implies:

$ P^{n+1} - I = 0 $

Thus, we have:

$ P^{n+1} = I $

This equation indicates that $ P $ raised to the power $ n+1 $ results in the identity matrix $ I $. Therefore, multiplying both sides by $ P^{-1} $, we deduce:

$ P^{n+1} \cdot P^{-1} = I \cdot P^{-1} $

Hence:

$ P^n = P^{-1} $

Therefore, the inverse of $ P $ is $ P^n $.

To solve for $|\alpha|$, we begin with the matrix $A$:

$ A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix} $

Given that $\operatorname{det}(A^2) = 25$, we can derive that $(\operatorname{det}(A))^2 = 25$. Hence, $|A| = \pm 5$.

Next, we compute the determinant of $A$:

The matrix $A$ is upper triangular. Thus, the determinant $|A|$ is the product of its diagonal elements:

$ |A| = 5 \times \alpha \times 5 = 25\alpha $

Given that $|A| = \pm 5$, we equate and solve for $\alpha$:

$ 25\alpha = \pm 5 $

This implies that $\alpha = \pm \frac{1}{5}$.

Therefore, the absolute value of $\alpha$ is:

$ |\alpha| = \frac{1}{5} $

$ \begin{aligned} P & =\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \operatorname{Tr}(P)=a+e+i \\ P^T & =\left[\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right] \operatorname{Tr}\left(P^T\right)=a+e+i \end{aligned} $

$ \Rightarrow \operatorname{Tr}(P)=\operatorname{Tr}\left(P^T\right) $

$ \operatorname{Tr}\left(P-P^T\right)+\operatorname{Tr}\left(P+P^\tau\right)+\frac{\operatorname{Tr}(P)}{\operatorname{Tr}\left(P^T\right)}+\operatorname{Tr}(P) $

$ \times \operatorname{Tr}\left(P^T\right)=0 $

$ \begin{aligned} \Rightarrow \operatorname{Tr}(P)-\operatorname{Tr}\left(P^T\right)+\operatorname{Tr}(P) & +\operatorname{Tr}\left(P^T\right)+\frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} \\ & +\operatorname{Tr}(P) \cdot \operatorname{Tr}(P)=0 \end{aligned} $

$ \Rightarrow(\operatorname{Tr}(P))^2+2 \operatorname{Tr}(P)+1=0 $

$(\operatorname{Tr}(P)+1)^2=0$

$\operatorname{Tr}(P)=-1$

To solve the given system of equations using the matrix inversion method, we need to ensure that the coefficient matrix is invertible. This requires that the determinant of the matrix is not equal to zero.

Consider the system of equations:

$ x + k y + 3z = -2 $

$ 4x + 3y + kz = 14 $

$ 2x + y + 2z = 3 $

The coefficient matrix for this system is:

$ \begin{bmatrix} 1 & k & 3 \\ 4 & 3 & k \\ 2 & 1 & 2 \end{bmatrix} $

To check the invertibility, we calculate the determinant of this matrix:

$ \text{Determinant} = 1(6 - k) - 4(2k - 3) + 2(k^2 - 9) $

Simplifying the expression:

$ 1(6 - k) = 6 - k $

$ -4(2k - 3) = -8k + 12 $

$ 2(k^2 - 9) = 2k^2 - 18 $

Combining these results gives:

$ 6 - k - 8k + 12 + 2k^2 - 18 = 2k^2 - 9k $

For the matrix to be invertible, we need the determinant to be non-zero:

$ 2k^2 - 9k \neq 0 $

This expression can be factored as:

$ k(2k - 9) \neq 0 $

Thus, the values of $ k $ for which the system is solvable by matrix inversion are:

$ k \neq 0 $

$ k \neq \frac{9}{2} $

Therefore, $ k $ should not be 0 and not $ \frac{9}{2} $.

$|A|=-21$

$\operatorname{tr}\left(A^3\right)=2024$

$ \begin{aligned} A & =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\ A^2 & =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\ & =\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+d c & b c+d^2 \end{array}\right] \\ A^3 & =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \cdot\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+d c & b c+d^2 \end{array}\right] \end{aligned} $

$ =\left[\begin{array}{c} a^3+2 a b c+b d c \\ a^2 c+a d c+b c^2+d^2 c \end{array}\right. $

$ \left.\begin{array}{c} a^2 b+a b d+b^2 c+b d^2 \\ a b c+2 b c d+d^3 \end{array}\right] $

Given, $|A|=a d-b c=-21$, and $\operatorname{tr}\left(A^3\right)=a^3+d^3+3 a b c+3 b c d=2024$

