Matrices and Determinants

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

Multiply by $A$

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

Multiply again by $A$

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

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

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

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

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

Given

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

So,

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

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

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

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

Now,

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

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

So,

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

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

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

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

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

• $\det(kM)=k^3\det(M)$

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

• $\det((a-1)A)=(a-1)^3\det(A)$

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

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

Now,

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

So,

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

Put $a=3$:

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

Hence,

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

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

Answer: 16 (Option D)

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

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

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

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

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

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

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

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

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

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

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

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

Evaluating each component:

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

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

$\log_4 5$ using change of base:

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

$\log_5 8$:

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

$\log_4 25$:

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

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

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

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

$ A_{11} = \log_4 25 $

$ A_{12} = -\log_5 8 $

$ A_{21} = -\log_4 5 $

$ A_{22} = \log_5 128 $

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

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

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

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

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

Thus, matrix $ C $ is:

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

To find $|C|$, we compute:

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

Finally, calculate $ 8|C| $:

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

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

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

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

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

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

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

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

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

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

Now post multiply by P

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We begin with the system:

$ \begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned} $

Step 1. Solve the first equation for $x$:

$ x = 6 - y - 2z. $

Step 2. Substitute $x = 6-y-2z$ into the second equation:

$ 2(6-y-2z) + 3y + az = a+1. $

Expanding and simplifying:

$ 12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11. $

Call this Equation (I).

Step 3. Substitute $x = 6-y-2z$ into the third equation:

$ -(6-y-2z) - 3y + bz = 2b. $

Expanding and simplifying:

$ -6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6. $

Call this Equation (II).

Step 4. For the system to have infinitely many solutions, the two equations in $y$ and $z$ must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant $k$ such that

$ -2 = k \cdot 1 \quad \Longrightarrow \quad k = -2. $

Apply this to the coefficient of $z$ and the constant term.

For the $z$-coefficient in Equations (I) and (II):

$ b+2 = k(a-4) = -2(a-4) = -2a+8. $

Thus,

$ b = -2a+6. $

For the constant term:

$ 2b+6 = k(a-11) = -2(a-11) = -2a + 22. $

Substitute $b = -2a+6$ into this equation:

$ 2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22. $

Simplify:

$ -4a + 18 = -2a + 22. $

Solve for $a$:

$ -4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22, $

$ 2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2. $

Substitute $a = -2$ into $b = -2a+6$:

$ b = -2(-2) + 6 = 4 + 6 = 10. $

Step 5. We now compute

$ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. $

Thus, the value of $7a+3b$ is

$ \boxed{16}. $

$ B = \operatorname{adj}(\operatorname{adj}(2A)) = \det(2A) \cdot (2A) $

Since $ A $ is a $ 3 \times 3 $ matrix with

$ \det(A) = \frac{1}{2}, $

the determinant of $ 2A $ is computed as

$ \det(2A) = 2^3 \det(A) = 8 \cdot \frac{1}{2} = 4. $

Thus,

$ B = 4 \cdot (2A) = 8A. $

Now, compute the determinant and the trace of $ B $:

Determinant of $ B $:

$ \det(B) = \det(8A) = 8^3 \det(A) = 512 \cdot \frac{1}{2} = 256. $

Trace of $ B $:

$ \operatorname{trace}(B) = \operatorname{trace}(8A) = 8 \cdot \operatorname{trace}(A) = 8 \cdot 3 = 24. $

Finally, adding these results:

$ \det(B) + \operatorname{trace}(B) = 256 + 24 = 280. $

Let us consider the matrices given:

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

$ P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $

$ Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} $

$ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $

where $ x, y, z, a, b, c, d $ are non-zero real numbers. We aim to verify properties of matrix $ Q $ given the relation $ Q R = R P $.

