Matrices and Determinants

$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3\end{array}\right|$

$ \begin{aligned} & =1(3 \beta-14)-1(3 \alpha-7)+1(2 \alpha-\beta) \\\\ & =3 \beta-14+7-3 \alpha+2 \alpha-\beta \\\\ & =2 \beta-\alpha-7 \end{aligned} $

So, for $\alpha=\beta \neq 7, \Delta \neq 0$ so unique solution.

$\alpha=\beta=7$, equation (i) and (ii) represent 2 parallel planes so no solution.

If $\alpha-2 \beta+7=0$, but $(\alpha, \beta) \neq(7,7)$, then no solution.

$A^{2}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]$

$ =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{array}\right]=A $

$\Rightarrow \mathrm{A}_{3}=\mathrm{A}_{4}=.......=\mathrm{A}$

Now,

$ \begin{aligned} (A+I)^{11} & ={ }^{11} C_{0} A^{11}+{ }^{11} C_{1} A^{10}+\ldots{ }^{11} C_{11} I \\\\ & =A\left({ }^{11} C_{0}+{ }^{11} C_{1} \ldots{ }^{11} C_{10}\right)+I \\\\ & =A\left(2^{11}-1\right)+I \end{aligned} $

Trace of

$ \begin{aligned} (A+I)^{11} & =2^{11}+4\left(2^{11}-1\right)+1-3\left(2^{11}-1\right)+1 \\\\ & =2 \times 2^{11}-4+3+2 \\\\ & =2^{12}+1 \\\\ & =4097 \end{aligned} $

$\Delta = \left| {\matrix{ 1 & { - 1} & 1 \cr 2 & 2 & \alpha \cr 3 & { - 1} & 4 \cr } } \right| = 0$

$ \Rightarrow \alpha = 4$

${\Delta _3} = 0$

$ = \left| {\matrix{ 1 & { - 1} & 5 \cr 2 & 2 & 8 \cr 3 & { - 1} & \beta \cr } } \right| = 0$

$ \Rightarrow \beta = 14$

$\therefore$ ${x^2} - 18x + 56 = 0$

$P = \left[ {\matrix{ {{a_1}} & {{b_1}} & {{c_1}} \cr {{a_2}} & {{b_2}} & {{c_2}} \cr {{a_3}} & {{b_3}} & {{c_3}} \cr } } \right]$

Given : ${P^T} = aP + (a - 1)I$

$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{b_1}} & {{b_2}} & {{b_3}} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr } } \right] = \left[ {\matrix{ {a{a_1} + a - 1} & {a{b_1}} & {a{c_1}} \cr {a{a_2}} & {a{b_2} + a - 1} & {a{c_2}} \cr {a{a_3}} & {a{b_3}} & {a{c_3} + a - 1} \cr } } \right]$

$ \Rightarrow {a_1} = a{a_1} + a - 1 \Rightarrow {a_1}(1 - a) = a - 1 \Rightarrow {a_1} = - 1$

Similarly, ${a_1} = {b_2} = {c_3} = - 1$

Now, $\left. \matrix{ {a_2} = a{b_1} \hfill \cr {b_1} = a{a_2} \hfill \cr} \right] \to {a_2} = {a^2}{a_2} \Rightarrow {a_2} = 0 \Rightarrow {b_1} = 0$

${c_1} = a{a_3}$

Similarly, all other elements will also be 0

${a_2} = {a_3} = {b_1} = {b_3} = {c_1} = {c_2} = 0$

$\therefore$ $P = \left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right]$

$|P| = - 1$

$|Adj(P){|_{n \times n}} = |A{|^{(n - 1)}}$

$ \Rightarrow |Adj(P)| = {( - 1)^2} = 1$

$x + y + kz = 2$ ............(i)

$2x + 3y - z = 1$ ..........(ii)

$3x + 4y + 2z = k$ ......(iii)

(1) + (2)

$3x + 4y + z(k - 1) = 3$

Comparing with (3)

$k = 3$

Now, $4x + 5y = 7$

$ \Rightarrow 3x + 3y = 3$

$7x + 8y = 10$

as ${4 \over 7} \ne {5 \over 8}$

$\therefore$ Unique solution satisfying $x + y = 1$

$\left| {A - d\left( {\matrix{ q & { - n} \cr { - p} & m \cr } } \right)} \right| = 0$

$\left| {\matrix{ {m - qd} & {n(1 + d)} \cr {p(1 + d)} & {q - md} \cr } } \right| = 0$

$(m - qd)(q - md) = np{(1 + d)^2}$

$mq - ({q^2} + {m^2})d + qm{d^2} = np(1 + {d^2}) + 2npd$

${d^2}(mq - np) + 1(mq - np) = (2np + {m^2} + {q^2})d$

$({d^2} + 1)(mq - np) = (2np + m + a)d$

${d^2} + 1 = 2np + {m^2} + {q^2}$

$2d = 2mq - 2np$

$ \Rightarrow {(1 + d)^2} = {(m + q)^2}$

If the matrix is invertible then its determinant should not be zero.

