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} $

To find the condition for infinitely many solutions, we first calculate the determinant $ D $ of the coefficients of $ x, y, z $:

$ D=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{array}\right| =\lambda - 6 $

For infinitely many solutions, $ D = 0 $. So, $ \lambda - 6 = 0 \Rightarrow \lambda = 6. $

Now, let us find $ D_1 $, the determinant obtained by replacing the first column with the constants:

$ D_1=\left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 5 \\ \mu & 3 & \lambda \end{array}\right| =2\lambda + 3\mu - 60 $

Next, the determinant $ D_2 $ is obtained by replacing the second column with the constants:

$ D_2=\left|\begin{array}{ccc} 1 & 6 & 1 \\ 1 & 10 & 5 \\ 2 & \mu & \lambda \end{array}\right| =4(\lambda - \mu + 10) $

Finally, $ D_3 $ is obtained by replacing the third column with the constants:

$ D_3=\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{array}\right| =\mu - 16 $

For the system to have infinitely many solutions, all the determinants must be zero:

$ D = D_1 = D_2 = D_3 = 0 $

Substitute $ \lambda = 6 $:

From $ D_3 = 0 $, $ \mu - 16 = 0 \Rightarrow \mu = 16. $

Thus,

$ \lambda + \mu = 6 + 16 = 22 $

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

$\Rightarrow|\boldsymbol{A}|=15 \boldsymbol{\alpha}-10+16=15 \boldsymbol{\alpha}+6$

Now adj $A=\left[\begin{array}{ccc}15 & 3 & -6 \\ -10 & 5 \alpha & 4 \\ 8 & -4 \alpha & 3 \alpha-2\end{array}\right]$

$\therefore \mathrm{B}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -5 \alpha & 0 \\ 0 & 4 \alpha & -2 \alpha\end{array}\right]+\operatorname{adj}(\mathrm{A})$

$B=\left[\begin{array}{ccc}16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha-2\end{array}\right]$

$\begin{aligned} & \&|\mathrm{~B}|=6(5 \alpha+6)=66 \text { (given in question) } \\ & \Rightarrow \alpha=1 \\ & \Rightarrow|\mathrm{~A}|=21\end{aligned}$

$\boldsymbol{A} \cdot \operatorname{Adj}(\boldsymbol{A})=|\boldsymbol{A}| \boldsymbol{I} \Rightarrow|\boldsymbol{A} \cdot \operatorname{Adj}(\boldsymbol{A})|=\| \boldsymbol{A}|\boldsymbol{I}|=\left(|\boldsymbol{A}|^3\right)|\boldsymbol{I}|=|\boldsymbol{A}|^3$

$\Rightarrow $ $|\boldsymbol{A}||\operatorname{Adj}(\boldsymbol{A})|=|\boldsymbol{A}|^3 \Rightarrow|\operatorname{Adj}(\boldsymbol{A})|=|\boldsymbol{A}|^2=(21)^2=441$

$\begin{aligned} & \left|\begin{array}{ccc}\cos 3 \theta & -8 & -12 \\ \cos 2 \theta & 3 & 3 \\ 1 & 1 & 3\end{array}\right|=0 \\ & \left|\begin{array}{ccc}\cos 3 \theta & -8 & -4 \\ \cos 2 \theta & 3 & 1 \\ 1 & 1 & 1\end{array}\right|=0\end{aligned}$

$\begin{aligned} & \left(C_1 \rightarrow C_1-C_2\right) \& C_2 \rightarrow C_2-C_3 \\ & \left|\begin{array}{ccc}\cos 3 \theta+8 & -4 & -3 \\ \cos 2 \theta-3 & 2 & 1 \\ 0 & 0 & 1\end{array}\right|=0\end{aligned}$

$\begin{aligned} & 2 \cos 3 \theta+16+4 \cos 2 \theta-12=0 \\ & \left(4 \cos ^3 \theta-3 \cos \theta\right)+2\left(2 \cos ^2 \theta-1\right)+2=0 \\ & 4 \cos ^3 \theta+4 \cos ^2 \theta-3 \cos \theta=0 \\ & \cos \theta\left(4 \cos ^2 \theta+4 \cos \theta-3\right)=0 \\ & \cos \theta(2 \cos \theta+3)(2 \cos \theta-1)=0 \\ & \cos \theta=0, \frac{1}{2},-\frac{3}{2} \text { (rejected) } \\ & \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{3}, \frac{5 \pi}{3} \\ & \text { Sum }=4 \pi\end{aligned}$