Steps and Calculations

Matrix Equation Analysis:

Given $ Q R = R P $, we equate both sides:

$ \begin{pmatrix} a x + c y & b x + d y \\ a z + 4c & b z + 4d \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix} $

From the matrix equality, we derive the following key equations:

$ a x + c y = 2a $

$ b x + d y = 3b $

$ a z + 4c = 2c $

$ b z + 4d = 3d $

Determinant Condition:

Comparing the equations:

Set $ a z = -2c $ and $ b z = -d $, resulting in $ \frac{a}{b} = \frac{2c}{d} $.

The requirement $ |R| \neq 0 $ implies $ |P| = |Q| $. So, equating determinants gives us:

$ |P| = |Q| \Rightarrow 6 = 4x - yz \quad \text{(Equation 1)} $

Condition for Non-Trivial Solutions:

From:

$ a(x - z) + c y = 0 $

$ a z + 2c = 0 $

implies a non-trivial solution matrix must have:

$ \left|\begin{array}{cc} x-2 & y \\ z & 2 \end{array}\right| = 0 \Rightarrow 2x - yz = 4 \quad \text{(Equation 2)} $

Solve for $ x $, $ yz $:

Solving Equations 1 and 2:

$ 4x - yz = 6 $

$ 2x - yz = 4 $

Subtract the second from the first:

$ (4x - yz) - (2x - yz) = 6 - 4 \Rightarrow 2x = 2 \Rightarrow x = 1 $

Plugging $ x = 1 $ back, we find:

$ 4(1) - yz = 6 \Rightarrow 4 - yz = 6 \Rightarrow yz = -2 $

Verify Determinants:

For $ Q - 2I $:

$ |Q - 2I| = \left|\begin{array}{cc} x-2 & y \\ z & 2 \end{array}\right| = 2x - yz = 0 $

For $ Q - 6I $:

$ |Q - 6I| = \left|\begin{array}{cc} x-6 & y \\ z & -2 \end{array}\right| = -2x + 12 - yz = 12 $

For $ Q - 3I $:

$ |Q - 3I| = \left|\begin{array}{cc} x-3 & y \\ z & 1 \end{array}\right| = x - 3 - yz = 0 $

Therefore, using the derived values, the statements for the determinants align correctly. The findings are:

$ |Q - 2I| = 0 $

$ |Q - 6I| = 12 $

$ yz = -2 $

These calculations validate the conditions for matrix $ Q $ concerning the predefined matrix properties.

To find the number of $3 \times 3$ invertible matrices $Q$ with integer entries that satisfy both $Q^{-1} = Q^T$ and $PQ = QP$, consider the given matrix $P$:

$ P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

The condition $PQ = QP$ implies:

$ \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

This results in:

$ \begin{pmatrix} 2a & 2b & 2c \\ 2d & 2e & 2f \\ 3g & 3h & 3i \end{pmatrix} = \begin{pmatrix} 2a & 2b & 3c \\ 2d & 2e & 3f \\ 2g & 2h & 3i \end{pmatrix} $

From this, we derive that:

$ c = 0, \quad f = 0, \quad h = 0, \quad g = 0 $

Thus, the matrix $Q$ is of the form:

$ Q = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & e \end{pmatrix} $

Next, using the condition $Q^T = Q^{-1}$, we have the equation:

$ Q Q^T = I $

This results in:

$ \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & i \end{pmatrix} \begin{pmatrix} a & c & 0 \\ b & d & 0 \\ 0 & 0 & i \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $

This simplifies to:

$ \begin{pmatrix} a^2 + b^2 & ac + bd & 0 \\ ac + bd & c^2 + d^2 & 0 \\ 0 & 0 & i^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $

From which we conclude:

$ a^2 + b^2 = 1, \quad ac + bd = 0, \quad c^2 + d^2 = 1, \quad i^2 = 1 $

Considering all possible integer solutions for these equations:

Case I:

$(a, b) = (0, 1)$ or $(0, -1)$,

$(c, d) = (1, 0)$ or $(-1, 0)$

Case II:

$(a, b) = (1, 0)$ or $(-1, 0)$,

$(c, d) = (0, 1)$ or $(0, -1)$

For each case, $e$ can be either 1 or -1 since $i^2=1$. Therefore, $e = \pm 1$.

