Matrices and Determinants

We have, $A^T=\alpha A+I$, where $A$ is $2 \times 2$ matrix and $\alpha \in R-\{-1,1\}$

$ \begin{aligned} \left(A^T\right)^T & =\alpha A^T+I \\\\ A & =\alpha A^T+I \\\\ A & =\alpha(\alpha A+I)+I \left[\because A^T=\alpha A+I\right]\\\\ A & =\alpha^2 A+(\alpha+1) I \\\\ A & \left(1-\alpha^2\right)=(\alpha+1) I \\\\ A & =\frac{(\alpha+1)}{(1-\alpha)(1+\alpha)} I \end{aligned} $

$ \begin{aligned} A & =\frac{1}{1-\alpha} I ..........(i)\\\\ |A| & =\frac{1}{(1-\alpha)^2} ...........(ii) \end{aligned} $

Also,

$ \begin{aligned} \left|A^2-A\right| & =4 \\\\ |A||A-I| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{1}{1-\alpha}-1\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2}\left|\left(\frac{\alpha}{1-\alpha}\right) I\right| & =4 \\\\ \frac{1}{(1-\alpha)^2} \times \frac{\alpha^2}{(1-\alpha)^2} & =4 \\\\ \frac{\alpha^2}{(1-\alpha)^4} & =4 \\\\ \frac{\alpha}{(1-\alpha)^2} & = \pm 2 \\\\ 2(1-\alpha)^2 & = \pm \alpha \end{aligned} $

If $2(1-\alpha)^2=\alpha$, then $2 \alpha^2-5 \alpha+2=0$

Sum of value of $\alpha=\frac{5}{2} \quad\left[\because\right.$ Sum of zero $\left.=\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\right]$

If $2(1-\alpha)^2=-\alpha$, then $2 \alpha^2-3 \alpha+2=0$

$\therefore$ No real value of $\alpha$

Hence, sum of all values of $\alpha=\frac{5}{2}$

Given that

$ \begin{aligned} & A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right] \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right| \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !} \times 5 ! 6 ! 7 !\left|\begin{array}{lll} 1 & 6 & 7 \times 6 \\ 1 & 7 & 8 \times 7 \\ 1 & 8 & 9 \times 8 \end{array}\right| \end{aligned} $

$ \begin{aligned} & R_3 \rightarrow R_3-R_2 \text { and } R_2 \rightarrow R_2-R_1 \\\\ & |A|=\left|\begin{array}{lll} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{array}\right|=2 \\\\ & |\operatorname{adj}(\operatorname{adj}(2 A))|=|2 A|^{(n-1)^2} \\\\ & =|2 A|^4=\left(2^3|A|\right)^4=2^{12}|A|^4=2^{16} \end{aligned} $

Given that $A$ is $3 \times 3$ matrix and $|A|=2$

$ \begin{aligned} & \text { Now, | 3adj }\left(|3 A| A^2\right) \text { | } \\\\ & =3^3\left|\operatorname{adj}\left(|3 A| A^2\right)\right| \\\\ & =3^3\left|\operatorname{adj}\left(54 A^2\right)\right| \\\\ & =3^3\left|54 A^2\right|^2 \\\\ & =3^3 \times\left(54^3\right)^2 \times|A|^4 \\\\ & =3^3 \times(54)^6 \times 2^4 ~~~~~ {[|A|=2 \text { given }]} \\\\ & =3^3 \times\left(3^3 \times 2\right)^6 \times 2^4 ~~~~~ {\left[\left(a^m\right)^n=a^{m n}\right]} \\\\ & =3^{11} \times 3^{10} \times 2^{10} ~~~~~~~ {\left[(a b)^m=a^m b^m\right]} \\\\ & =(3)^{11} \times(6)^{10} \end{aligned} $

