Matrices and Determinants

We are given:

• $B=(I+A)^{-1} \Longrightarrow B^{-1}=I+A$

• $A+C=I \Longrightarrow A=I-C$

Substitute $A$ into the first equation:

$ B^{-1}=I+(I-C) $

$ B^{-1}=2 I-C........\text { (1) } $

pre-multiply $B: B \cdot B^{-1}=B(2 I-C) I=2 B-B C \Longrightarrow B C=2 B-I$ ........ (2)

post multiply in eqn (1): $B^{-1} B=(2 I-C) \cdot B I=2 B-C B \Longrightarrow C B=2 B-I$ .......(3)

from (2) & (3), $B C=C B$

3. Set up the System of Equations Since $B C=C B$, we use the given matrix for $B C$ :

$ C B=\left[\begin{array}{cc} 1 & -5 \\ -1 & 2 \end{array}\right] $

Now, substitute this into the vector equation :

$ \left[\begin{array}{cc} 1 & -5 \\ -1 & 2 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \end{array}\right]=\left[\begin{array}{c} 12 \\ -6 \end{array}\right] $

This results in two linear equations :

• $x_1-5 x_2=12$

• $-x_1+2 x_2=-6$

1. Solve for $x_1$ and $x_2$
   
   Adding the two equations :

$ \left(x_1-x_1\right)+\left(-5 x_2+2 x_2\right)=12-6 $

$\Rightarrow $ $x_2=-2$

Substitute $x_2=-2$ into the second equation :

$ \begin{gathered} -x_1+2(-2)=-6 \\ -x_1-4=-6 \Longrightarrow-x_1=-2 \Longrightarrow \mathbf{x}_1=\mathbf{2} \end{gathered} $

Final result $x_1+x_2=2+(-2)=\mathbf{0}$

Given two $3 \times 3$ matrices $P=\left[p_{i j}\right], Q=\left[q_{i j}\right]$ are given

$ q_{i j}=2^{(i+j-1)} p_{i j}, \operatorname{det}(Q)=2^{10} . $

To calculate $\operatorname{det}(\operatorname{adj}(\operatorname{adj} P))$

We know $\operatorname{det}(\operatorname{adj}(\operatorname{adj} P))=(\operatorname{det}(P))^{(n-1)^2}$

where $n$ is order of matrix which is 3

so, $\operatorname{det}(\operatorname{adj}(\operatorname{adj} P))=(\operatorname{det}(P))^4-(1)$

Let P matrix

$P=\left[\begin{array}{lll}p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33}\end{array}\right]$

and $Q=\left[\begin{array}{lll}q_{11} & q_{12} & q_{13} \\ q_{21} & q_{22} & q_{23} \\ q_{31} & q_{32} & q_{33}\end{array}\right]$

use $q_{i j}=2^{i+j-1} p_{i j}$ to find Q matrix

$ \text { e.g. } q_{11}=2^{1+1-1} p_{11}=2 p_{11}, q_{23}=2^{2+3-1} p_{23}=2^4 p_{23} $

$ Q=\left[\begin{array}{ccc} 2 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{array}\right] $

$\begin{aligned} & Q=\left[\begin{array}{ccc}2 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33}\end{array}\right] \\ & \operatorname{det}(Q)=\left|\begin{array}{ccc}2 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33}\end{array}\right|=2^{10}\end{aligned}$

taking common 2 from $R_1, 2^2$ from $R_2 \& 2^3$ from $R_3$

$ \operatorname{det}(Q)=2 \cdot 2^2 \cdot 2^3\left|\begin{array}{lll} p_{11} & 2 p_{12} & 2^2 p_{13} \\ p_{21} & 2 p_{22} & 2^2 p_{23} \\ p_{31} & 2 p_{32} & 2^2 p_{33} \end{array}\right|=2^{10} $

taking common 2 from $C_2$ and $2^2$ from $C_3$

$ 2 \cdot 2^2 \cdot 2^3 \cdot 2 \cdot 2^2\left|\begin{array}{lll} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{array}\right|=2^{10} $

$\begin{aligned} & \Rightarrow 2^9 \cdot \operatorname{det}(P)=2^{10} \\ & \Rightarrow \operatorname{det}(P)=2 \\ & \text { from }(1) \operatorname{det}(\operatorname{adj}(\operatorname{adj} P))=(\operatorname{det}(P))^4=2^4=16\end{aligned}$

$ \begin{aligned} & \mathrm{f}(\mathrm{x})=\int \frac{7 x^{10}+9 x^8}{\left(1+x^2+2 x^9\right)^2} d x \\ & =\int \frac{7 x^{10}+9 x^8}{x^{18}\left(\frac{1}{x^9}+\frac{1}{x^7}+2\right)^2}=\int \frac{\frac{7}{x^8}+\frac{9}{x^{10}}}{\left(\frac{1}{x^9}+\frac{1}{x^7}+2\right)^2} \end{aligned} $

substitute $\frac{1}{x^9}+\frac{1}{x^7}+2=t$

$ \begin{aligned} & \left(\frac{-9}{x^{10}}-\frac{7}{x^8}\right) d x=d t \Rightarrow\left(\frac{7}{x^8}+\frac{9}{x^{10}}\right) d x=-d t \\ & f(t)=-\int \frac{d t}{t^2}=\frac{1}{t}+C \end{aligned} $

put value of t

$ \begin{aligned} &\begin{aligned} & f(x)=\frac{1}{\frac{1}{x^9}+\frac{1}{x^7}+2}+C \\ & f(x)=\frac{x^9}{2 x^9+x^2+1}+C \end{aligned}\\ &\text { It is given that } \lim _{x \rightarrow 0} f(x)=0\\ &\begin{aligned} & \lim _{x \rightarrow 0} \frac{x^9}{2 x^9+x^2+1}+C=0 \\ & 0+C=0 \Rightarrow C=0 \\ & \text { so } f(x)=\frac{x^9}{2 x^9+x^2+1} \\ & f^{\prime}(x)=\frac{9 x^8\left(2 x^9+x^2+1\right)-x^9\left(18 x^8+2 x\right)}{\left(2 x^9+x^2+1\right)^2} \\ & f^{\prime}(1)=\frac{9(1)^8\left(2(1)^9+(1)^2+1\right)-(1)^9\left(18(1)^8+2(1)\right)}{\left(2(1)^9+(1)^2+1\right)^2} \\ & =\frac{9 \times 4-20}{16}=\frac{16}{16}=1 \end{aligned} \end{aligned} $