Combining these options, we find a total of 16 possible matrices $Q$.

Given, $A^3-5 A^2+7 A+I=0$

$ \Rightarrow A^5-5 A^4+7 A^3+A^2=0 $

Also, $A^5-6 A^4+12 A^3-6 A^2+2 A+2 I$

$ \begin{array}{r} =l A+m I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ A^5-5 A^4+7 A^3+A^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Subtracting Eqs. (i) and (ii), we get

$ -A^4+5 A^3-7 A^2+2 A+2 I=I A+m I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ A^4-5 A^3+7 A^2+A=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

On adding Eqs. (iii) and (iv), we get

$ \begin{aligned} & 3 A+2 I=l A+m I \\ \Rightarrow \quad & I=3 \text { and } m=2 \end{aligned} $

So, $\quad l+m=5$

$ \begin{aligned} &\text { Given, }\\ &A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{array}\right], A^{-1}=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & y \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] \end{aligned} $

$ \begin{aligned} & A A^{-1}=\left[\begin{array}{ccc} 1 & 0 & y+1 \\ 0 & 1 & 2 y+2 \\ 4-4 x & 3 x-3 & x y+2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow y+1=0 \text { and } 4-4 x=0 \\ & \Rightarrow y=-1 \text { and } x=1 \\ & \text { So, } \log _{2 / 5}\left(2^1+2^{-1}\right)=-1 \\ & \Rightarrow(x, y) \text { satisfies } y=\log _{2 / 5}\left(2^x+2^{-x}\right) \end{aligned} $

$ \begin{aligned} & \text { Let } A=\left[\begin{array}{lll} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0 \end{array}\right] \\ & A X=0 \\ & {\left[\begin{array}{lll} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0 \end{array}\right]\left[\begin{array}{l} l \\ m \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & {\left[\begin{array}{l} a_1 l+b_1 m \\ a_2 l+b_2 m \\ a_3 l+b_3 m \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & a_1 l+b_1 m=0 \\ & a_2 l+b_2 m=0 \\ & a_3 l+b_3 m=0 \\ & \Rightarrow a_1: a_2: a_3=b_1: b_2: b_3 \Rightarrow|A|=0 \end{aligned} $

⇒ Determinant of each submatrix of order $2 \times 2$ is zero $\Rightarrow$ Rank $=1$

Since, the system of equations has non-trivial solution

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 2 & 3 & a \\ 1 & a & -2 \\ 3 & 1 & 3 \end{array}\right|=0 \\ & \Rightarrow 2(3 a+2)-3(3+6)+a(1-3 a)=0 \\ & \Rightarrow 6 a+4-9-18+a-3 a^2=0 \\ & \Rightarrow 6 a-23+a-3 a^2=0 \\ & \Rightarrow 3 a^2-7 a+23=0 \\ & \Rightarrow D<0 \end{aligned} $

So, no real values of $a$ is possible.

$ \text { } \begin{aligned} 2 x+3 y+z & =-1 \\ 3 x+y+z & =4 \\ x-3 y-2 z & =1 \\ \Rightarrow \quad x-2 y & =5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{array}{lr} \Rightarrow 5 x+3 y=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow 5(5+2 y)+3 y=-1 \\ \Rightarrow 25+13 y=-1 \\ \Rightarrow 13 y=-26 \\ \Rightarrow y=-2 \\ \text { So, } y=-2 \end{array} $

For non-trivial solution,

$ \begin{aligned} & \quad\left|\begin{array}{rrr} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{array}\right|=0 \\ & 1(2+3)-a(a+2)+1(3 a-4)=0 \\ & \Rightarrow \quad 5-a^2-2 a+3 a-4=0 \\ & \Rightarrow \quad a^2-a-1=0 \\ & \Rightarrow \quad a=\frac{1 \pm \sqrt{5}}{2} \end{aligned} $