$\begin{aligned} & \mathrm{A}=\mathrm{P}+\mathrm{I} \\ & \mathrm{A}=\left[\begin{array}{lll}0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0\end{array}\right]+\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ & \mathrm{A}=\mathrm{P}+\mathrm{I} \text { where } \mathrm{P}^3=0 \\ & \mathrm{~A}^{99}=\mathrm{I}+99 \mathrm{P}+{ }^{99} \mathrm{C}_2 \mathrm{P}^2\end{aligned}$

$\begin{aligned} & \mathrm{A}^{99}-\mathrm{I}=99\left[\begin{array}{lll}0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0\end{array}\right]+{ }^{99} \mathrm{C}_2\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & 0 & 0 \\ 99 \times 3 & 0 & 0 \\ 44550 & 99 \times 3 & 0\end{array}\right]\end{aligned}$

$b_{31}=44550, b_{21}=99 \times 3, b_{32}=99 \times 3$

$\therefore $ $\frac{\mathrm{b}_{31}-\mathrm{b}_{21}}{\mathrm{~b}_{32}}$ = $\frac{44550-297}{297}=149$

$ f(n)=\left|\begin{array}{ccc} n & -1 & -5 \\ -2 n^2 & 3(2 k+1) & 2 k+1 \\ -3 n^3 & 3 k(2 k+1) & 3 k(k+2)+1 \end{array}\right|$

$\sum\limits_{n=1}^k f(n)=\left|\begin{array}{ccc}\frac{k(k+1)}{2} & -1 & -5 \\ -2 \frac{k(k+1)(2 k+1)}{6} & 3(2 k+1) & 2 k+1 \\ -3 \frac{k^2(k+1)^2}{4} & 3 k(2 k+1) & 3 k(k+2)+1\end{array}\right|=98$

$\begin{aligned} & \Rightarrow \frac{\mathrm{k}(\mathrm{k}+1)(2 \mathrm{k}+1)}{2} \cdot \frac{7}{3}=98 \\ & \Rightarrow \mathrm{k}(\mathrm{k}+1)(2 \mathrm{k}+1)=84 \\ & \Rightarrow \mathrm{k}=3\end{aligned}$

Since the images of the standard basis vectors are given, the columns of matrix $M$ are exactly these vectors.

So,

$ M= \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} $

Now we are given

$ M\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\7\\11\end{pmatrix} $

So,

$ \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\7\\11\end{pmatrix} $

This gives the system:

$ x-z=1 $

$ 2x+y+z=7 $

$ 3x+2y+z=11 $

Now solve step by step.

From the first equation,

$ x=1+z $

Substitute into the second equation:

$ 2(1+z)+y+z=7 $

$ 2+3z+y=7 $

$ y+3z=5 \quad \Rightarrow \quad y=5-3z $

Now substitute $x=1+z$ and $y=5-3z$ into the third equation:

$ 3(1+z)+2(5-3z)+z=11 $

$ 3+3z+10-6z+z=11 $

$ 13-2z=11 $

$ -2z=-2 $

$ z=1 $

Then,

$ x=1+z=2 $

and

$ y=5-3z=2 $

Therefore,

$ x+y+z=2+2+1=5 $

So the correct answer is:

$ \boxed{5} $

Option B

Let

$ u=\begin{bmatrix}1\\0\\1\end{bmatrix}, \qquad v=\begin{bmatrix}0\\0\\1\end{bmatrix} $

We are given

$ A^T u=\begin{bmatrix}5\\2\\2\end{bmatrix}, \qquad A^T v=\begin{bmatrix}3\\1\\1\end{bmatrix} $

and

$ Au=\begin{bmatrix}3\\4\\4\end{bmatrix}, \qquad Av=\begin{bmatrix}1\\3\\1\end{bmatrix} $