On putting $b c=a d+21$

$ \begin{aligned} a^3+d^3+3 a(a d+21)+3(a d+ & 21) d =2024 \end{aligned} $

$ \begin{aligned} \Rightarrow a^3+d^3+3 a^2 d+63 a+3 a d^2 & +63 d =2024 \\ \Rightarrow a^3+d^3+3 a^2 d+3 a d^2+63(a+ & d) =2024 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad(a+d)^3+63(a+d)=2024 \\ & \Rightarrow \quad(a+d)\left[(a+d)^2+63\right]=2024 \\ & \therefore \operatorname{tr}(A)=a+d \end{aligned} $

On putting $a+d=6$

$ \begin{aligned} \Rightarrow \mathrm{LHS} & =6[36+63]=6 \times 99 \\ & =594 \neq \mathrm{RHS} \end{aligned} $

On putting $a+d=11$

$ \begin{aligned} \Rightarrow \mathrm{LHS} & =11[121+63] \\ & =11 \times 184=2024=\mathrm{RHS} \end{aligned} $

Therefore, $\operatorname{tr}(A)$ i.e. $a+d$ will be 11.

Given $A=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]$

∴ Given that $A$ is a skew-symmetric matrix.

$ \begin{aligned} & A^T=-A \\ & \Rightarrow \quad\left[\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]=-\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]=\left[\begin{array}{lll} -a & -b & -c \\ -d & -e & -f \\ -g & -h & -i \end{array}\right] \end{aligned} $

On comparing corresponding elements, we get

$ \begin{aligned} a & =-a \\ \Rightarrow \quad 2 a & =0 \Rightarrow a=0 \\ e & =-e \Rightarrow 2 e=0 \Rightarrow e=0 \\ i & =-i \Rightarrow 2 i=0 \Rightarrow i=0 \end{aligned} $

$ \begin{aligned} &\text { Now, } d=-b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ &\begin{aligned} & g=-c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & b=-d \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & h=-f \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & c=-g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & f=-h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{aligned} \end{aligned} $

Eq. (iii) divide by Eq. (v),

$ \begin{aligned} \frac{b}{c} & =\frac{-d}{-g} \\ \Rightarrow \quad \frac{b}{c} & =\frac{-d}{g} \times(-1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii)\end{aligned} $

Eq. (iv) gives $\frac{h}{f}=-1$ and Eq. (vi) gives

$ \frac{f}{h}=-1 $

But for $h=0, \frac{f}{h}$ is not defined.

On putting $(-1)=\frac{h}{f}$ in Eq. (viii),

$ \frac{b}{c}=\frac{-d}{g} \times \frac{h}{f}=\frac{-d h}{g f} $

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 $

Given that,

$ a_{i j}= \begin{cases}1, & \text { if } i+j=7 \\ 0, & \text { if } i+j \neq 7\end{cases} $

Now, $\left(A(\operatorname{adj} A) A^{-1}\right) A^2$

$ \begin{array}{lrr} =\left(A(\operatorname{Adj} A) A^{-1}\right) A^2 & \\ =\left((|A| I) A^{-1}\right) A^2 & {[\because A \operatorname{adj} A=|A| I]} \\ =-1\left(A^{-1} A\right) A & {[\because|A|=-1]} \\ =-I A & {\left[\because A^{-1} A=I\right]} \\ =-A & \end{array} $

Given system of equations has a non-trivial solution

$ \begin{aligned} & \Pi_1 \equiv x+a y+a z=0 \\ & \Pi_2 \equiv b x+y+b z=0 \\ & \Pi_3 \equiv c x+c y+z=0 \end{aligned} $

It can be written as $A X=0$

Where, $A=\left[\begin{array}{lll}1 & a & a \\ b & 1 & b \\ c & c & 1\end{array}\right]$

So, the system has non-trivial solutions hence it has infinitely many solutions or unique solutions.

$ \therefore \quad|A|=0 $

Now, let the system has infinitely many solutions

So, let $y=t ; t \in \mathbf{R}$

Then from Eqs. (i) and (ii), we get

$ x+a z=-a t \Rightarrow b x+b z=-t $

Solving these, we get

$ z=\frac{t(1-a b)}{b(a-1)}, x=\frac{a}{b} t \frac{(1-b)}{(1-a)} $

Hence, if $a \neq b \neq c$, then

$ \begin{aligned} & x=\frac{a}{b} t \frac{(1-b)}{(1-a)}, y=t \\ & z=\frac{t}{b} \frac{(1-a b)}{(a-1)}, t \in \mathbf{R} \end{aligned} $

If $a=b=c$

Let $y=t_1, z=t_2, t_1, t_2 \in \mathbf{R}$

Then by Eq. (i)

$ x=-a\left(t_1+t_2\right) $

So, the system of equations has unique solution.

Given that, Rank of matrix $A=$ rank of matrix $B$

$ f(A)=f(B) \text { and } f(B)=3 $

∴ Order of matrix $B \geq$ Rank of matrix $B$.