So, $ \left|\begin{array}{ccc} e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right| \neq 0 $

$ \Rightarrow e^t \times e^{-t} \times e^{-t}\left|\begin{array}{ccc} 1 & \sin t-2 \cos t & -2 \sin t-\cos t \\ 1 & 2 \sin t+\cos t & \sin t-2 \cos t \\ 1 & \cos t & \sin t \end{array}\right| \neq 0 $

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

$ \begin{aligned} & e^{-t}\left|\begin{array}{ccc} 0 & -\sin t-3 \cos t & -3 \sin t+\cos t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right| \neq 0 \\\\ & \Rightarrow e^{-t}\left(2 \sin t \cos t+6 \cos ^2 t+6 \sin ^2 t-2 \sin t \cos t\right) \neq 0 \\\\ & \Rightarrow 6e^{-t} \neq 0, \text { for } \forall t \in R . \end{aligned} $

$\mathrm{A}^{2}=3 \mathrm{~A}+\alpha \mathrm{I}$

$A^{3}=3 A^{2}+\alpha A$

$\mathrm{A}^{3}=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A}$

$\mathrm{A}^{3}=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I}$

$\mathrm{A}^{4}=(9+\alpha) \mathrm{A}^{2}+3 \alpha \mathrm{A}$

$=(9+\alpha)(3 \mathrm{~A}+\alpha \mathrm{I})+3 \alpha \mathrm{A}$

$=\mathrm{A}(27+6 \alpha)+\alpha(9+\alpha)$

$\Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1$

$\Rightarrow \beta=\alpha(9+\alpha)=-8$

$D=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{array}\right|=0 \Rightarrow \alpha=-1,3$

$D_{x}=\left|\begin{array}{ccc}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{array}\right|=0 \Rightarrow \beta=2$

$D_{y}=\left|\begin{array}{ccc}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{array}\right|=0$

$D_{z}=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{array}\right|=0$

$\beta=2, \alpha=-1$

$\alpha=-1, \beta=2$ Infinite solution

$A^{T}=A, B^{T}=-B, C^{T}=-C$

$ \begin{aligned} P & =A^{13} B^{26}-B^{26} A^{13} \\\\ P^{T} & =\left(A^{13} B^{26}-B^{26} A^{13}\right)^{T}=\left(A^{13} B^{26}\right)^{T}-\left(B^{26} A^{B}\right)^{T} \\\\ & =\left(B^{26}\right)^{T}\left(A^{13}\right)^{T}-\left(A^{13}\right)^{T}\left(B^{26}\right)^{T} \\\\ & =\left(B^{T}\right)^{26}\left(A^{T}\right)^{13}-\left(A^{T}\right)^{13}\left(A^{T}\right)^{26} \\\\ & =B^{26} A^{13}-A^{13} B^{26}=-\left(A^{13} B^{26}-B^{26} A^{13}\right)=-P \end{aligned} $

$P$ is skew-symmetric matrix $\Rightarrow S_{1}$ is false

$Q=A^{26} C^{13}-C^{13} A^{26}=Q^{T}=\left(A^{26} C^{13}-C^{13} A^{26}\right)^{T}$

$Q=\left(A^{26} C^{13}\right)^{T}-\left(C^{13} A^{26}\right)^{T}=\left(C^{13}\right)^{T}\left(A^{26}\right)^{T}-\left(A^{26}\right)^{T}\left(C^{13}\right)^{T}$

$=\left(C^{T}\right)^{13}\left(A^{T}\right)^{26}-\left(A^{T}\right)^{26}\left(C^{T}\right)^{13}=-C^{13} A^{26}+A^{26} C^{13}$

$=A^{26} C^{13}+C^{13} A^{26}$

$\Rightarrow Q^{T}=Q \Rightarrow Q$ is symmetric matrix $\Rightarrow S_{2}$ is true.