Given system of linear equation is

$ \begin{gathered} 2 x-y+3 z=5 \\\\ 3 x+2 y-z=7 \\\\ 4 x+5 y+\alpha z=\beta \\\\ \text { Now, } \Delta=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{array}\right|=7(\alpha+5) \end{gathered} $

So, this system of equation has unique solution, if $\alpha \neq-5$

and $\Delta_1=\left|\begin{array}{ccc}5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & \alpha\end{array}\right|=17 \alpha-5 \beta+30$

and $\Delta_2=\left|\begin{array}{ccc}2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & \alpha\end{array}\right|=-11 \beta+\alpha+104$

and $\Delta_3=\left|\begin{array}{ccc}2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta\end{array}\right|=7(\beta-9)$

For infinitely many solutions,

$ \Delta=\Delta_1=\Delta_2=\Delta_3=0 $

$ \begin{aligned} & \text { So, } \Delta=0 \\\\ & \Rightarrow 7(\alpha+5)=0 \\\\ & \Rightarrow \alpha=-5 \end{aligned} $

$ \begin{aligned} & \Delta_3 =0 \\\\ & \Rightarrow 7(\beta-9) =0 \\\\ & \Rightarrow \beta =9 \end{aligned} $

If $\alpha=-5$ and $\beta=8$, then $\Delta$ equals zero but $\Delta_3$ does not, which would imply the system is inconsistent for $\alpha=-5$ and $\beta=8$.

Therefore, the option "The system is inconsistent for $\alpha=-5$ and $\beta=8$ " is correct.

$ \begin{aligned} & \mathrm{A}=\left[\begin{array}{cc} 1 & 5 \\ \lambda & 10 \end{array}\right] \\\\ & \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\ & \Rightarrow\left|\begin{array}{cc} 1-x & 5 \\ \lambda & 10-x \end{array}\right|=0 \\\\ & \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\ & \Rightarrow 10-11 x+x^2-5 \lambda=0 \end{aligned} $

Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$

$ \Rightarrow \alpha A^2+\beta A-I=0 $

and $A^2-11 A+(10-5 \lambda) I=0$

On solving, we get

$ \alpha=\frac{1}{5}, \beta=-\frac{11}{5} $

$ \begin{aligned} & \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\ & \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\ & =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\ & =\frac{125}{25}+9=14 \end{aligned} $

Since, the given system has a non trivial solution,

$ \text { So, } \Delta=0 $

$ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow 1\left(\tan ^2 \theta-\sqrt{7}\right)-1(-\tan \theta-\sqrt{7})+\sqrt{3}(-1-\tan \theta)=0 \\\\ & \Rightarrow \tan ^2 \theta-\sqrt{7}+\tan \theta+\sqrt{7}-\sqrt{3}-\sqrt{3} \tan \theta=0 \\\\ & \Rightarrow \tan \theta(\tan \theta-\sqrt{3})+1(\tan \theta-\sqrt{3})=0 \\\\ & \Rightarrow \tan \theta=\sqrt{3} \text { or } \tan \theta=-1 \\\\ & \therefore \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\\\ & \text { So, } \frac{120}{\pi} \sum \theta=\frac{120}{\pi}\left\{\frac{4 \pi-8 \pi-3 \pi+9 \pi}{12}\right\} \\\\ & =\frac{120}{\pi}\left[\frac{2 \pi}{12}\right]=20 \end{aligned} $

We have,

$ \begin{aligned} & |\mathrm{A}|=\left|\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right|=2(4-1)-1(2-0)+0 \\\\ & =6-2=4 \\\\ & \text { So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\ & \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\ & =(32)^{2^3}=32^8 \\\\ & \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\ & \Rightarrow 16^n=16^{2+8} \Rightarrow n=10 \end{aligned} $

Concepts :

(a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$

(b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$

$ \text { Here, } P=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right], A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] $

$ \text { Here, } \mathrm{PP}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right] $

$ =\left[\begin{array}{cc} \frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} $