It is given that $\mathrm{A}=\left[\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^2 & 4 & 1\end{array}\right]$

$ \begin{aligned} & \text { substitute } f^{\prime}(1)=1 \Rightarrow A=\left[\begin{array}{ccc} 0 & 0 & 1 \\ \frac{1}{4} & 1 & 1 \\ \alpha^2 & 4 & 1 \end{array}\right] \\ & |A|=1\left(4 \times \frac{1}{4}-\alpha^2\right)=1-\alpha^2 \end{aligned} $

It is given that $\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A}) \&|B|=81$

$|B|=|A|^{(n-1)^2}, \mathrm{n}=3$ as A is $3 \times 3$ matrix

$ \begin{aligned} & |B|=|A|^{(3-1)^2}=81 \\ & |A|^4=81=3^4 \\ & |A|= \pm 3 \end{aligned} $

$1-\alpha^2= \pm 3\left\{|A|=1-\alpha^2\right\}$

$ 1-\alpha^2=-3 \text { or } 1-\alpha^2=3 $

$\alpha^2=4$ or $\alpha^2=-2$ not possible

so value of $\alpha^2$ is 4

For the system to have infinitely many solutions, the determinant of the coefficient matrix ( $\Delta$ ) must be zero.

$ \begin{aligned} & \Delta=\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{array}\right|=0 \\ & 1(15-2 a)-1(6-a)+1(4-5)=0 \\ & 15-2 a-6+a-1=0 \\ & 8-a=0 \Rightarrow a=8 \end{aligned} $

Now, for infinitely many solutions, $\Delta_x=\Delta_y=\Delta_z=0$.

Check $\Delta_x$ :

$ \begin{aligned} & \Delta_x=\left|\begin{array}{ccc} 6 & 1 & 1 \\ 36 & 5 & 8 \\ b & 2 & 3 \end{array}\right|=0 \\ & 6(15-16)-1(108-8 b)+1(72-5 b)=0 \\ & 6(-1)-108+8 b+72-5 b=0 \\ & -6-108+72+3 b=0 \\ & -42+3 b=0 \Rightarrow 3 b=42 \Rightarrow b=14 \\ & \Delta_y=\left|\begin{array}{ccc} 1 & 6 & 1 \\ 2 & 36 & 8 \\ 1 & 14 & 3 \end{array}\right|=1(108-112)-6(6-8)+1(28-36)=-4+12-8=0 \\ & \Delta_z=\left|\begin{array}{ccc} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & 14 \end{array}\right|=1(70-72)-1(28-36)+6(4-5)=-2+8-6=0 \end{aligned} $

If $a=8$ and $b \neq 14$, the system has no solution.

If $a=8$ and $b=14$, the system has infinitely many solutions.

If $a \neq 8$, the system has a unique solution regardless of $b$.

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{array}\right|=1\left(\begin{array}{c} \left.1-\cos ^2 \gamma\right)-\cos \alpha(\cos \alpha-\cos \beta \cos \gamma) +\cos \beta(\cos \alpha \cos \gamma-\cos \beta) \end{array}\right. \\ & =1-\cos ^2 \gamma-\cos ^2 \alpha-\cos ^2 \beta+2 \cos \beta+2+\cos \alpha \cos \beta \cos \gamma \\ & \left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|=-\cos \alpha(-\cos \beta \cos \gamma)+\cos \beta(\cos \alpha \cos \gamma) =2 \cos \alpha \cos \beta-\cos \gamma \\ & \Rightarrow \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \because q=\left|\begin{array}{ccc} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{array}\right|=-10-2=-12 \\ & \text { And }-p+q=\left|\begin{array}{ccc} 0 & 0 & -3 \\ -2 & -3 & -6 \\ 2 & -3 & -3 \end{array}\right|=-36 \\ & \Rightarrow p=q+36=24 \\ & \Rightarrow p^2 \neq 196 q^2 \end{aligned}\\ &\text { Option (4) is correct. } \end{aligned} $

$ \begin{aligned} & x-n y+z=6 \\ & x-(n-2) y+(n+1) z=8 \\ & (n-1) y+z=1 \end{aligned} $

$ \left|\begin{array}{ccc} 1 & -n & 1 \\ 1 & (n-2) & n+1 \\ 0 & n-1 & 1 \end{array}\right|=0 \Rightarrow n=1,2 \text { or } n=-1 \text { (rejected) } $

For unique solution $\mathrm{n}=3,4,5,6$

Now P (Probability when system of equations has unique solution) $=\frac{4}{6}$

So $\mathrm{k}=4$

Now required sum $=4+(3+4+5+6)=22$

$ \begin{aligned} X & =A^{-1} B=\left(\frac{\operatorname{adj} A}{|A|}\right) B= \pm \frac{1}{10}\left(\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right)\left(\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right)= \pm \frac{1}{10}\left(\begin{array}{l} 20 \\ -10 \\ 10 \end{array}\right)= \pm\left(\begin{array}{l} 2 \\ -1 \\ 1 \end{array}\right) \\ & \therefore|x+y+z|=2 \end{aligned} $