We have to find

$ \det\left(\operatorname{adj}(A^2+A)\right) $

given that

$ \det(A)=1 $

Step 1: Use the relation $A(u-v)$

First note that

$ u-v=\begin{bmatrix}1\\0\\1\end{bmatrix}-\begin{bmatrix}0\\0\\1\end{bmatrix} =\begin{bmatrix}1\\0\\0\end{bmatrix} $

So,

$ A\begin{bmatrix}1\\0\\0\end{bmatrix}=A(u-v)=Au-Av $

Now,

$ Au-Av= \begin{bmatrix}3\\4\\4\end{bmatrix}- \begin{bmatrix}1\\3\\1\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix} $

Hence the first column of $A$ is

$ A\begin{bmatrix}1\\0\\0\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix} $

Also, since

$ A\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix} $

the third column of $A$ is

$ \begin{bmatrix}1\\3\\1\end{bmatrix} $

Now use

$ A\begin{bmatrix}1\\0\\1\end{bmatrix} = A\begin{bmatrix}1\\0\\0\end{bmatrix} + A\begin{bmatrix}0\\0\\1\end{bmatrix} $

which gives

$ \begin{bmatrix}3\\4\\4\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix} + \begin{bmatrix}1\\3\\1\end{bmatrix} $

This is consistent.

So let

$ A= \begin{bmatrix} 2 & a & 1\\ 1 & b & 3\\ 3 & c & 1 \end{bmatrix} $

Now use $\det(A)=1$.

Step 2: Find the second column using determinant

Compute determinant:

$ \det(A)= \begin{vmatrix} 2 & a & 1\\ 1 & b & 3\\ 3 & c & 1 \end{vmatrix} $

Expanding along the first row,

$ \det(A)=2\begin{vmatrix}b&3\\ c&1\end{vmatrix} -a\begin{vmatrix}1&3\\ 3&1\end{vmatrix} +1\begin{vmatrix}1&b\\ 3&c\end{vmatrix} $

$ =2(b-3c)-a(1-9)+(c-3b) $

$ =2b-6c+8a+c-3b $

$ =8a-b-5c $

Given $\det(A)=1$,

$ 8a-b-5c=1 $

But we still need more information. For that use the conditions involving $A^T$.

Step 3: Use transpose conditions

Since

$ A^T\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix} $

this means the third column of $A^T$ is $\begin{bmatrix}3\\1\\1\end{bmatrix}$, so the third row of $A$ is

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

Thus,

$ c=1 $

Now $A$ becomes

$ A= \begin{bmatrix} 2 & a & 1\\ 1 & b & 3\\ 3 & 1 & 1 \end{bmatrix} $

Next, use

$ A^T\begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix} $

But

$ A^T\begin{bmatrix}1\\0\\1\end{bmatrix} = A^T\begin{bmatrix}1\\0\\0\end{bmatrix} + A^T\begin{bmatrix}0\\0\\1\end{bmatrix} $

Now $A^T\begin{bmatrix}1\\0\\0\end{bmatrix}$ is the first column of $A^T$, i.e. the first row of $A$, which is

$ \begin{bmatrix}2\\a\\1\end{bmatrix} $

And

$ A^T\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix} $

Therefore,

$ \begin{bmatrix}2\\a\\1\end{bmatrix} + \begin{bmatrix}3\\1\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix} $

So,

$ \begin{bmatrix}5\\a+1\\2\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix} $

Hence,

$ a=1 $

Now use $\det(A)=1$:

$ 8a-b-5c=1 $

with $a=1,\ c=1$,

$ 8-b-5=1 $

$ 3-b=1 $

$ b=2 $

Thus,

$ A= \begin{bmatrix} 2&1&1\\ 1&2&3\\ 3&1&1 \end{bmatrix} $

Step 4: Find $A^2+A$

First compute $A^2$:

$ A^2= \begin{bmatrix} 2&1&1\\ 1&2&3\\ 3&1&1 \end{bmatrix} \begin{bmatrix} 2&1&1\\ 1&2&3\\ 3&1&1 \end{bmatrix} $

Multiplying row by column:

$ A^2= \begin{bmatrix} 8&5&6\\ 13&8&10\\ 10&6&7 \end{bmatrix} $

So,

$ A^2+A= \begin{bmatrix} 8&5&6\\ 13&8&10\\ 10&6&7 \end{bmatrix} + \begin{bmatrix} 2&1&1\\ 1&2&3\\ 3&1&1 \end{bmatrix} = \begin{bmatrix} 10&6&7\\ 14&10&13\\ 13&7&8 \end{bmatrix} $