Similarly $P^T P=1$

$ \because $ $Q=P A P^{T}$

Now, $Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007$ times $=P A^{2007} P^T$

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

$ A^3=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] $
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$ A^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] $

$ \begin{aligned} & \text { Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow a=1, b=2007, c=0, d=1 \\\\ & \therefore 2 a+b-3 c-4 d=2(1)+2007-3(0)-4(1) \\\\ & =2+2007-4=2005 \end{aligned} $

We have, $P^2=I-P$

$ \begin{aligned} \Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\ & =2 I-3 P \text { [Using Eq. (i)] } \end{aligned} $

$ \begin{aligned} \Rightarrow P^8 & =(2 I-3 P)^2 \\\\ & =4 I+9 P^2-12 P \\\\ & =13 I-21 P \text { [Using Eq. (i)] } \end{aligned} $

and

$ \begin{aligned} P^6 & =(I-P)(2 I-3 P) \\\\ & =2 I-3 P-2 P+3 P^2 \\\\ & =5 I-8 P \text { [Using Eq. (i)] } \end{aligned} $

$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\\\ & \text { and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$

Given system of equations,

$ \begin{aligned} x+y+z & =6 ........(i)\\\\ x+2 y+\alpha z & =10 ........(ii)\\\\ x+3 y+5 z & =\beta ........(iii) \end{aligned} $

Here,

$ \begin{aligned} \Delta & =\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{array}\right| \\\\ & =1(10-3 \alpha)-1(5-\alpha)+1(3-2) \\\\ & =10-3 \alpha-5+\alpha+1 \\\\ & =6-2 \alpha \end{aligned} $

For unique solution, $\Delta \neq 0$

$ \Rightarrow 6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3 $

When, $\alpha=3$

$ \begin{aligned} \Delta_1 & =\left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \beta & 3 & 5 \end{array}\right| \\\\ & =\left|\begin{array}{ccc} 0 & 1 & 0 \\ -2 & 2 & 1 \\ \beta-18 & 3 & 2 \end{array}\right|=-1(-4-\beta+18) \\\\ & =\beta-14 \end{aligned} $

and

$\begin{aligned} \Delta_2 & =\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5\end{array}\right| \\\\ & =1(50-3 \beta)-6(5-3)+1(\beta-10) \\\\ & =50-3 \beta-12+\beta-10 \\\\ & =28-2 \beta=2(14-\beta)\end{aligned}$

and

$ \begin{aligned} \Delta_3 & =\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 3 & \beta \end{array}\right| \\ & =1(2 \beta-30)-1(\beta-10)+6(3-2) \\\\ & =2 \beta-30-\beta+10+6 \\\\ & =\beta-14 \end{aligned} $

Thus, at $\beta=14, \Delta_1=\Delta_2=\Delta_3=0$

$ \Rightarrow \alpha=3, \beta=14$

So, system has infinite solutions.

Given system of equations,

$ \text { and } \quad \begin{aligned} x+y+a z & =b \\ 2 x+5 y+2 z & =6 \\ x+2 y+3 z & =3 \end{aligned} $

Since, given system of equation has infinitely many solutions

$ \therefore D=0 \text { and } D_1=D_2=D_3=0 $

$ \begin{aligned} & \text { Here, } D=\left|\begin{array}{lll} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow 1(15-4)-1(6-2)+a(4-5)=0 \\\\ & \Rightarrow -a+11-4=0 \\\\ & \Rightarrow a=7 \end{aligned} $

$ \begin{aligned} & \text { and } D_1=\left|\begin{array}{ccc} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow b(15-4)-1(18-6)+a(12-15)=0 \\\\ & \Rightarrow 11 b-12-3 a=0 \\\\ & \Rightarrow 11 b-12-21=0 ~~~~~~~(\because a=7)\\\\ & \Rightarrow 11 b-33=0 \\\\ & \Rightarrow b=3 \\\\ & \therefore 2 a+3 b=2(7)+3(3)=14+9=23 \end{aligned} $