$ \begin{aligned} &A=\left(\begin{array}{ll} 2 & 3 \\ 3 & 5 \end{array}\right),|A|=1\\ &\left|A^{2025}-3 A^{2024}+A^{2023}\right|=\left|A^{2023}\right|\left|A^2-3 A+I\right|=16 \end{aligned} $

$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{array}\right| \\ & \operatorname{det}(\Delta)=0 \\ & \alpha=19 \end{aligned} $

$ \begin{aligned} &A=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right], B=\left[\begin{array}{ll} 23 & 49 \\ 45 & 21 \end{array}\right]\\ &\text { Characteristic equation of A, }\\ &\begin{aligned} & (3-\lambda)(-1-\lambda)+4=0 \\ & (\lambda-3)(\lambda+1)+4=\lambda^2-2 \lambda+=0 \\ & \Rightarrow A^2-2 A+I=0 \\ & A^2=2 A-I, A^3=2 A^2-A=2(2 A-I)-A=3 A-2 I \\ & A^4=4 A^2+I-4 A=4(2 A-I)-4 A+I \\ & =4(2 A-I)-3 A=5 A-4 I \\ & \left(A^5\right)^3=(5 A-4 I)^3 \\ & =125 A^3-3 \times 25 A^2(4)+3(5 A)\left(4^2\right)-64 I \\ & =125(3 A-2 I)-300(2 A-I)+240 A-64 I \\ & =A(375-600+240)+I(-250+300-64) \\ & =15 A-14 I \\ & \Rightarrow A^{15}+B=15 A-14 I+B \\ & =\left[\begin{array}{ll} 2 & -11 \\ 1 & -11 \end{array}\right] \quad\left(A^{15}+B\right)\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \Rightarrow 2 x=11 y \end{aligned} \end{aligned} $

Let us assume that the new matrix we get after performing elementary row operations on the identity matrix is $A$.

This means,

$A = E_n E_{n-1} \ldots E_1 I$

Here, each $E_i$ represents an elementary matrix (a matrix obtained by performing one elementary row operation on the identity matrix).

Now, the determinant of $A$ can be written as:

$\det(A) = \prod\limits_{k=1}^{n} \det(E_k)$

For an elementary matrix, $\det(E_k)$ can be either $1$, $-1$, or any non-zero number $k$. Therefore,

$\det(A) \ne 0$

This tells us that any matrix obtained from the identity matrix through elementary row transformations must have a non-zero determinant.

Now, let’s check the determinant of each given option:

(A) $\det\left(\begin{bmatrix}1 & 1 & 1\\1 & 1 & 1\\1 & 1 & 1\end{bmatrix}\right) = 0$

(B) $\det\left(\begin{bmatrix}1 & 1 & 1\\2 & 3 & 4\\1 & 2 & 1\end{bmatrix}\right) = -2 \ne 0$

(C) $\det\left(\begin{bmatrix}1 & 1 & 1\\2 & 3 & 4\\2 & 5 & 8\end{bmatrix}\right) = 0$

(D) $\det\left(\begin{bmatrix}1 & 1 & 1\\-1 & 1 & 2\\0 & 2 & 3\end{bmatrix}\right) = 0$

Since only option (B) has a non-zero determinant, this means that it can be obtained from the identity matrix through elementary row operations.

The steps to obtain matrix (B) from the $3 \times 3$ identity matrix are shown below:

Start with:

$I = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$

Perform $R_2 \rightarrow R_2 + 2R_3$ :

$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}$

Next, $R_1 \rightarrow R_1 - R_3$ :

$\begin{bmatrix}1 & 0 & -1\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}$

Then, $R_3 \rightarrow -2R_3$ :

$\begin{bmatrix}1 & 0 & -1\\0 & 1 & 2\\0 & 0 & -2\end{bmatrix}$

Now, $R_3 \rightarrow R_3 + R_2$ :

$\begin{bmatrix}1 & 0 & -1\\0 & 1 & 2\\0 & 1 & 0\end{bmatrix}$

Next, $R_1 \rightarrow R_1 + R_2$ :

$\begin{bmatrix}1 & 1 & 1\\0 & 1 & 2\\0 & 1 & 0\end{bmatrix}$

Then, $R_3 \rightarrow R_3 + R_1$ :

$\begin{bmatrix}1 & 1 & 1\\0 & 1 & 2\\1 & 2 & 1\end{bmatrix}$

Finally, $R_2 \rightarrow R_2 + 2R_1$ :

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

Hence, option (B) can indeed be obtained by performing elementary row transformations on the $3 \times 3$ identity matrix.

We can write the matrix as

$ M = \begin{bmatrix}2 & -1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} $

So, let $ B = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} $. Then $ M = B + I $.

Find $ B^2 $

$ B^2 = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} $

This means $ B $ is a nilpotent matrix, i.e., its square is a zero matrix.