$ \begin{aligned} & \mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]\left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = I \\\\\ & \mathrm{B}^2=\left[\begin{array}{cc} 1 & -\mathrm{i} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\mathrm{i} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{B}^3=\left[\begin{array}{cc} 1 & -3 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned} $

Similarly,

$ \begin{aligned} & \mathrm{B}^{2023}=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \\\\ & \mathrm{M}^2=\mathrm{M} \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{~A} \\\\ & \mathrm{M}^3=\mathrm{M}^2 \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{AA}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{~A} \end{aligned} $

Similarly,

$ \begin{aligned} & \mathrm{M}^{2023}= \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{~A} \\\\ & \mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{AA}^{\mathrm{T}} = \mathrm{I}.\mathrm{B}^{2023}.\mathrm{I}=\mathrm{B}^{2023} \\\\ & =\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned} $

$ \therefore $ $ \text { Inverse of }\left(\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}\right) \text { is }\left[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}\right] $

$ \begin{aligned} & |A|=\frac{1}{\log x \log y \log z}\left|\begin{array}{ccc} \log x & \log y & \log z \\ \log x & 2 \log y & \log z \\ \log x & \log y & 3 \log z \end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{array}\right|=2 \\\\ & \Rightarrow\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|=\left|\operatorname{adj}\left(A^2\right)\right|^2=\left(\left|A^2\right|^2\right)^2=|A|^8=2^8 \end{aligned} $

Given system of equations

$ \begin{aligned} & a x+2 a y-3 a z=1 \\\\ & (2 a+1) x+(2 a+3) y+(a+1) z=2 \\\\ & (3 a+5) x+(a+5) y+(a+2) z=3 \\\\ & \text { Let } A=\left|\begin{array}{ccc} a & 2 a & -3 a \\\\ 2 a+1 & 2 a+3 & a+1 \\\\ 3 a+5 & a+5 & a+2 \end{array}\right| \\\\ & =a\left|\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 a+1 & 1-2 a & 7 a+4 \\\\ 3 a+5 & -5 a-5 & 10 a+17 \end{array}\right| \\\\ & =a\left(15 a^{2}+31 a+37\right) \\\\ & \text { Now } A=0 \\\\ & \Rightarrow a=0 \end{aligned} $

So, $S_{1}=R-\{0\}$ and at $a=0$

System has infinite solution but $a \in R-\{0\}$

$\therefore S_{2}=\Phi$

$|A|^{(n-1)^{3}}=12^{4}$

$ \begin{aligned} &|A|^{8}=12^{4} \\\\ &|A|=\sqrt{12} \\\\ &\left|A^{-1} \operatorname{adj} A\right|=\left|A^{-1}\right| \cdot|A|^{2} \\\\ &=|A| = 2\sqrt3 \end{aligned} $

For infinite many solution, $\Delta=0$ and $\Delta_x=0$

$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{array}\right|=0 \\\\ & \Rightarrow 1(-3 \lambda+16)-2(-4 \lambda+32)+3(16-24)=0 \\\\ & \Rightarrow 16-3 \lambda+8 \lambda-64-24=0 \Rightarrow 5 \lambda=72 \\\\ & \therefore \lambda=\frac{72}{5} \\\\ & \Delta_x=\left|\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 3 & -4 \\ 9+\mu & 4 & -\lambda \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 3(-3 \lambda+16)-2(-4 \lambda+36+4 \mu)+3(16-27-3 \mu)=0 \\\\ & \Rightarrow-9 \lambda+48+8 \lambda-72-8 \mu-33-9 \mu=0 \\\\ & \Rightarrow-\lambda-17 \mu=57 \\\\ & \Rightarrow-17 \mu=57+\lambda \\\\ & \therefore -\mu=\frac{57+\frac{72}{5}}{17}\\\\ & \Rightarrow \mu=\frac{-357}{85}=\frac{-21}{5} \end{aligned} $

$ \text { Thus, }(\lambda, \mu) \equiv\left(\frac{72}{5}, \frac{-21}{5}\right) $

Given : $A^{2}+B=A^{2} B\quad...(i)$

$\Rightarrow A^{2}+B-I=A^{2} B-I$

$\Rightarrow A^{2} B-A^{2}-B+I=I$

$\Rightarrow A^{2}(B-I)-I(B-I)=I$

$\Rightarrow\left(A^{2}-I\right)(B-I)=I$

$\therefore A^{2}-I$ is the inverse matrix of $B-I$ and vice versa.