Given, $A^2=I$

and $b=|A|$

Let $ A=\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] $

$ \begin{aligned} \therefore \quad A^2 & =\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right]\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] \\\\ & =\left[\begin{array}{cc} a_1^2+b_1 a_2 & a_1 b_1+b_1 b_2 \\ a_1 a_2+a_2 b_2 & b_1 a_2+b_2^2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \left(\because A^2=I\right) \end{aligned} $

By equality of matrices, we get

$ \begin{aligned} & a_1 a_2+a_2 b_2=0 \text { and } a_1 b_1+b_1 b_2=0 \\\\ &\Rightarrow a_2\left(a_1+b_2\right)=0 \text { and } b_1\left(a_1+b_2\right)=0 \\\\ &\Rightarrow a_1+b_2=0 \text { (since, } b_1, a_2 \neq 0 \text { ) }\\\\ &\Rightarrow \text { Sum of diagonal elements }=0 \\\\ &\Rightarrow a=0 \end{aligned} $

$ \begin{aligned} & \text { Now, } A^2=I \\\\ & \Rightarrow \left|A^2\right|=1 \\\\ & \Rightarrow |A|^2=1 ~~~~~~~\left(\because\left|A^n\right|=|A|^n\right)\\\\ & \Rightarrow b^2=1 ~~~~~~~~~~~(\because|A|=b)\\\\ & \therefore 3 a^2+4 b^2=3(0)+4(1)=4 \end{aligned} $

For infinite solution $\Delta=\Delta_x=\Delta_y=\Delta_z=0$

$ \Delta=\left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2 $

If $\beta=1$, then all planes are overlapping

$\therefore$ Option (A) is correct.

Option (B) :

If $\alpha=2 \Rightarrow \Delta \neq 0$

$\therefore $ Unique solution exist

$\therefore$ Option (B) is incorrect.

Option (C) :

$ \begin{aligned} & \alpha=2, \beta=1 \\\\ & 2 x+y+z=1 \\\\ & x+2 y+z=1 \\\\ & x+y+2 z=1 \end{aligned} $

Adding all three equations,

$ x+y+z=\frac{3}{4} $

$\therefore$ option (C) is correct.

Option (D) :

If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$

$\therefore$ No solution.

$\therefore$ Option (D) is correct.

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

Let $\theta=\frac{\pi}{3}$

$ \begin{aligned} A^2 & =\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right] \\\\ A^3 & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 3 \theta & \sin 3 \theta \\\\ -\sin 3 \theta & \cos 3 \theta \end{array}\right] \end{aligned} $

$\begin{aligned} & \therefore \quad A^{30}=\left[\begin{array}{cc}\cos 30 \theta & \sin 30 \theta \\\\ -\sin 30 \theta & \cos 30 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\ & A^{25}=\left[\begin{array}{cc}\cos 25 \theta & \sin 25 \theta \\\\ -\sin 25 \theta & \cos 25 \theta\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]=A \\\\ & \therefore \quad A^{30}+A^{25}-A=I\end{aligned}$

$\left|\begin{array}{lll}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$

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

$\Rightarrow$ $\lambda=1,-2$

For $\lambda=1 \Rightarrow \infty$ solution

$\lambda=-2 \Rightarrow$ no solution

$\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6$

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

$ \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} $

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

For option A :

${R_1} \to {R_1} + {R_2}$

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

$\therefore$ Option A can be obtained.

For option B :

${R_1} \leftrightarrow {R_2}$

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

$\therefore$ Option B can be obtained.

Option C :

Not possible by a single elementary row operation.

Option D :

${R_2} \to {R_2} + 2{R_1}$

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

$\therefore$ Option D can be obtained.

Given,

$x + y + z = 6$ ...... (1)

$2x + 5y + \alpha z = \beta $ ..... (2)

$x + 2y + 3z = 14$ ...... (3)

System of equation have infinite many solutions.