Expand $ M^{26} $ using the Binomial Theorem

$ M^{26} = (B + I)^{26} = {^{26}C_0} I + {^{26}C_1} B + {^{26}C_2} B^2 + \ldots + {^{26}C_{26}} B^{26} $

As $ B^2 = 0 $, all higher powers of $ B $ are zero. So,

$ M^{26} = I + 26B $

Substitute the values

$ M^{26} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + 26 \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} = \begin{bmatrix}27 & -26 \\ 26 & -25\end{bmatrix} $

So, $ p = 27, \, q = -26, \, r = 26, \, s = -25. $

Find the sum $ D = \sum_{k=1}^{26} M^k $

$ D = M + M^2 + M^3 + \ldots + M^{26} $

Substitute $ M = B + I $:

$ D = (B + I) + (B + I)^2 + (B + I)^3 + \ldots + (B + I)^{26} $

Since $ (B + I)^k = I + kB $, we have

$ D = (I + B) + (I + 2B) + (I + 3B) + \ldots + (I + 26B) $

Combine like terms:

$ D = 26I + B(1 + 2 + 3 + \ldots + 26) $

The sum $ 1 + 2 + 3 + \ldots + 26 = 351 $, so

$ D = 26I + 351B $

Substitute the matrices:

$ \sum_{k=1}^{26} M^k = \begin{bmatrix}377 & -351 \\ 351 & -325\end{bmatrix} = \begin{bmatrix}a & b \\ c & d\end{bmatrix} $

Hence, $ a = 377, \, b = -351, \, c = 351, \, d = -325. $

So, option (B) is incorrect.

Check option (C)

Given system of equations:

$ 27x - 26y = m $

$ 26x - 25y = n $

Determinant of coefficients:

$ \left|\begin{array}{rr} 27 & -26 \\ 26 & -25 \end{array}\right| \neq 0 $

Since the determinant is non-zero, a unique solution exists.

On solving, we get:

$ x = -24n - 25m, \quad y = -23n - 26m $

Both $ x $ and $ y $ are integers. Hence, option (C) is correct.

Check option (D)

We have:

$ \left|\begin{array}{cc} a + t & b \\ c & d + t \end{array}\right| = ad + at + dt + t^2 - bc $

Substitute $ a = 377, b = -351, c = 351, d = -325 $:

$ = (377)(-325) + t(52) + t^2 + (351)(351) = t^2 + 52t + 676 = (t + 26)^2 $

Since $ (t + 26)^2 \neq 0 $ for all $ t > 0 $, the determinant is non-zero, so a unique solution exists. Option (D) is correct.

Check option (A)

Let $ N = \begin{bmatrix}j & k \\ l & m\end{bmatrix} $.

Compute $ MN $:

$ MN = \begin{bmatrix}2 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}j & k \\ l & m\end{bmatrix} = \begin{bmatrix}2j - l & 2k - m \\ j & k\end{bmatrix} $

Compute $ N\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} $:

$ N\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}j & j + k \\ l & l + m\end{bmatrix} $

Comparing both, we get $ j = l $ and $ 2k - m = j + k $ ⟹ $ m - k = j $.

Determinant of $ N $:

$ |N| = jm - lk = j(m - k) $

This is not always zero if $ j \neq 0 $ and $ m \neq k $, which means $ N $ can be invertible. Hence, option (A) is correct.

Let $ \alpha $ be a solution of the equation $ x^2 + x + 1 = 0 $. This implies that $ \alpha $ is a cube root of unity, denoted as $ \omega $, where $ \alpha^2 + \alpha + 1 = 0 $.

We are given matrix equations with parameters $ a $ and $ b $ as follows:

$ \begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} $

Solving these equations:

$ \begin{aligned} & \left[4 - a - 2b, 64 - a - 14b, 52 + 2a - 8b\right] = \left[0, 0, 0\right] \\ & \Rightarrow a + 2b = 4 \\ & \Rightarrow a + 14b = 64 \end{aligned} $

From these, solve for $ a $ and $ b $:

$ \begin{aligned} & 12b = 60 \Rightarrow b = 5 \\ & a = -6 \end{aligned} $

We now substitute $ a $ and $ b $ into the given function:

$ \frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3 $

We substitute $ \alpha = \omega $ and simplify:

$ \begin{aligned} & \frac{4}{\omega} + \frac{m}{1} + \frac{n}{\omega^2} = 3 \\ & \Rightarrow 4\omega^2 + m + n\omega = 3 \end{aligned} $

Substitute $ \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i $ and $ \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i $:

$ \begin{aligned} & 4\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) + m + n\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 3 \\ & \Rightarrow -2 + m - \frac{n}{2} = 3 \\ & \text{and } \frac{-4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0 \end{aligned} $

From the above, solve for $ n $ and $ m $:

$ \begin{aligned} & n = 4 \\ & m = 7 \end{aligned} $

Thus, $ m + n = 11 $.

$|A|=\left|\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right|$

$\begin{aligned} &\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}\\ &\text { Then } \mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)\\ &\Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 2 & 2+\mathrm{p} \\ 6 & 6 & 8+3 \mathrm{p} \end{array}\right| \end{aligned}$

$\begin{aligned} & \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \\ & |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \\ & =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=24 \end{aligned}$

$\begin{aligned} &\text { For infinitely many solution }\\ &\begin{aligned} & \Delta=0 \\ & \left|\begin{array}{ccc} 1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & \lambda \end{array}\right|=0 \\ & \Rightarrow 1(3 \lambda+3)-5(4 \lambda+72)-1(4-72)=0 \\ & \Rightarrow-17 \lambda+3-4 \times 72-4=0 \\ & \Rightarrow 17 \lambda=-289 \\ & \Rightarrow \lambda=-17 \\ & \Delta 1=0 \end{aligned} \end{aligned}$

$\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 1 & 5 & -1 \\ 7 & 3 & -3 \\ \mu & 1 & -17 \end{array}\right|=0 \\ & \Rightarrow 1(-51+3)-5(-119+3 \mu)-1(7-3 \mu)=0 \\ & \Rightarrow-48+595-15 \mu-7+3 \mu=0 \\ & \Rightarrow 12 \mu=540 \\ & \mu=45 \\ & x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y-17 z=45 \\ & \text { Let } z=1 \\ & x+5 y=1+\lambda] \times 4 \\ & 4 x+3 y=7+3 \lambda \end{aligned}$