So, $\quad a=\frac{1+\sqrt{5}}{2}$

$ \text { } \begin{aligned} |\operatorname{adj} A| & =|A|^{n-1} \\ \left|\operatorname{adj}\left(A^2\right)\right| & =\left|A^2\right|^{n-1} \\ & =\left|A^2\right|^2 \\ \Rightarrow \quad A^2 & =\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right] \\ & =\left[\begin{array}{lll} 7 & 8 & 6 \\ 5 & 7 & 6 \\ 6 & 6 & 5 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow|A|^2=7(35-36)-8(25-36)+6(30-42) \\ & \quad=-7+88-72=9 \\ & \therefore\left|\operatorname{adj}\left(A^2\right)\right|=\left|A^2\right|^2=9^2=81 \end{aligned} $

$ \begin{array}{r} x-2 y+z=0 \\ 2 x+3 y+z=6 \\ x+2 y+p z=q \end{array} $

$ \begin{aligned} &\text { System has infinite solution, }\\ &\begin{array}{ll} \therefore & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ & \Delta=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & p \end{array}\right|=0 \\ \Rightarrow & 1(3 p-2)+2(2 p-1)+1(4-3)=0 \\ \Rightarrow & 3 p-2+4 p-2+1=0 \\ \Rightarrow & 7 p=3 \Rightarrow p=\frac{3}{7} \\ & \Delta_z=\left|\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 3 & 6 \\ 1 & 2 & q \end{array}\right|=0 \\ \Rightarrow & 1(3 q-12)+2(2 q-6)=0 \\ \Rightarrow & 3 q-12+4 q-12=0 \\ \Rightarrow & 7 q=24 \Rightarrow q=\frac{24}{7} \\ \therefore & q-p=\frac{24}{7}-\frac{3}{7}=3 \end{array} \end{aligned} $

We have,

$ \begin{aligned} & (\sin \theta) x-y+z=0 \\ & x-(\cos \theta) y+z=0 \\ & x+y+(\sin \theta) z=0 \end{aligned} $

System has non-trivial solution.

$ \therefore \quad \Delta=0 $

$ \begin{aligned} & \left|\begin{array}{ccc} \sin \theta & -1 & 1 \\ 1 & -\cos \theta & 1 \\ 1 & 1 & \sin \theta \end{array}\right|=0 \\ & \Rightarrow \quad \sin \theta(-\sin \theta \cos \theta-1) +1(\sin \theta-1)+1(1+\cos \theta)=0 \\ & \Rightarrow \quad-\sin ^2 \theta \cos \theta-\sin \theta+\sin \theta-1+1 +\cos \theta=0 \\ & \Rightarrow \quad \cos \theta\left(1-\sin ^2 \theta\right)=0 \\ & \Rightarrow \quad \cos ^3 \theta=0 \Rightarrow \cos \theta=0 \Rightarrow \theta=\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} A & =\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{array}\right] \\ |A| & =1(1-3)-2(2-1)+3(6-1) \\ & =-2-2+15=11 \\ B & =\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{array}\right] \\ \Rightarrow \quad & |B|=2(4-8)-3(6-4)+4(12-4) \\ & =-8-6+32=18 \\ \therefore \quad & \sqrt{|\operatorname{adj} A B|}=\sqrt{|\operatorname{adj} A||\operatorname{adj} B|} \\ & =\sqrt{|A|^{3-1}|B|^{3-1}} \\ & =|A||B|=11(18)=198 \end{aligned} \end{aligned} $

Step 1: Find the determinant of $ A $

Matrix $ A $ is: $ A=\left[\begin{array}{lll} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{array}\right] $

So, $ |A| = 1(3-18) - 5(12-6) + 2(24-2) $

Calculate each part:

• $1 \times (3 - 18) = 1 \times (-15) = -15$
• $-5 \times (12 - 6) = -5 \times 6 = -30$
• $2 \times (24 - 2) = 2 \times 22 = 44$