Step 5: Use determinant property

For a $3\times 3$ matrix $M$,

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

So we first find

$ \det(A^2+A) $

Now,

$ A^2+A=A(A+I) $

Hence,

$ \det(A^2+A)=\det(A)\det(A+I) $

Given $\det(A)=1$, so

$ \det(A^2+A)=\det(A+I) $

Now

$ A+I= \begin{bmatrix} 3&1&1\\ 1&3&3\\ 3&1&2 \end{bmatrix} $

Its determinant is

$ \det(A+I)= \begin{vmatrix} 3&1&1\\ 1&3&3\\ 3&1&2 \end{vmatrix} $

Expanding along the first row:

$ =3\begin{vmatrix}3&3\\1&2\end{vmatrix} -1\begin{vmatrix}1&3\\3&2\end{vmatrix} +1\begin{vmatrix}1&3\\3&1\end{vmatrix} $

$ =3(6-3)-(2-9)+(1-9) $

$ =3\cdot 3-(-7)-8 $

$ =9+7-8=8 $

So,

$ \det(A^2+A)=8 $

Therefore,

$ \det(\operatorname{adj}(A^2+A))=\det(A^2+A)^2=8^2=64 $

Final Answer

$ \boxed{64} $

So, the correct option is D.

For a homogeneous system of three linear equations in $x,y,z$ to have infinitely many solutions, the coefficient matrix must be singular for every $t \in \mathbb{R}$.

So, we form the coefficient matrix:

$ A= \begin{pmatrix} 1 & 2 & t\\ 6 & 1 & 5t\\ 3 & t^2 & f(t) \end{pmatrix} $

Since the system has infinitely many solutions for all $t$, we must have

$ \det(A)=0 \quad \text{for all } t \in \mathbb{R}. $

Now compute the determinant.

$ \det(A)= \begin{vmatrix} 1 & 2 & t\\ 6 & 1 & 5t\\ 3 & t^2 & f(t) \end{vmatrix} $

Expanding along the first row,

$ \det(A)= 1\begin{vmatrix}1&5t\\ t^2&f(t)\end{vmatrix} -2\begin{vmatrix}6&5t\\ 3&f(t)\end{vmatrix} +t\begin{vmatrix}6&1\\ 3&t^2\end{vmatrix} $

Now evaluate each minor:

$ \begin{vmatrix}1&5t\\ t^2&f(t)\end{vmatrix} = f(t)-5t^3 $

$ \begin{vmatrix}6&5t\\ 3&f(t)\end{vmatrix} = 6f(t)-15t $

$ \begin{vmatrix}6&1\\ 3&t^2\end{vmatrix} = 6t^2-3 $

So,

$ \det(A)=\big(f(t)-5t^3\big)-2\big(6f(t)-15t\big)+t(6t^2-3) $

$ = f(t)-5t^3-12f(t)+30t+6t^3-3t $

$ = -11f(t)+t^3+27t $

Since this is zero for all $t$,

$ -11f(t)+t^3+27t=0 $

Hence,

$ f(t)=\frac{t^3+27t}{11} $

Now differentiate:

$ f'(t)=\frac{3t^2+27}{11} =\frac{3(t^2+9)}{11} $

Since $t^2+9>0$ for all $t\in\mathbb{R}$, we get

$ f'(t)>0 \quad \text{for all } t\in\mathbb{R}. $

Therefore, $f$ is strictly increasing on $\mathbb{R}$.

Hence, the correct option is:

$ \boxed{\text{Option B}} $

$ \begin{aligned} & D=\lambda-5 \\ & D_1=\lambda+\mu-18 \\ & D_2=4 \lambda-2 \mu+6 \\ & D_3=\mu-13 \\ & \text { Here } x=\frac{D_1}{D} ; y=\frac{D_2}{D} ; z=\frac{D_3}{D} \end{aligned} $

For the system to have infinitely many solutions, the determinant of the coefficient matrix and all the corresponding determinants must be zero.