$\frac{\underset{-}{4 \mathrm{x}}+20 \mathrm{y}=\underset{-}{4+4 \lambda}}{-17 \mathrm{y}=3-\lambda}$

$\begin{aligned} \begin{aligned} \mathrm{y} & =\frac{\lambda-3}{17}, \mathrm{x}=1+\lambda-\frac{5 \lambda-15}{17} \\ & =\frac{32-12 \lambda}{17} \\ 7 & \leq \frac{\lambda-3}{17}+\frac{32+12 \lambda}{17}+\lambda \leq 77 \\ 7 & \leq \frac{30 \lambda+29}{17} \leq 77 \\ 3 & \leq \lambda \leq 42 \\ \lambda & =3,20,37 \end{aligned} \end{aligned}$

$\begin{aligned} & |\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))|=81 \\ & =|A|^{(n-1)^3}=(3)^4 \Rightarrow|A|^8=3^4 \Rightarrow|A|=3^{1 / 2} \\ & |\operatorname{adj}(\operatorname{adj} A)|^{\frac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4\right)} \\ & {\left[|A|^{(n-1)^2}\right]^{\frac{(n-1)^2}{2}}=|A|^{3 n^2-5 n-4}} \\ & |A|^{2(n-1)^2}=|A|^{3 n^2-5 n-4} \\ & \Rightarrow 2(n-1)^2=3 n^2-5 n-4 \\ & \quad n^2-n-6=0 \\ & \Rightarrow n=-2,3 \\ & \sum_{x \leftarrow 5}\left|A^{n^2+n}\right|=\left|A^2\right|+\left|A^{12}\right| \\ & =3+3^6=732 \end{aligned}$

$\begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{array}\right| \\ & =2(-12-3 \lambda+6)-3(-28-7 \lambda+24)+5(21-36) \\ & =-12-6 \lambda+12+21 \lambda-75 \\ & =15 \lambda-75 \\ & \Rightarrow 15 \lambda-75=0 \\ & \Rightarrow \lambda=5 \\ & \Delta_1=\left|\begin{array}{ccc} 9 & 3 & 5 \\ 8 & 3 & -2 \\ 16-\mu & 3 & -9 \end{array}\right| \\ & =9(-27+6)-3(-72+32-2 \mu)+5(24-48+3 \mu) \\ & =-189+120+6 \mu-120+15 \mu \\ & =21 \mu-189=0 \\ & \Rightarrow \mu=9 \end{aligned}$

$\begin{aligned} & \therefore r=\left|\frac{4(5)-3(9)}{\sqrt{(4)^2+(3)^2}}\right| \\ & r=\frac{7}{5} \end{aligned}$

$\begin{aligned} &\begin{aligned} & A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \\ & A^2=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] \\ & A^3=A+A^2-I \\ & A^3=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \\ & A^4=A^2+A^2-I=2 A^2-I \\ & A^4=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] \text { and } A^5=\left[\begin{array}{lll} 1 & 0 & 0 \\ 3 & 0 & 1 \\ 2 & 1 & 0 \end{array}\right] \\ & A^{50}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{array}\right] \end{aligned}\\ &\text { Sum of elements }=53 \end{aligned}$

To find the expression $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|$, we break it down as follows:

Recognize that:

$ |2 \operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |3A (\operatorname{adj}(2A))|^2 $

Apply properties of determinants:

$ = 2^3 (3^3)^2 |A|^2 \left|\operatorname{adj}(2A)\right|^2 $

Further simplify using $|\operatorname{adj}(B)| = |B|^{n-1}$ for a $3 \times 3$ matrix:

$ = 2^3 \cdot 3^6 \cdot 5^2 \cdot (|2A|^2)^2 $

Simplify $|2A|$:

$ = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot |A|^4 $

Continue to simplify:

$ = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot 5^4 $

Expand and combine powers:

$ = 2^{15} \cdot 3^6 \cdot 5^6 $

Therefore, $\alpha = 15$, $\beta = 6$, and $\gamma = 6$. So, $\alpha + \beta + \gamma = 15 + 6 + 6 = 27$.

$\begin{aligned} &\begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned}\\ &\text { For infinite solutions } \Rightarrow \Delta=0=\Delta_1=\Delta_2=\Delta_3 \end{aligned}$

$\begin{aligned} & \Delta=\left|\begin{array}{lll} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{array}\right|=0 \\ & \Rightarrow-4 \lambda-3 \lambda \mu+4 \mu+31=0 \\ & \Delta_1=\left|\begin{array}{ccc} 5 & \lambda & 3 \\ 7 & 2 & -1 \\ 9 & 5 & \mu \end{array}\right|=0 \Rightarrow-9 \lambda-7 \lambda \mu+10 \mu+76=0 \\ & \Delta_2=\left|\begin{array}{ccc} 2 & 3 & 5 \\ 3 & -1 & 7 \\ 4 & \mu & 9 \end{array}\right|=0 \Rightarrow \mu+5=0 \Rightarrow \mu=-5 \\ & \Delta_3=\left|\begin{array}{lll} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{array}\right|=0 \Rightarrow \lambda+1=0 \Rightarrow \lambda=-1 \end{aligned}$

$\begin{aligned} &\therefore \text { For infinite solution } \mu=-5 \text { and } \lambda=-1\\ &\begin{aligned} \text { Now } \mu^2+\lambda^2 & =25+1 \\ & =26 \end{aligned} \end{aligned}$