$\therefore$ ${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$ and $\Delta = 0$

Now, $\Delta = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 5 & \alpha \cr 1 & 2 & 3 \cr } } \right| = 0$

${C_1} \to {C_1} - {C_3}$

${C_2} \to {C_2} - {C_3}$

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

$ \Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$

$ \Rightarrow 8 - \alpha = 0$

$ \Rightarrow \alpha = 8$

Now, $x + y + z = 6$

$2x + 5y + 8z = \beta $

$x + 2y + 3z = 14$

$\therefore$ ${\Delta _x} = \left| {\matrix{ 6 & 1 & 1 \cr \beta & 5 & 8 \cr {14} & 2 & 3 \cr } } \right| = 0$

${C_1} \to {C_1} - 6{C_3}$

${C_2} \to {C_2} - {C_3}$

$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {\beta - 48} & { - 3} & 8 \cr { - 4} & { - 1} & 3 \cr } } \right| = 0$

$ \Rightarrow - \beta + 48 - 12 = 0$

$ \Rightarrow \beta = 36$

$\therefore$ $\alpha + \beta = 8 + 36 = 44$

AB is zero matrix

$ \Rightarrow |A| = |B| = 0$

So neither A nor B is invertible

If $|A| = 0$

$ \Rightarrow |\mathrm{adj}\,A| = 0$ so $\mathrm{adj}\,A$

$AX = 0$ is homogeneous system and $|A| = 0$

So, it is having infinitely many solutions

(A) $M = {A^4} - {B^4}$

${M^T} = {({A^4} - {B^4})^T} = {({A^T})^4} - {({B^T})^4}$

$ = {A^4} - {( - B)^4} = {A^4} - {B^4} = M$

(B) $M = AB - BA$

${M^T} = {(AB - BA)^T} = {(AB)^T} - {(BA)^T}$

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

$ = - BA - A( - B)$

$ = AB - BA = M$

(C) $M = {B^5} - {A^5}$

${M^T} = {({B^T})^5} - {({A^T})^5} = - ({B^5} + {A^5}) \ne - M$

(D) $M = AB + BA$

${M^T} = {(AB)^T} + {(BA)^T}$

$ = {B^T}{A^T} + {A^T}{B^T} = - BA - AB = - M$

$A = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right]$

$ \Rightarrow {A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] \times \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]$

$ \Rightarrow {A^3} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = l$

Now ${B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}$

${B_0} = A + 2{A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & 2 \cr 2 & 0 & 0 \cr 0 & 2 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 1 & 2 \cr 2 & 0 & 1 \cr 1 & 2 & 0 \cr } } \right]$

$|{B_0}| = 9$

Since, ${B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}$

Hence $|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}$

$ = |{3^2}{|^{16}} = {3^{32}}$

Characteristic equation of A is given by

$\left| {A - \lambda I} \right| = 0$

$\left| {\matrix{ {4 - \lambda } & { - 2} \cr \alpha & {\beta - \lambda } \cr } } \right| = 0$

$ \Rightarrow {\lambda ^2} - (4 + \beta )\lambda + (4\beta + 2\alpha ) = 0$

So, ${A^2} - (4 + \beta )A + (4\beta + 2\alpha )I = 0$

$|A| = 4\beta + 2\alpha = 18$

${A^2} = \left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right]\left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right] = \left[ {\matrix{ { - 3} & { - 8} \cr 8 & {21} \cr } } \right]$

$\alpha {A^2} + \beta A = \left[ {\matrix{ { - 3\alpha } & { - 8\alpha } \cr {8\alpha } & {21\alpha } \cr } } \right] + \left[ {\matrix{ \beta & {2\beta } \cr { - 2\beta } & { - 5\beta } \cr } } \right]$