$ \begin{aligned} & D=D_1=D_2=D_3=0 \end{aligned} $

First, set $D=0$:

$ \begin{aligned} & \lambda-5=0 \Rightarrow \lambda=5 \end{aligned} $

Next, set $D_3=0$:

$ \begin{aligned} & \mu-13=0 \Rightarrow \mu=13 \end{aligned} $

Now check $D_1=0$ using $\lambda=5$ and $\mu=13$:

$ \begin{aligned} & \lambda+\mu-18=5+13-18=0 \end{aligned} $

Also check $D_2=0$:

$ \begin{aligned} & 4\lambda-2\mu+6=4(5)-2(13)+6=20-26+6=0 \end{aligned} $

So the values are consistent. Therefore,

$ \begin{aligned} & \therefore \lambda+\mu=5+13=18 \end{aligned} $

We need the largest real value of $\alpha$ such that

$ \det(A-\alpha I)=0 $

That means $\alpha$ is an eigenvalue of the matrix

$ A=\begin{bmatrix} 1&2&7\\ 4&-2&8\\ 3&8&-7 \end{bmatrix}. $

So first we find the eigenvalues of $A$.

Step 1: Form $A-\alpha I$

$ A-\alpha I= \begin{bmatrix} 1-\alpha & 2 & 7\\ 4 & -2-\alpha & 8\\ 3 & 8 & -7-\alpha \end{bmatrix} $

Now,

$ \det(A-\alpha I)= \begin{vmatrix} 1-\alpha & 2 & 7\\ 4 & -2-\alpha & 8\\ 3 & 8 & -7-\alpha \end{vmatrix} =0 $

Step 2: Expand the determinant

Expanding along the first row:

$ \det(A-\alpha I) =(1-\alpha)\begin{vmatrix}-2-\alpha & 8\\ 8 & -7-\alpha\end{vmatrix} -2\begin{vmatrix}4 & 8\\ 3 & -7-\alpha\end{vmatrix} +7\begin{vmatrix}4 & -2-\alpha\\ 3 & 8\end{vmatrix} $

Now calculate each minor:

$ \begin{vmatrix}-2-\alpha & 8\\ 8 & -7-\alpha\end{vmatrix} =(-2-\alpha)(-7-\alpha)-64 $

$ =(\alpha+2)(\alpha+7)-64 =\alpha^2+9\alpha+14-64 =\alpha^2+9\alpha-50 $

Next,

$ \begin{vmatrix}4 & 8\\ 3 & -7-\alpha\end{vmatrix} =4(-7-\alpha)-24 =-28-4\alpha-24 =-52-4\alpha $

And,

$ \begin{vmatrix}4 & -2-\alpha\\ 3 & 8\end{vmatrix} =32-3(-2-\alpha) =32+6+3\alpha =38+3\alpha $

So,

$ \det(A-\alpha I) =(1-\alpha)(\alpha^2+9\alpha-50)-2(-52-4\alpha)+7(38+3\alpha) $

Now simplify:

$ (1-\alpha)(\alpha^2+9\alpha-50) =\alpha^2+9\alpha-50-\alpha^3-9\alpha^2+50\alpha $

$ =-\alpha^3-8\alpha^2+59\alpha-50 $

Also,

$ -2(-52-4\alpha)=104+8\alpha $

and

$ 7(38+3\alpha)=266+21\alpha $

Therefore,

$ \det(A-\alpha I) =-\alpha^3-8\alpha^2+59\alpha-50+104+8\alpha+266+21\alpha $

$ =-\alpha^3-8\alpha^2+88\alpha+320 $

So the equation becomes

$ -\alpha^3-8\alpha^2+88\alpha+320=0 $

Multiplying by $-1$:

$ \alpha^3+8\alpha^2-88\alpha-320=0 $

Step 3: Factorise

Try integer roots. On putting $\alpha=8$:

$ 8^3+8(8^2)-88(8)-320=512+512-704-320=0 $

So $(\alpha-8)$ is a factor.