$\begin{aligned} & A^2(A-2 I)-4(A-I)=0 \\ & A^3-2 A^2-4 A+4 I=0 \end{aligned}$

Multiply by $A$

$\begin{aligned} & A^4=2 A^3+4 A^2-4 A \\ & A^4=2\left(2 A^2+4 A-4 I\right)+4 A^2-4 A \\ & A^4=8 A^2+4 A-8 I \end{aligned}$

Multiply again by $A$

$\begin{aligned} & \Rightarrow A^5=8 A^3+4 A^2-8 A \\ & \Rightarrow A^5=8\left(2 A^2+4 A-4 I\right)+4 A^2-8 A \\ & \Rightarrow A^5=20 A^2+24 A-32 I \end{aligned}$

Comparing with $A^5=\alpha A^2+\beta A+\gamma I$

$\begin{aligned} & \alpha=20, \beta=24, \gamma=-32 \\ & \therefore \alpha+\beta+\gamma=20+24-32 \\ & =44-32 \\ & =12 \end{aligned}$

Let $|A|=0 \Rightarrow \alpha \beta-(-6)=0 \Rightarrow \alpha \beta=-6$ and $\alpha+\beta=1 \Rightarrow \alpha \beta$ are roots of the equation

$\begin{aligned} & x^2-x-6=0 \Rightarrow x=3,-2 . \text { Since } \alpha>0 \\ & \Rightarrow \quad \alpha=3, \beta=-2 \\ & \Rightarrow \quad A=\left[\begin{array}{ll} 3 & -1 \\ 6 & -2 \end{array}\right] \Rightarrow I+A=\left[\begin{array}{ll} 4 & -1 \\ 6 & -1 \end{array}\right] \\ & (I+A)^2=\left[\begin{array}{ll} 10 & -3 \\ 18 & -5 \end{array}\right] \Rightarrow(I+A)^4=\left[\begin{array}{ll} 46 & -15 \\ 90 & -29 \end{array}\right] \\ & \Rightarrow \quad(I+A)^8=\left[\begin{array}{cc} 766 & -255 \\ 1530 & -509 \end{array}\right] \end{aligned}$

Given

$A+I=\begin{bmatrix}1 & a & 1\\ 2 & 1 & 0\\ a & 1 & 2\end{bmatrix}$

So,

$A=\begin{bmatrix}1-1 & a & 1\\ 2 & 1-1 & 0\\ a & 1 & 2-1\end{bmatrix} =\begin{bmatrix}0 & a & 1\\ 2 & 0 & 0\\ a & 1 & 1\end{bmatrix}$

1) Find $a$ using $\det(A)=-4$

Compute $\det(A)$ by expanding along the first row:

$ \det(A)=0\cdot C_{11}+a\cdot C_{12}+1\cdot C_{13} $

Now,

$ C_{12}=(-1)^{1+2}\begin{vmatrix}2&0\\ a&1\end{vmatrix} =-\,(2\cdot 1-0\cdot a)=-2 $

$ C_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\ a&1\end{vmatrix} =+\,(2\cdot 1-0\cdot a)=2 $

So,

$ \det(A)=a(-2)+1(2)=2-2a=2(1-a) $

Given $\det(A)=-4$:

$ 2(1-a)=-4 \implies 1-a=-2 \implies a=3 $

2) Evaluate $\det\left((a+1)\,\operatorname{adj}((a-1)A)\right)$

Use properties for a $3\times 3$ matrix:

• $\det(kM)=k^3\det(M)$

• $\det(\operatorname{adj}(M))=(\det M)^{3-1}=(\det M)^2$

• $\det((a-1)A)=(a-1)^3\det(A)$

Let $B=(a-1)A$. Then

$ \det\left((a+1)\operatorname{adj}(B)\right)=(a+1)^3\det(\operatorname{adj}(B)) =(a+1)^3(\det B)^2 $

Now,

$ \det B=((a-1)^3)(\det A)=(a-1)^3(-4) $

So,

$ \det\left((a+1)\operatorname{adj}((a-1)A)\right) =(a+1)^3\left[(-4)(a-1)^3\right]^2 =(a+1)^3\cdot 16\cdot (a-1)^6 $

Put $a=3$:

$ (a+1)^3=4^3=2^6,\quad (a-1)^6=2^6,\quad 16=2^4 $

Hence,

$ \text{Determinant}=2^6\cdot 2^4\cdot 2^6=2^{16}=2^m3^n $

So $m=16,\; n=0 \implies m+n=16$.

Answer: 16 (Option D)

$\begin{aligned} &\begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned}\\ &\text { has infinite solution }\\ &\begin{aligned} & \Rightarrow \Delta=0, \Delta_1=\Delta_2=\Delta_3 \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{array}\right|=0 \\ & \Delta_2=0 \Rightarrow\left|\begin{array}{ccc} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{array}\right|=0 \\ & \Rightarrow 3(-3+4)-3(2+1)+\beta(8+3)=0 \\ & \Rightarrow 3-9+11 \beta=0 \\ & \Rightarrow \quad \beta=\frac{6}{11} \\ & \Delta_3=0 \Rightarrow\left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{array}\right|=0 \\ & \Rightarrow \quad 3(4 \alpha+6)-1(8+3)+3(4-\alpha)=0 \\ & \quad 12 \alpha+18-11+12-3 \alpha=0 \\ & \quad 9 \alpha=-19 \\ & \quad \alpha=\frac{-19}{9} \end{aligned} \end{aligned}$

$\therefore \quad 22 \beta-9 \alpha=31$

$\begin{aligned} & |A|=\left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right| \\ & =a_{11} a_{22}-a_{21} a_{12} \\ & =\{-1,0,1\} \end{aligned}$