So, $(B-I)\left(A^{2}-I\right)=I$

$\Rightarrow B A^{2}-B-A^{2}+I=I$

$\therefore A^{2}+B=B A^{2} \quad...(ii)$

So, by (i) and (ii)

$A^{2} B=B A^{2}$

$ \begin{aligned} & \Delta=0=\left|\begin{array}{ccc} \alpha^2 & \alpha & 1 \\\\ 1 & 1 & 1 \\\\ \mathrm{a} & \mathrm{b} & \mathrm{c} \end{array}\right| \\\\ & \Rightarrow \alpha^2(\mathrm{c}-\mathrm{b})-\alpha(\mathrm{c}-\mathrm{a})+(\mathrm{b}-\mathrm{a})=0 \end{aligned} $

It is singular when $\alpha=1$

$ \begin{aligned} & \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} \\\\ & \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\\\ & =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3 \end{aligned} $

$M=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$

$|M|=-1+1=0 \Rightarrow M$ is singular so non-invertible $ \Rightarrow $ [A] is wrong.

(B) $M\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right] \Rightarrow\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right]$

$ \left.\begin{array}{l} a_1+a_2+a_3=-a_1 \\ a_1+a_3=-a_2 \\ a_2=-a_3 \end{array}\right\} \Rightarrow a_1=0 \text { and } a_2+a_3=0 \text { infinite solutions exists [B] is correct. } $

Option (D) :

$ \begin{aligned} & M-2 I=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]-2\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{array}\right] \\\\ & |M-2 I|=0 \Rightarrow[D] \text { is wrong } \end{aligned} $

Option (C) :

$ \begin{aligned} & \mathrm{MX}=0 \Rightarrow\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\\\ & x+y+z=0 \\\\ & x+z=0 \\\\ & y=0 \\\\ & \therefore \text { Infinite solution } \end{aligned} $

$[\mathrm{C}]$ is correct

Given

$\begin{gathered}x+2 y+z=7 \\\\ x+\alpha z=11 \\\\ 2 x-3 y+\beta z=\gamma\end{gathered}$

Using Cramer's rule

$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{array}\right| \\\\ & =1(3 \alpha)-2(\beta-2 \alpha)+1(-3) \\\\ & =3 \alpha-2 \beta+4 \alpha-3 \\\\ & =7 \alpha-2 \beta-3 \end{aligned} $

$\begin{aligned} \Delta_x & =\left|\begin{array}{ccc}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta\end{array}\right| \\\\ & =7(3 \alpha)-2(11 \beta-\gamma \alpha)+1(-33) \\\\ & =21 \alpha-22 \beta+22 \gamma-33\end{aligned}$

$\begin{aligned} \Delta_y & =\left|\begin{array}{ccc}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{array}\right| \\\\ & =1(11 \beta-\alpha \gamma)-7(\beta-2 \alpha)+1(\gamma-22) \\\\ & =11 \beta-\alpha \gamma-7 \beta+14 \alpha+\gamma-22 \\\\ & =14 \alpha+4 \beta+\gamma-\alpha \gamma-22\end{aligned}$

$\begin{aligned} \Delta_z & =\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right| \\\\ & =1(33)-2(\gamma-22)+7(-3) \\\\ & =33-2 \gamma+44-21 \\\\ & =-2 \gamma+56\end{aligned}$

For unique solution $\Delta \neq 0$

For infinite solution

$ \Delta=\Delta x=\Delta y=\Delta z=0 $

For no solution $\Delta=0$ and atleast one in $\Delta x, \Delta y, \Delta z$ is non zero.

$ \Delta=0 $

$ \Rightarrow \beta=\frac{1}{2}(7 \alpha-3) $

(P) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$

then $ \Delta=0, \Delta x=\Delta y=\Delta z=0$

$\therefore$ Infinite solution

(Q) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$

$\therefore \Delta=0$ and $\Delta_2 \neq 0$

$\Rightarrow$ No solution.

$ \begin{aligned} & \text { (R) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma \neq 28 \\\\ & \Rightarrow \Delta \neq 0=\text { unique solution } \\\\ & \begin{aligned} & \text { (S) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma=28 \\\\ & \therefore \Delta \neq 0, \Delta=4-2 \beta \\\\ & \Delta x=44-22 \beta \\\\ & \Delta y=4 \beta-8 \\\\ & \Delta z=0 \end{aligned} \end{aligned} $

$\therefore x=11, y=-2, z=0$ is the solution.