Divide:

$ \alpha^3+8\alpha^2-88\alpha-320=(\alpha-8)(\alpha^2+16\alpha+40) $

Now solve

$ \alpha^2+16\alpha+40=0 $

$ \alpha=\frac{-16\pm\sqrt{256-160}}{2} =\frac{-16\pm\sqrt{96}}{2} =\frac{-16\pm 4\sqrt6}{2} =-8\pm 2\sqrt6 $

Hence the eigenvalues are

$ 8,\quad -8+2\sqrt6,\quad -8-2\sqrt6 $

So the largest possible value is

$ p=8 $

Step 4: Put $p=8$ in the circle

Given circle:

$ (x-p)^2+(y-2p)^2=320 $

Substitute $p=8$:

$ (x-8)^2+(y-16)^2=320 $

Its centre is $(8,16)$ and radius is

$ r=\sqrt{320}=8\sqrt5 $

Step 5: Find intersections with the coordinate axes

On the $x$-axis

For points on the $x$-axis, $y=0$.

$ (x-8)^2+(0-16)^2=320 $

$ (x-8)^2+256=320 $

$ (x-8)^2=64 $

$ x-8=\pm 8 $

So,

$ x=16 \quad \text{or} \quad x=0 $

Thus the circle meets the $x$-axis at

$ (16,0),\ (0,0) $

On the $y$-axis

For points on the $y$-axis, $x=0$.

$ (0-8)^2+(y-16)^2=320 $

$ 64+(y-16)^2=320 $

$ (y-16)^2=256 $

$ y-16=\pm 16 $

So,

$ y=32 \quad \text{or} \quad y=0 $

Thus the circle meets the $y$-axis at

$ (0,32),\ (0,0) $

Step 6: Count distinct intersection points

The distinct points are

$ (16,0),\ (0,0),\ (0,32) $

So the circle intersects the coordinate axes at

$ \boxed{3\text{ points}} $

Hence, the correct option is:

$ \boxed{\text{Option C}} $

Let

$ A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}, \qquad a,b,c,d \in \{0,1,2,3,4\} $

and it satisfies

$ A^2-4A+3I=0. $

We have to count those matrices for which the sum of diagonal elements is

$ a+d=4. $

So we must find the number of matrices satisfying both:

1. $A^2-4A+3I=0$

2. $\operatorname{tr}(A)=a+d=4$

Step 1: Factor the matrix equation

We write

$ A^2-4A+3I=(A-I)(A-3I)=0. $

So the matrix $A$ satisfies the polynomial

$ (x-1)(x-3)=0. $

Hence the only possible eigenvalues are $1$ and $3$.

Now the trace is given to be $4$. Since trace is sum of eigenvalues, the eigenvalues must be

$ 1 \text{ and } 3. $

So every such matrix has distinct eigenvalues $1$ and $3$, hence it is diagonalizable.

Step 2: Use trace and determinant

For a $2\times 2$ matrix with eigenvalues $1$ and $3$,

• trace $=1+3=4$

• determinant $=1\cdot 3=3$

So we need all matrices with entries in $\{0,1,2,3,4\}$ such that

$ a+d=4,\qquad ad-bc=3. $

Let $d=4-a$. Then

$ a(4-a)-bc=3 $

that is,

$ bc=a(4-a)-3. $

Now check all possible values of $a\in\{0,1,2,3,4\}$.

Step 3: Case-wise counting

Case 1: $a=0$

Then $d=4$, and

$ bc=0\cdot 4-3=-3 $

not possible.

Case 2: $a=1$

Then $d=3$, and

$ bc=1\cdot 3-3=0 $

So we need ordered pairs $(b,c)$ with $b,c\in\{0,1,2,3,4\}$ and $bc=0$.

That means at least one of $b,c$ is $0$.

Number of such ordered pairs:

• $b=0$: $5$ choices for $c$

• $c=0$: $5$ choices for $b$

• but $(0,0)$ counted twice

So total

$ 5+5-1=9 $

matrices.

Case 3: $a=2$

Then $d=2$, and

$ bc=2\cdot 2-3=1 $

So

$ bc=1. $

Since $b,c\in\{0,1,2,3,4\}$, this is possible only for

$ (b,c)=(1,1). $

So total matrices in this case:

$ 1 $

Case 4: $a=3$

Then $d=1$, and

$ bc=3\cdot 1-3=0 $

Again, number of ordered pairs $(b,c)$ with $bc=0$ is

$ 9. $

Case 5: $a=4$

Then $d=0$, and

$ bc=4\cdot 0-3=-3 $

not possible.