Add them up: $ |A| = -15 - 30 + 44 = -1 $

Step 2: Find the value of $ \left| (\operatorname{adj} A)^{-1} \right| $

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

The order of matrix $A$ is 3, so: $ |\operatorname{adj} A| = |A|^{3-1} = |A|^2 $

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

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

Given, $x+\lambda y-2 z=1$

$ x-y+\lambda z=2 $

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

∵ Given, system is inconsistent

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

$ \therefore \lambda_1+\lambda_2=1 $

Given, $(\sin \theta) x+y-2 z=0$

$ 2 x-y+(\cos \theta) z=0 $

And $-3 x+(\sec \theta) y+3 z=0$

∵ system has non-trivial solution.

$ \begin{aligned} & \Delta=0 \\ & \left|\begin{array}{ccc} \sin \theta & 1 & -2 \\ 2 & -1 & \cos \theta \\ -3 & \sec \theta & 3 \end{array}\right|=0 \\ & \sin \theta(-3-1)-1(6+3 \cos \theta)-2(2 \sec \theta-3) =0\\ & \Rightarrow-4 \sin \theta-6-3 \cos \theta-4 \sec \theta+6=0 \\ & \Rightarrow 4 \sec \theta+3 \cos \theta+4 \sin \theta=0 \\ & \Rightarrow 4+3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\ & \Rightarrow 4+3\left(\frac{1+\cos 2 \theta}{2}\right)+2 \sin 2 \theta=0 \\ & \Rightarrow 8+3+3 \cos 2 \theta+4 \sin 2 \theta=0 \\ & \Rightarrow 11+3 \cos 2 \theta+4 \sin 2 \theta=0 \end{aligned} $

This equation has no solution because the maximum value of $4 \sin 2 \theta+3 \cos 2 \theta$ is 5 . Since $5+11 \neq 0$, there is no value of $\theta$ that satisfies the equation.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \\ & \because \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A \\ & \operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))=|\operatorname{adj} A|^{2-2} \cdot \operatorname{adj} A \\ & \quad=\operatorname{adj} A=|A| A^{-1}\left(\because A^{-1}=\frac{1}{|A|} \operatorname{adj} A\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \left|\begin{array}{ccc} x & -3 & 2 \\ -1 & -2 & x-1 \\ 1 & x-2 & 3 \end{array}\right|=0 \\ & \Rightarrow x(-6-(x-1)(x-2))+3(-3-(x-1)) +2(-(x-2)+2)=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow x\left(-6-x^2+3 x-2\right)+3(-3-x+1) +2(-x+2+2)=0 \\ & \Rightarrow x\left(-x^2+3 x-8\right)+3(-2-x) +2(4-x)=0 \\ & \Rightarrow-x^3+3 x^2-8 x-6-3 x+8-2 x=0 \\ & \therefore \text { Sum of roots }=\frac{-3}{-1}=3 . \end{aligned} $

We have, $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|=A \lambda^3+B \lambda^2+C \lambda+D$

Now, $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|$

$ \begin{aligned} & =1(-3-3 \lambda-2)+(1-\lambda)(4+(1+\lambda)(3-\lambda)) \\ & =(-3 \lambda-5)+(1-\lambda)\left(4+3-\lambda+3 \lambda-\lambda^2\right) \\ & =(-3 \lambda-5)+(1-\lambda)\left(7+2 \lambda-\lambda^2\right) \\ & =(-3 \lambda-5)+\left(7+2 \lambda-\lambda^2-7 \lambda-2 \lambda^2+\lambda^3\right) \\ & =\lambda^3-3 \lambda^2-8 \lambda+2 \end{aligned} $

$ \begin{aligned} &\therefore \lambda^3-3 \lambda^2-8 \lambda+2=A \lambda^3+B \lambda^2+C \lambda+D\\ &\text { On comparing both sides, we get }\\ &\begin{aligned} A & =1, B=-3, C=-8, D=2 \\ \therefore \quad D+A & =2+1=3 \end{aligned} \end{aligned} $