$ = \left[ {\matrix{ { - 3\alpha + \beta } & { - 8\alpha + 2\beta } \cr {8\alpha - 2\beta } & {21\alpha - 5\beta } \cr } } \right] = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]$

On Comparing

$8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2$

$ \Rightarrow \alpha = 2,\,\beta = 8$

So, $\alpha + \beta = 10$

$A'BA = \left[ {\matrix{ 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ {{9^2}} & { - {{10}^2}} & {{{11}^2}} \cr {{{12}^2}} & {{{13}^2}} & { - {{14}^2}} \cr { - {{15}^2}} & {{{16}^2}} & {{{17}^2}} \cr } } \right]A$

$ = \left[ {\matrix{ {{9^2} + {{12}^2} - {{15}^2}} & { - {{10}^2} + {{13}^2} + {{16}^2}} & {{{11}^2} - {{14}^2} + {{17}^2}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$ = \left[ {{9^2} + {{12}^2} - {{15}^2} - {{10}^2} + {{13}^2} + {{16}^2} + {{11}^2} - {{14}^2} + {{17}^2}} \right]$

$ = [({9^2} - {10^2}) + ({11^2} + {12^2}) + ({13^2} - {14^2}) + ({16^2} - {15^2}) + {17^2}]$

$ = [ - 19 + 265 + ( - 27) + 31 + 289]$

$ = [585 - 46] = [539]$

$\Delta = \left| {\matrix{ 8 & 1 & 4 \cr 1 & 1 & 1 \cr \lambda & { - 3} & 0 \cr } } \right|$

$ = 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$

$ = 24 + \lambda - 12 - 4\lambda $

$ = 12 - 3\lambda $

So for $\lambda = 4$, it is having infinitely many solutions.

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

$ = - 2(3) - 1( - \mu )+4( - \mu )$

$ = - 6 - 3\mu = 0$

For $\mu = - 2$

Distance of $(4, - 2,{{ - 1} \over 2})$ from $8x + y + 4z + 2 = 0$

$ = {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$ units

$|(A + I)(adj\,A + I)| = 4$

$ \Rightarrow |A\,adj\,A + A + adj\,A + I| = 4$

$ \Rightarrow |(A)I + A + adj\,A + I| = 4$

$|A| = - 1 \Rightarrow |A + adj\,A| = 4$

$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]\,adj\,A = \left[ {\matrix{ a & { - b} \cr { - c} & d \cr } } \right]$

$ \Rightarrow \left| {\matrix{ {(a + d)} & 0 \cr 0 & {(a + d)} \cr } } \right| = 4$

$ \Rightarrow a + d = \, \pm \,2$

$\Delta = \left| {\matrix{ 2 & { - 3} & 5 \cr 1 & 3 & { - 1} \cr 3 & { - 1} & {{\lambda ^2} - |\lambda |} \cr } } \right| = 2\left( {3{\lambda ^2} - 3|\lambda | - 1} \right) + 3\left( {{\lambda ^2} - |\lambda | + 3} \right) + 5( - 1 - 9)$

$ = 9{\lambda ^2} - 9|\lambda | - 43$

$ = 9|\lambda {|^2} - 9|\lambda | - 43$

$\Delta = 0$ for 2 values of $|\lambda |$ out of which one is $-\mathrm{ve}$ and other is $+\mathrm{ve}$

So, 2 values of $\lambda$ satisfy the system of equations to obtain no solution.