Step 4: Total count

Total number of matrices is

$ 9+1+9=19. $

Hence the required number of matrices is

$ \boxed{19} $

So, the correct option is:

$ \boxed{\text{Option D}} $

Given

$ A=\begin{bmatrix} 1 & 1 & 2\\ -2 & 0 & 1\\ 1 & 3 & 5 \end{bmatrix} $

We have to find the sum of all elements of

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

$ \det A= 1\begin{vmatrix} 0 & 1\\ 3 & 5 \end{vmatrix} -1\begin{vmatrix} -2 & 1\\ 1 & 5 \end{vmatrix} +2\begin{vmatrix} -2 & 0\\ 1 & 3 \end{vmatrix} $

$ =1(0\cdot 5-1\cdot 3)-1((-2)\cdot 5-1\cdot 1)+2((-2)\cdot 3-0\cdot 1) $

$ =-3-(-11)+2(-6) $

$ =-3+11-12=-4 $

So,

$ \det A=-4 $

For an $n\times n$ matrix,

$ \det(\operatorname{adj}A)=(\det A)^{n-1} $

Since $A$ is $3\times 3$,

$ \det(\operatorname{adj}A)=(\det A)^2=(-4)^2=16 $

Now let

$ B=2(\operatorname{adj}A)^{-1} $

We need $\operatorname{adj}(\operatorname{adj}B)$.

For a $3\times 3$ matrix, the property is:

$ \operatorname{adj}(\operatorname{adj}B)=(\det B)\,B $

Since

$ B=2(\operatorname{adj}A)^{-1} $

and for a $3\times 3$ matrix,

$ \det(2M)=2^3\det(M)=8\det(M) $

therefore,

$ \det B=8\det\left((\operatorname{adj}A)^{-1}\right) $

Also,

$ \det\left((\operatorname{adj}A)^{-1}\right)=\frac{1}{\det(\operatorname{adj}A)}=\frac{1}{16} $

Hence,

$ \det B=8\cdot \frac{1}{16}=\frac{1}{2} $

So,

$ \operatorname{adj}(\operatorname{adj}B)=(\det B)B=\frac12\,B $

But

$ B=2(\operatorname{adj}A)^{-1} $

Thus,

$ \operatorname{adj}(\operatorname{adj}B)=\frac12 \cdot 2(\operatorname{adj}A)^{-1}=(\operatorname{adj}A)^{-1} $

So the required matrix is simply

$ (\operatorname{adj}A)^{-1} $

We know,

$ \operatorname{adj}A=(\det A)A^{-1} $

Hence,

$ (\operatorname{adj}A)^{-1}=\frac{1}{\det A}A $

Since $\det A=-4$,

$ (\operatorname{adj}A)^{-1}=\frac{1}{-4}A=-\frac14 A $

Therefore the required matrix is

$ -\frac14 A $

First find the sum of elements of $A$:

$ 1+1+2-2+0+1+1+3+5=12 $

Therefore sum of elements of $-\frac14 A$ is

$ -\frac14 \cdot 12=-3 $

Final Answer

$ \boxed{-3} $

So, the correct option is D.

The given system of equations is:

$x + 5y + 6z = 4$

$2x + 3y + 4z = 7$

$x + 6y + az = b$

Find the determinant of the coefficient matrix ($\Delta$)

The coefficient matrix is formed by the coefficients of $x$, $y$, and $z$.

$ \Delta = \begin{vmatrix} 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a \end{vmatrix} $

Let us expand this determinant along the first row:

$ \Delta = 1(3 \cdot a - 6 \cdot 4) - 5(2 \cdot a - 1 \cdot 4) + 6(2 \cdot 6 - 1 \cdot 3) $

$ \Delta = 1(3a - 24) - 5(2a - 4) + 6(12 - 3) $

$ \Delta = 3a - 24 - 10a + 20 + 6(9) $

$ \Delta = -7a - 4 + 54 $

$ \Delta = -7a + 50 $

For the system to have infinitely many solutions, we equate $\Delta$ to $0$:

$ -7a + 50 = 0 \implies 7a = 50 \implies a = \frac{50}{7} $

Find the determinant $\Delta_z$

To find $\Delta_z$, we replace the $z$-column (third column) with the constant terms from the right-hand side of the equations.