Given, $X_{4 \times 3}, Y_{4 \times 3}$ and $P_{2 \times 3}$ are the matrices.

$ \text { } \begin{aligned} Since, & {\left[P_{2 \times 3}\left(X_{3 \times 4}^T Y_{4 \times 3}\right)^{-1} P_{3 \times 2}^T\right]^T } \\ &=\left[P_{2 \times 3}\left(X^T Y\right)_{3 \times 3}^{-1} P_{3 \times 2}^T\right]^T \\ &=\left(P\left(X^T Y\right)^{-1} P^T\right)_{2 \times 2}^T \end{aligned} $

$ \begin{aligned} & \text { Here, } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right] \\ & A^2=\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]=\left[\begin{array}{ll} 7 & 12 \\ 18 & 31 \end{array}\right] \\ & \alpha A^2-\beta A=2 I \\ & \Rightarrow \alpha\left[\begin{array}{ll} 7 & 12 \\ 18 & 31 \end{array}\right]-\beta\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]=2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow \quad 7 \alpha-\beta=2 \\ & \,\,\,\,\,\,\, \quad 18 \alpha-3 \beta=0 \\ & \Rightarrow \quad \beta=6 \alpha \\\Rightarrow & \alpha=2, \beta=12 \\ & \alpha^2+\beta=4+12=16 \end{aligned} $

At $\alpha=-2$, determinant becomes

$ \left|\begin{array}{ccc} 1 & 9 & 25 \\ 0 & 4 & 16 \\ 1 & 1 & 9 \end{array}\right|=k $

On expanding with $C_1$,

$ \begin{aligned} (36-16)+0+1(144-100) & =k \\\Rightarrow && k =64 \end{aligned} $

Given, the system of equations,

$ x+y+z=5, x+2 y+2 z=6 $

and  $ x+3 y+\lambda z=\mu $

Since, equation is solvable, then $D \neq 0$

$ \begin{aligned} & \Rightarrow \quad\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{array}\right| \neq 0 \\ & \Rightarrow(2 \lambda-6)-1(\lambda-2)+1(3-2) \neq 0 \\ & \Rightarrow \lambda \neq 3, \mu \in R \end{aligned} $

As we know,

If $n$ is the order of a square matrix.

Then, $|\operatorname{adj}(\operatorname{adj} A)|=|A|^{(n-1)^2}$

So, $\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|=\left|A^2\right|^{(n-1)^2}$

∵ Order of $A=n=3$

$ \text { } \begin{aligned}Then, \left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right| & =\left|A^2\right|^{(3-1)^2} \\ & =\left|A^2\right|^4=|A|^8 \end{aligned} $

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} $.

To determine the value of $\lambda$ for which the given system of linear equations has no solution, we need to analyze the consistency of these equations. The given system is:

$3x - 2y + z = 2$ (Equation 1)

$4x - 3y + 3z = -5$ (Equation 2)

$7x - 5y + \lambda z = 9$ (Equation 3)

For a system of linear equations to have no solution, the equations must be inconsistent, meaning they do not intersect at any point. This happens when the determinant of the coefficient matrix is zero, and the augmented matrix creates a contradiction.

First, let's form the coefficient matrix and calculate its determinant. The coefficient matrix $A$ of the system is:

$A = \begin{pmatrix} 3 & -2 & 1 \\ 4 & -3 & 3 \\ 7 & -5 & \lambda \end{pmatrix}$

The determinant of matrix $A$ is given by:

$\text{det}(A) = \begin{vmatrix} 3 & -2 & 1 \\ 4 & -3 & 3 \\ 7 & -5 & \lambda \end{vmatrix}$

Expanding this determinant along the first row, we get:

$\text{det}(A) = 3\begin{vmatrix} -3 & 3 \\ -5 & \lambda \end{vmatrix} - (-2)\begin{vmatrix} 4 & 3 \\ 7 & \lambda \end{vmatrix} + 1\begin{vmatrix} 4 & -3 \\ 7 & -5 \end{vmatrix}$

Now, calculate these 2x2 determinants:

$\begin{vmatrix} -3 & 3 \\ -5 & \lambda \end{vmatrix} = (-3)(\lambda) - (3)(-5) = -3\lambda + 15$