We have

$ A+2 B=\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And $\quad 2 A-B=\left[\begin{array}{rrr}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, $2 A+4 B=\left[\begin{array}{rcr}2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2\end{array}\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Substract Eqs. (ii) from (iii), we get

$ \begin{aligned} &\begin{aligned} & 5 B=\left[\begin{array}{rrr} 2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2 \end{array}\right]-\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]=\left[\begin{array}{rrr} 0 & 5 & -5 \\ 10 & -5 & 0 \\ -10 & 5 & 0 \end{array}\right] \\ \Rightarrow & B=\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right] \end{aligned}\\ &\therefore \text { From Eqs. (ii), we get }\\ &\begin{aligned} & 2 A=\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]+\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right] \\ \Rightarrow & 2 A=\left[\begin{array}{rrr} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{array}\right] \Rightarrow A=\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{array}\right] \\ \therefore & \operatorname{tr}(A)-\operatorname{tr}(B) \\ & =(1-1+1)-(0-1+0)=1+1=2 \end{aligned} \end{aligned} $

We have,

$A, C$ are $3 \times 3$ matrices

$B \times D$ are $3 \times 1$ matrices

$A X=B$ has unique solution

$C X=D$ has infinite solution.

Since, $A X=B$ have a unique solution.

$\Rightarrow \operatorname{rank}(A)=$ number of unknowns $=3$

$\therefore \quad \operatorname{rank}([A: B])=\operatorname{rank}(A)=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Since, $C X=D$ have infinite many solutions

$\therefore \quad \operatorname{Rank}(C)=$ not unique $\Rightarrow \operatorname{rank}(C)<3$

$\Rightarrow \quad \operatorname{Rank}([C: D])<3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$\operatorname{Rank}([A: D]) \geq \operatorname{rank}[(C: B)]$

We have, $A$ and $B$ are two non-square matrices such that

$ P=A+B, Q=A^T B, R=A B^T $

Since, $A+B$ is defined,

So order of $A$ and $B$ are same.

Let order of $A$ and $B$ be $m \times n$

$\therefore$ Order of $A^T$ is $n \times m$ and order of $B^T$ is $n \times m$

Now, order of $P=$ order of $A+B=m \times n$

Order of $Q=$ order of $A^T B=n \times n$

And order of $R=$ order of $A B^T=m \times m$

Now, order of $P Q=m \times n$

Order of $Q R=$ not defined

Order of $R Q=$ not defined

Order of $Q P=$ not defined

Order of $R P=m \times n$

$\therefore$ Order of $A=$ order of $P Q$ and $R P$

Given equations are $x+y-z=1$,

$ 2 x+4 y-z=0 \text { and } 3 x+4 y+5 z=18 $

The augmented matrix for the system is

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 2 & 4 & -1 & 0 \\ 3 & 4 & 5 & 18 \end{array}\right] $

We need to transform this matrix into

$ B=\left[\begin{array}{cccc} 1 & a & 0 & -1 \\ 0 & 2 & 1 & b \\ 0 & 0 & c & 32 \end{array}\right] $

Now, perform the row operation for matrix $A$, we get

$ \begin{gathered} R_2 \rightarrow R_2-2 R_1 \text { and } R_3 \rightarrow R_3-3 R_1 \\ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 1 & 8 & 15 \end{array}\right] \end{gathered} $

Now, $R_3 \rightarrow R_3-\frac{1}{2} R_2$

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 7.5 & 16 \end{array}\right] $

Apply $R_3 \rightarrow 2 R_3$, we get

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 15 & 32 \end{array}\right] $

Again apply $R_1 \rightarrow R_1+R_2$, we get

$ A=\left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 15 & 32 \end{array}\right] $