$ \Delta_z = \begin{vmatrix} 1 & 5 & 4 \\ 2 & 3 & 7 \\ 1 & 6 & b \end{vmatrix} $

We again expand along the first row:

$ \Delta_z = 1(3 \cdot b - 6 \cdot 7) - 5(2 \cdot b - 1 \cdot 7) + 4(2 \cdot 6 - 1 \cdot 3) $

$ \Delta_z = 1(3b - 42) - 5(2b - 7) + 4(12 - 3) $

$ \Delta_z = 3b - 42 - 10b + 35 + 4(9) $

$ \Delta_z = -7b - 7 + 36 $

$ \Delta_z = -7b + 29 $

Equating $\Delta_z$ to zero for infinitely many solutions, we get:

$ -7b + 29 = 0 \implies 7b = 29 \implies b = \frac{29}{7} $

Check the options for the point $(a, b)$

We have obtained the point coordinates as $(a, b) = \left( \frac{50}{7}, \frac{29}{7} \right)$.

Let us evaluate the difference $a - b$ to find out which linear profile equation matches our outcome:

$ a - b = \frac{50}{7} - \frac{29}{7} $

$ a - b = \frac{50 - 29}{7} $

$ a - b = \frac{21}{7} = 3 $

Writing this relation in terms of $x$ and $y$ (where $x = a$ and $y = b$), it perfectly satisfies the line equation:

$ x - y = 3 $

Thus, the point $(a, b)$ lies on the line given in Option B.

$ \begin{aligned} & x+2 y+z=5 \\ & 2 x+y+\alpha z=5 \\ & 8 x+4 y+\beta z=18 \end{aligned} $

For a system of three linear equations to have no solution, it is necessary that the coefficient matrix be singular, i.e. its determinant is zero: $\,\Delta=0$.

$ \begin{aligned} & \Delta=\left|\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{array}\right|=0 \\ & \Delta=1(\beta-4 \alpha)-2(2 \beta-8 \alpha)+1(0)=0 \\ & =\beta-4 \alpha-4 \beta+16 \alpha=0 \\ & \Rightarrow 4 \alpha=\beta \end{aligned} $

Now substitute $\beta=4\alpha$ in the third equation. Then the left side of the third equation becomes $8x+4y+4\alpha z$, which is exactly $4$ times the left side of the second equation $2x+y+\alpha z$.

But the constants do not match: $4\times 5=20$, whereas the third equation has $18$. So we get $4\times(2x+y+\alpha z)=20$ but $8x+4y+4\alpha z=18$, which is a contradiction. Hence, the system is inconsistent and has no solution.

Therefore, $\dfrac{\beta}{\alpha}=\dfrac{4\alpha}{\alpha}=4$.

$ \begin{aligned} & \mathrm{A}=\left[\begin{array}{ll} 1 & 2 \\ 1 & \alpha \end{array}\right] \\ & \mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=0 \end{aligned} $

trace of matrix

$ \begin{aligned} & \alpha+1=4 \\ & \alpha=3 \\ & \mathrm{~B}=\left[\begin{array}{ll} 3 & 3 \\ \beta & 2 \end{array}\right] \\ & \mathrm{B}^2-5 \mathrm{~B}-6 \mathrm{I}=0 \end{aligned} $

$\begin{aligned} & \operatorname{det} .(B)=6-3 \beta=-6 \\ & 3 \beta=12 \\ & \beta=4 \\ & B-A=\left[\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right] ; B+A=\left[\begin{array}{ll}4 & 5 \\ 5 & 5\end{array}\right] \\ & (B-A)(B+A)=\left[\begin{array}{cc}13 & 15 \\ 7 & 10\end{array}\right] \\ & \operatorname{adj}(A)=\left[\begin{array}{cc}3 & -2 \\ -1 & 1\end{array}\right], \operatorname{adjB}=\left[\begin{array}{cc}2 & -3 \\ -4 & 3\end{array}\right] \\ & \operatorname{adj}(A+B)=\left[\begin{array}{cc}5 & -5 \\ -5 & 4\end{array}\right]\end{aligned}$

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

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