$\begin{array}{c|c|c|c} \mathrm{x} & \mathrm{P}_{\mathrm{i}} & \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}} & \mathrm{P}_1 \mathrm{X}_{\mathrm{i}}{ }^2 \\ -1 & \frac{3}{16} & -\frac{3}{16} & \frac{3}{16} \\ 0 & \frac{10}{16} & 0 & 0 \\ 1 & \frac{3}{16} & \frac{3}{16} & \frac{3}{16} \\ \hline & & \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=0 & \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}{ }^2=\frac{3}{8} \end{array}$

$\begin{aligned} & \therefore \operatorname{var}(\mathrm{x})=\sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}^2-\left(\sum \mathrm{P}_{\mathrm{i}} X_{\mathrm{i}}\right)^2 \\ & =\frac{3}{8}-0=\frac{3}{8} \end{aligned}$

$\begin{aligned} &\begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{array}\right|=1(20-16)-1(10-4)+1(4-2) \\ & =4-6+2=0 \end{aligned}\\ &\text { For infinite solutions }\\ &\begin{aligned} & \Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \\ & \mathrm{~m}^2-3 \mathrm{x}+2=0 \\ & \mathrm{~m}=1,2 \\ & \alpha=1, \beta=2 \\ & \therefore \sum_{\mathrm{n}=1}^{10}\left(\mathrm{n}^\alpha+\mathrm{n}^\beta\right)=\sum_{\mathrm{n}=1}^{10} \mathrm{n}^1+\sum_{\mathrm{n}=1}^{10} \mathrm{n}^2 \\ &= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \\ &= 55+385 \\ &= 440 \end{aligned} \end{aligned}$

$\begin{aligned} & A=\left[\begin{array}{lll} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{array}\right] \\ & A=\left[\begin{array}{ccc} 2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8 \end{array}\right] \\ & A^2=2^2\left[\begin{array}{ccc} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{array}\right] \end{aligned}$

$\begin{aligned} &=4\left[\begin{array}{ccc} - & - & - \\ - & - & - \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16) \end{array}\right]\\ &\text { Sum of elements of } 3^{\text {rd }} \text { row }=4(14+14 \sqrt{2}+28)\\ &\begin{aligned} & =4(42+14 \sqrt{2}) \\ & =168+56 \sqrt{2} \\ & \alpha+\beta \sqrt{2} \\ \therefore \quad & \alpha+\beta=168+56=224 \end{aligned} \end{aligned}$

To solve the problem, we need to determine the determinant of matrix $ A $:

$ A = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix} $

The determinant of $ A $, denoted as $|A|$, is calculated as:

$ |A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8) $

Evaluating each component:

$\log_5 128$ can be simplified using change of base formula:

$\log_5 128 = \frac{\log_{10}128}{\log_{10}5}$

$\log_4 5$ using change of base:

$\log_4 5 = \frac{\log_{10}5}{\log_{10}4}$

$\log_5 8$:

$\log_5 8 = \frac{\log_{10}8}{\log_{10}5}$

$\log_4 25$:

$\log_4 25 = \frac{\log_{10}25}{\log_{10}4}$

Now, substitute these values into $|A|$:

$ |A| = \left(\frac{\log_{10}128}{\log_{10}5}\right) \left(\frac{\log_{10}25}{\log_{10}4}\right) - \left(\frac{\log_{10}5}{\log_{10}4}\right) \left(\frac{\log_{10}8}{\log_{10}5}\right) $

Next, we find the cofactors of matrix $ A $:

$ A_{11} = \log_4 25 $

$ A_{12} = -\log_5 8 $

$ A_{21} = -\log_4 5 $

$ A_{22} = \log_5 128 $

Then, calculate matrix $ C $ whose elements are given by $ C_{ij} = \sum\limits_{k=1}^{2} a_{ik} A_{jk} $:

$ C_{11} = a_{11}A_{11} + a_{12}A_{12} = |A| = \frac{11}{2} $

$ C_{12} = a_{11}A_{21} + a_{12}A_{22} = 0 $

$ C_{21} = a_{21}A_{11} + a_{22}A_{12} = 0 $

$ C_{22} = a_{21}A_{21} + a_{22}A_{22} = |A| = \frac{11}{2} $

Thus, matrix $ C $ is:

$ C = \begin{bmatrix} \frac{11}{2} & 0 \\ 0 & \frac{11}{2} \end{bmatrix} $

To find $|C|$, we compute:

$ |C| = \left(\frac{11}{2}\right)\left(\frac{11}{2}\right) = \frac{121}{4} $

Finally, calculate $ 8|C| $:

$ 8|C| = 8 \times \frac{121}{4} = 242 $

$\begin{aligned} & \left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}\right|, x \in R \\ & R_2 \rightarrow R_2-R_1 \& R_3 \rightarrow R_3-R_1 \\ & f(x)\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right| \end{aligned}$

Expand about $\mathrm{R}_1$, we get

$f(x)=2+4 \sin 4 x$

$\therefore M=\max$ value of $f(x)=6$

$\mathrm{m}=\mathrm{min}$ value of $\mathrm{f}(\mathrm{x})=-2$

$\therefore \mathrm{M}^4-\mathrm{m}^4=1280$

$\begin{aligned} & \mathrm{P}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \\ & \mathrm{~B}=\mathrm{PAPT} \end{aligned}$

Pre multiply by $\mathrm{P}^{\mathrm{T}}$ ( Given)

$\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{PA} \mathrm{P}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$

Now post multiply by P

$\mathrm{P}^{\mathrm{T}} \mathrm{BP}=\mathrm{AP}^{\mathrm{T}} \mathrm{P}=\mathrm{A}$