Given,

$3(\sin 3\theta )x - y + z = 2$

$3(\cos 2\theta )x + 4y + 3z = 3$

$6x + 7y + 7z = 9$

For no solutions determinant of coefficient will be = 0

$\therefore$ $D = \left| {\matrix{ {3\sin 3\theta } & { - 1} & 1 \cr {3\cos 2\theta } & 4 & 3 \cr 6 & 7 & 7 \cr } } \right| = 0$

$ \Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$

$ \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0$

$ \Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0$

$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta + 2(1 - 2{\sin ^2}\theta ) - 2 = 0$

$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta - 4{\sin ^2}\theta = 0$

$ \Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0$

$ \Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0$

$\therefore$ $\sin \theta = 0$

$ \Rightarrow \theta = \pi ,2\pi ,3\pi $ when $\theta \in (0,4\pi )$

or,

$4{\sin ^2}\theta + 4\sin \theta - 3 = 0$

$ \Rightarrow 4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0$

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

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

$\therefore$ $\sin \theta = {1 \over 2}$

or,

$\sin \theta = - {3 \over 2}$ [not possible as $\sin \in [ - 1,1]$]

$\therefore$ $\sin \theta = {1 \over 2}$

$ \Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$

$\therefore$ Possible values of $\theta = \pi ,2\pi ,3\pi ,{\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$

$\therefore$ Total 7 values of $\theta$ possible.

Given,

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

and $B = \left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right]$

$AB = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right]\left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right]$

$ = \left[ {\matrix{ {4 + 4\alpha } & { - 4\alpha - 4} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right]$

Given,

$|AB| = 0$

$\therefore$ $\left| {\matrix{ {4 + 4\alpha } & { - 4\alpha - 4} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right| = 0$

$ \Rightarrow (4\alpha + 4)\left| {\matrix{ 1 & { - 1} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right| = 0$

$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 9 + 2\alpha - 6) = 0$

$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 2\alpha + 3) = 0$

$\therefore$ $\alpha - = - 1$

or ${\alpha ^2} + 2\alpha + 3 = 0$

${\alpha _1} + {\alpha _2} = - 2$

$\therefore$ Sum of all values of $\alpha = - 1 - 2 = - 3$

$\therefore$ Absolute value of $\alpha = | - 3| = 3$

Given,

$\det (A) = 2$,

$\det (B) = 3$

and $\det \left( {\left( {\det \left( {5\left( {\det A} \right)B} \right)} \right){A^2}} \right) = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\det \left( {5\left( {\det A} \right)B} \right){A^2}} \right| = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\left| {5\left( {\det A} \right)\left. B \right|{A^2}} \right.} \right| = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\left| {5\left| {A\left| B \right.\left| {{A^2}} \right.} \right.} \right.} \right| = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\left| {5\,.\,2\,.\,B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$\Rightarrow \left| {\left| {10B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow \left| {{{10}^2}\,.\,\left| B \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

As $\left| {k\,.\,A} \right| = {k^n}|A|$

$ \Rightarrow \left| {100 \times 3{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {(300)^2}\,.\,|{A^2}| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {(300)^2}\,.\,|A{|^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {(300)^2}\,.\,{2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow 9 \times 100 \times 100 \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {3^2} \times {2^2} \times {5^2} \times {2^2} \times {5^2} \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {2^6}\,.\,{3^2}\,.\,{5^4} = {2^a}\,.\,{3^b}\,.\,{5^c}$

Comparing both sides, we get

$a = 6$, $b = 2$, $c = 4$

$\therefore$ $a + b + c = 6 + 2 + 4 = 12$

Given $A = \left[ {\matrix{ 2 & { - 1} \cr 0 & 2 \cr } } \right]$

and

$B = I - {5_{{C_1}}}(adj\,A) + {5_{{C_2}}}{(adj\,A)^2} - {5_{{C_3}}}{(adj\,A)^3} + {5_{{C_4}}}{(adj\,A)^4} - {5_{{C_5}}}{(adj\,A)^5}$

$ = {\left( {I - (adj\,A)} \right)^5}$

Cofactor of $A = \left[ {\matrix{ {{{( - 1)}^{1 + 1}}\,.\,2} & {{{( - 1)}^{1 + 2}}\,.\,0} \cr {{{( - 1)}^{2 + 1}}\,.\,( - 1)} & {{{( - 1)}^{2 + 2}}\,.\,2} \cr } } \right]$