$ \begin{aligned} &\text { Thus, } a=3, b=-2 \text { and } c=15\\ &\begin{array}{ll} \Rightarrow & c=15 \\ \therefore & \sqrt{a+b+c}=\sqrt{3+(-2)+15} \\ & =\sqrt{1+15}=\sqrt{16}=4 \end{array} \end{aligned} $

Given, $K=\left|\begin{array}{ccc}9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36\end{array}\right|$

$ \begin{aligned} & =9\left|\begin{array}{ll} 36 & 25 \\ 49 & 36 \end{array}\right|-25\left|\begin{array}{ll} 16 & 25 \\ 25 & 36 \end{array}\right|+16\left|\begin{array}{ll} 16 & 36 \\ 25 & 49 \end{array}\right| \\ & =9(1296-1225)-25(576-625) +16(784-900) \\ & =9(71)-25(-49)+16(-116) \\ & =639+1225-1856=8 \\ & \therefore K=8 \end{aligned} $

So, the roots are $K$ and $K+1$, i.e., 8 and 9

$ \begin{aligned} \text { Sum of roots } & =8+9=17 \\ & =\frac{-b}{a}=\alpha+\beta \end{aligned} $

And product of roots $=8 \times 9=72$

$ =\frac{c}{a}=\alpha \beta $

So, the general form of a quadratic equation is $x^2-(\alpha+\beta) x+\alpha \beta=0$

$ \Rightarrow \quad x^2-17 x+72=0 $

Given, $A=\left[\begin{array}{ccc}1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13\end{array}\right]$

$ \begin{aligned} & \quad|A|=\left|\begin{array}{ccc} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{array}\right| \\ & =1(4 \times 13-(-6) \times(-11))-(-3)(-2 \times 13 -(-6) \times 7)-5(-2 \times(-11)-7 \times 4) \\ & =1(52-66)+3(-26+42)-5(22-28) \\ & =-14+48+30=64 \end{aligned} $

Since, $|\operatorname{adj} A|=|A|^{n-1}$, for $n \times n$ matrix $A$

So, for $3 \times 3$ matrix,

$ \therefore \quad \begin{aligned} |\operatorname{adj} A| & =|A|^{3-1}=|A|^2 \\ \therefore \quad|\operatorname{adj} A| & =|A|^2=(64)^2=4096 \\ \sqrt{|\operatorname{adj} A|} & =\sqrt{4096}=64 \end{aligned} $

$ \text {} \begin{aligned} \Delta r & =\left|\begin{array}{cc} \frac{1}{3 r-2} & \frac{2}{3 r-5} \\ 0 & \frac{3}{3 r+1} \end{array}\right| \\ & =\frac{1}{(3 r-2)} \times \frac{3}{(3 r+1)}-\frac{2}{3 r-5} \times 0 \\ & =\frac{3}{(3 r-2)(3 r+1)} \end{aligned} $

Using partial fractions

$ \begin{aligned} \Delta r & =\frac{A}{3 r-2}+\frac{B}{3 r+1} \\ \Rightarrow \quad 3 & =A(3 r+1)+B(3 r-2) \end{aligned} $

If $\quad r=\frac{2}{3}$, then

$ \begin{aligned} & \Rightarrow \quad 3=A\left(3 \times \frac{2}{3}+1\right)+B\left(3 \times \frac{2}{3}-2\right) \\ & \Rightarrow 3 A=3 \\ & \Rightarrow \quad A=1 \end{aligned} $

If $\quad r=-\frac{1}{3}$, then

$ \begin{aligned} & \quad 3=B\left(3 \times\left(-\frac{1}{3}\right)-2\right)=B(-1-2) \\ & \Rightarrow \quad B=-1 \\ & \text { So, } \Delta r=\frac{1}{3 r-2}-\frac{1}{3 r+1} \\ & \therefore \sum_{r=1}^{33} \Delta r=\sum_{r=1}^{33}\left(\frac{1}{3 r-2}-\frac{1}{3 r+1}\right) \\ & \Rightarrow\left(\frac{1}{1}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+ \\ & \quad \ldots+\left(\frac{1}{97}-\frac{1}{100}\right) \\ & \Rightarrow 1-\frac{1}{100}=\frac{99}{100}=0.99 \end{aligned} $