$\begin{vmatrix} 4 & 3 \\ 7 & \lambda \end{vmatrix} = (4)(\lambda) - (3)(7) = 4\lambda - 21$

$\begin{vmatrix} 4 & -3 \\ 7 & -5 \end{vmatrix} = (4)(-5) - (-3)(7) = -20 + 21 = 1$

Substitute these into the expansion to get the determinant:

$\text{det}(A) = 3(-3\lambda + 15) + 2(4\lambda - 21) + 1(1)$

$= 3(-3\lambda + 15) + 2(4\lambda - 21) + 1$

$= 3(-3\lambda) + 3(15) + 2(4\lambda) - 2(21) + 1$

$= -9\lambda + 45 + 8\lambda - 42 + 1$

$= -\lambda + 4$

For the system to have no solution, $\text{det}(A)$ must be zero, because that's when the coefficient matrix does not have full rank and the system is inconsistent:

$-\lambda + 4 = 0$

$\lambda = 4$

Therefore, the value of $\lambda$ that makes the system of linear equations have no solution is $4$.

Answer: (A) 4

Since, $B$ is the inverse of $A$.

$\begin{aligned} & \therefore \quad A \times B=I \\ & \Rightarrow\left[\begin{array}{lll} 3 & 2 & 3 \\ 4 & 1 & 0 \\ 2 & 5 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 / 49 & 13 / 49 & -3 / 49 \\ -4 / 49 & -3 / 49 & 12 / 49 \\ \alpha / 49 & -11 / 49 & -5 / 49 \end{array}\right] \\ & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} (3-8+3 \alpha / 49 & (39-6-33) / 49 & (-9+24-15) / 49 \\ (4-4+0) / 49 & (52-3+0) / 49 & (-12+12+0) / 49 \\ (2-20+\alpha) / 49 & (26-15-11) / 49 & (-6+60-5) / 49 \end{array}\right] \\ & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} (-5+3 \alpha) / 49 & 0 & 0 \\ 0 & 1 & 0 \\ (-18+\alpha) / 49 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{-5+3 \alpha}{49}=1 \text { and } \frac{-18+\alpha}{49}=0 \\ & \Rightarrow \quad 3 \alpha=54 \text { and } \alpha=18 \\ & \Rightarrow \quad \alpha=18 \text { and } \alpha=18 \end{aligned}$

Thus, $\alpha=18$

$\text { Given, } A=\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]$

$ \begin{aligned} & \Rightarrow \quad \operatorname{adj} A=\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ -(-\cos \theta) & \sin \theta \end{array}\right] \\ & {\left[\because \text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \text {, then adj } A=\left[\begin{array}{cc} d & -c \\ -b & a \end{array}\right]\right] } \end{aligned} $

$\begin{aligned} & \Rightarrow \operatorname{adj} A=A \\ & \begin{aligned} \operatorname{Now,} A\left(\operatorname{adj}(A)^{-1}\right) & =A(A)^{-1} \\ = & I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad\left[\because A A^{-1}=A^{-1} A=I\right] \end{aligned} \end{aligned}$

Given, system of equations has a non-trivial solution

$\therefore \quad\left|\begin{array}{ccc} a-1 & -1 & -1 \\ -1 & b-1 & -1 \\ -1 & -1 & c-1 \end{array}\right|=0$

On applying $C_2 \rightarrow C_2-C_3$, we get

$\left|\begin{array}{ccc} a-1 & 0 & -1 \\ -1 & b & -1 \\ -1 & -c & c-1 \end{array}\right|=0$

On applying $R_2 \rightarrow R_2-R_3$, we get

$\left|\begin{array}{ccc} a-1 & -1 & -1 \\ 0 & b+c & -c \\ -1 & -c & c-1 \end{array}\right|=0$

On applying $R_3 \rightarrow R_3-R_1$, we get

$\left|\begin{array}{ccc} a-1 & 0 & -1 \\ 0 & b+c & -c \\ a & -c & c \end{array}\right|=0$

On expanding along $R_1$, we get

$\begin{aligned} & (a-1)\left(b c+c^2-c^2\right)-1(a b+a c)=0 \\ & \Rightarrow \quad(a-1)(b c)-a b-a c=0 \\ & \Rightarrow \quad a b c-b c-a b-a c=0 \\ & \Rightarrow \quad a b+b c+a c=a b c \\ & \end{aligned}$