$ = \left[ {\matrix{ 2 & 0 \cr 1 & 2 \cr } } \right]$

Transpose of cofactor of $A = \left[ {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right]$

$\therefore$ $adj\,A = \left[ {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right]$

Now, $I - adj\,A$

$ = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] - \left[ {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right]$

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

Now let,

$P = I - adj\,A = \left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]$

$\therefore$ ${P^2} = \left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]\left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]$

$ = \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$

${P^4} = {P^2}\,.\,{P^2} = \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 4 \cr 0 & 1 \cr } } \right]$

${P^5} = {P^4}\,.\,P = \left[ {\matrix{ 1 & 4 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right] = \left[ {\matrix{ { - 1} & { - 5} \cr 0 & { - 1} \cr } } \right]$

$\therefore$ $B = \left[ {\matrix{ { - 1} & { - 5} \cr 0 & { - 1} \cr } } \right]$

Now sum of elements $ = - 1 - 5 - 1 + 0 = - 7$

$2x + y - z = 7$

$x - 3y + 2z = 1$

$x + 4y + \delta z = k$

$\Delta = \left| {\matrix{ 2 & 1 & { - 1} \cr 1 & { - 3} & 2 \cr 1 & 4 & \delta \cr } } \right| = - 7\delta - 21 = 0$

$\delta = - 3$

${\Delta _1} = \left| {\matrix{ 7 & 1 & { - 1} \cr 1 & { - 3} & 2 \cr k & 4 & { - 3} \cr } } \right|$

$ \Rightarrow 6 - k = 0 \Rightarrow k = 6$

$\delta + k = - 3 + 6 = 3$

Given, ${a_{ij}} = {2^{j - i}}$

Now, $A = \left[ {\matrix{ {{2^0}} & {{2^1}} & {{2^2}} \cr {{2^{ - 1}}} & {{2^0}} & {{2^1}} \cr {{2^{ - 2}}} & {{2^{ - 1}}} & {{2^0}} \cr } } \right]$

$ = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$

${A^2} = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$

$ = \left[ {\matrix{ {1 + 1 + 1} & {2 + 2 + 2} & {4 + 4 + 4} \cr {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} & {2 + 2 + 2} \cr {{1 \over 4} + {1 \over 4} + {1 \over 4}} & {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} \cr } } \right]$

$ = \left[ {\matrix{ 3 & 6 & {12} \cr {{3 \over 2}} & 3 & 6 \cr {{3 \over 4}} & {{3 \over 2}} & 3 \cr } } \right]$

$ = 3\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$

$ = 3A$

Similarly, ${A^3} = {3^2}A$

${A^4} = {3^3}A$

$\therefore$ ${A^2} + {A^3} + \,\,......\,\, + \,\,{A^{10}}$

$ = 3A + {3^2}A + {3^3}A + \,\,......\,\, + \,\,{3^9}A$

$ = A(3 + {3^2} + {3^3} + \,\,......\,\, + \,\,{3^9})$

$ = A\left( {{{3({3^9} - 1)} \over {3 - 1}}} \right) = {{3({3^9} - 1)} \over 2}A$ = $\left( {{{{3^{10}} - 3} \over 2}} \right)A$

$\Delta = \left| {\matrix{ 2 & 3 & { - 1} \cr 1 & 1 & 1 \cr 1 & { - 1} & {|\lambda |} \cr } } \right| = 0 \Rightarrow |\lambda | = 7$

But at $\lambda = 7,\,{D_x} = {D_y} = {D_z} = 0$

${P_1}:2x + 3y - z = - 2$

${P_2}:x + y + z = 4$

${P_3}:x - y + |\lambda |z = 4\lambda - 4$

So clearly $5{P_2} - 2{P_1} = {P_3}$, so at $\lambda = 7$, system of equation is having infinite solutions.

So $\lambda = - 7$ is correct answer.