$\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}$

Similarly $A^{10}=P^T B^{10} P=C$

$\begin{aligned} & A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array}\right] \text { (Given) } \\ & \Rightarrow A^2=\left[\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \\ 0 & 1 \end{array}\right] \end{aligned}$

Similarly check $\mathrm{A}^3$ and so on since $\mathrm{C}=\mathrm{A}^{10}$

$\Rightarrow$ Sum of diagonal elements of C is $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$

$\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{~g} \mathrm{~cd}(\mathrm{~m}, \mathrm{n})=1 \text { (Given) } \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0} f(x)=\left|\begin{array}{ccc} a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1 \end{array}\right| \\ & =(a+1)(2(b+1)-b)+1(a b-a(b+1))+b a \\ & =(a+1)(b+2)-a+a b \\ & =b+a+2=\lambda+\mu a+v b \\ & \lambda=2, \mu=1, v=1 \Rightarrow(\lambda+\mu+v)^2=16 \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{D}=\left|\begin{array}{rrr} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{array}\right|=0, \lambda \mu+42 \lambda-4 \mu+107=0 \\ & \mathrm{D}_1=2 \lambda \mu+99 \lambda-10 \mu+255 \\ & \mathrm{D}_2=13-\mu \\ & D_3=5 \lambda+5 \\ & D_2=0 \Rightarrow \mu=13 \& D_3=0 \Rightarrow \lambda=-1 \end{aligned}\\ &\text { check & verify for these values } \mathrm{D} \& \mathrm{D}_2=0 \end{aligned}$

$\begin{aligned} & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 2 & -1 & 1 \\ 5 & \lambda & 3 \\ 100 & -47 & \mu \end{array}\right|=0 \\ & 2(\lambda \mu+141)+(5 \mu-300)-235-100 \lambda=0 \ldots (1)\\ & \Delta_3=0 \Rightarrow\left|\begin{array}{ccc} 2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212 \end{array}\right|=0 \\ & 6 \lambda=-12 \Rightarrow \lambda=-2 \\ & \text { Put } \lambda=2 \text { in }(1) \\ & 2(-2 \mu+141)+5 \mu-300-235+200=0 \\ & \mu=53 \\ & \therefore 57 \end{aligned}$

$\begin{aligned} & \mathrm{D}=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda \end{array}\right|=0 \\ & \lambda=17 \\ & D_z=\left|\begin{array}{lll} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu \end{array}\right| \neq 0 \\ & \mu \neq 18 \end{aligned}$

$\begin{aligned} & \text { Let } A=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right] \\ & A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \Rightarrow\left[\begin{array}{l} a_{12} \\ a_{22} \\ a_{32} \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right] \Rightarrow \begin{array}{l} a_{22}=0 ; a_{12}=0 \\ a_{32}=1 \end{array} \end{aligned}$

$\begin{aligned} A\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right] \Rightarrow \begin{array}{l} 4 a_{11}+a_{12}+3 a_{13}=0 \\ 4 a_{21}+a_{22}+3 a_{23}=1 \Rightarrow 4 a_{21}+3 a_{23}=1 \\ 4 a_{31}+a_{32}+3 a_{33}=0 \end{array} \\ A\left[\begin{array}{l} 2 \\ 1 \\ 2 \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \Rightarrow \begin{array}{l} 2 a_{11}+a_{12}+2 a_{13}=1 \\ 2 a_{21}+a_{22}+2 a_{23}=0 \Rightarrow a_{21}+a_{23}=0 \\ 2 a_{31}+a_{32}+2 a_{33}=0 \end{array} \\ -4 a_{23}+3 a_{23}=1 \Rightarrow a_{23}=-1 \end{aligned}$

$\begin{aligned} &\begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned}\\ &\text { For infinitely many solutions }\\ &\begin{aligned} & \mathrm{D}=\left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{array}\right|=0 \\ & (\lambda-3)(2 \lambda+1)=0 \\ & \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 5 & \lambda-4 & \lambda \\ 7 & \lambda-1 & \lambda-4 \\ 9 & \lambda+2 & -(\lambda+2) \end{array}\right|=0 \\ & 2(3-\lambda)(23-2 \lambda)=0 \\ & \lambda=3 \\ & \therefore \lambda^2+\lambda=9+3=12 \end{aligned} \end{aligned}$

$\begin{aligned} & {\left[\mathrm{A}\left(\operatorname{adj}\left(\mathrm{~A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)^{-1} \cdot \mathrm{~B}\right]^{-1}} \\ & \mathrm{~B}^{-1} \cdot\left(\operatorname{adj}\left(\mathrm{~A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right) \cdot \mathrm{A}^{-1} \\ & \mathrm{~B}^{-1} \operatorname{adj}\left(\mathrm{~A}^{-1}\right) \mathrm{A}^{-1}+\mathrm{B}^{-1}\left(\operatorname{adj}\left(\mathrm{~B}^{-1}\right)\right) \cdot \mathrm{A}^{-1} \\ & \mathrm{~B}^{-1}\left|\mathrm{~A}^{-1}\right| \mathrm{I}+\left|\mathrm{B}^{-1}\right| \mathrm{IA}^{-1} \\ & \frac{\mathrm{~B}^{-1}}{|\mathrm{~A}|}+\frac{\mathrm{A}^{-1}}{|\mathrm{~B}|} \\ & \Rightarrow \frac{\operatorname{adjB}}{|\mathrm{B}||\mathrm{A}|}+\frac{\operatorname{adj}}{|\mathrm{A}||\mathrm{B}|} \\ & =\frac{1}{|\mathrm{~A}||\mathrm{B}|}(\operatorname{adjB}+\operatorname{adjA}) \end{aligned}$