$A=\left|\begin{array}{ccc}-1 & x & -3 \\ 2 & 4 & z \\ y & 5 & -6\end{array}\right|$ is a symmetric

So, $A=A^T$

that means, $x=2, y=-3, z=5$

and $B=\left[\begin{array}{ccc}0 & 2 & q \\ p & 0 & -4 \\ -3 & r & s\end{array}\right]$ is a skew-matrix

So, $B=-B^T$

that means, $p=-2 q=3, r=4, s=0$

Therefore, $|A|=\left|\begin{array}{ccc}-1 & 2 & -3 \\ 2 & 4 & 5 \\ -3 & 5 & -6\end{array}\right|$

$ =-1(-49)-2(3)-3(22)=-23 $

and $|B|=\left|\begin{array}{ccc}0 & 2 & 3 \\ -2 & 0 & -4 \\ -3 & 4 & 0\end{array}\right|$

$ =-2(-12)+3(-8)=0 $

Now, $|A B|=|A||B|=-23 \times 0=-23$

$ x y z+q+r=2 \times(-3) \times(5)+3+4=-23 $

Hence, $|A|+|B|-|A B|=x y z+q+r$

$A=\left[\begin{array}{ccc}-x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x\end{array}\right]$

and $B=\left[\begin{array}{ccc}2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1\end{array}\right]$

If $A^{-1}=B \Rightarrow$ So, $A B=I$

$ \begin{aligned} & \Rightarrow\left[\begin{array}{ccc} -x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{array}\right]=5 \\ & \Rightarrow-2 x+7 x=1 \Rightarrow x=1 / 5 \end{aligned} $

Now, $\left[\begin{array}{ccc}x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4\end{array}\right]$

$R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_2$

$ =\left[\begin{array}{ccc} x & x+1 & x+2 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] $

therefore determinant of the given matrix $=0$

Hence, from the given options put $x=1 / 5$ in $5 x-1$ we get, $5 \times \frac{1}{5}-1=0$

$ \begin{aligned} &\text { that means, }\\ &\left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right|=5 x-1 \end{aligned} $

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

$ x+2 y+\alpha z=1 $

and $2 x-y+z=\beta$

So, $A=\left[\begin{array}{ccc}2 & 3 & -3 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{array}\right]$

for infinite many solutions put $|A|=0$

$ \begin{aligned} & \Rightarrow 2(2+\alpha)-3(1-2 \alpha)-3(-1-4)=0 \\ & \Rightarrow 1+8 \alpha+15=0 \Rightarrow \alpha=-2 \end{aligned} $

Now, $D_y=\left|\begin{array}{ccc}2 & 3 & -3 \\ 1 & 1 & -2 \\ 2 & \beta & 1\end{array}\right|$

$ \begin{aligned} & \text { put } D_y=0 \\ & \begin{aligned} \Rightarrow 2(1+2 \beta)-3(1+4)-3(\beta-2)=0 \\ \Rightarrow \beta-7=0 \Rightarrow \beta=7 \\ \text { Therefore, } \frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{-2}{7}-\frac{7}{(-2)} \\ \qquad=\frac{-4+49}{14}=\frac{45}{14} \end{aligned} \end{aligned} $

We have,

$ \left|\begin{array}{ccc} 1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta \end{array}\right|=0 $

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

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta \end{array}\right|=0 \\ & R_3 \rightarrow R_3-\left(\sin ^2 \theta\right) R_1 \end{aligned} $

$ \begin{aligned} &\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 1 & 1+4 \sin 4 \theta \end{array}\right|=0\\ &\text { Expand w.r.t along } C_1 \text {, we get }\\ &\begin{aligned} & 1+4 \sin 4 \theta+1=0 \\ & 4 \sin 4 \theta=-2 \\ & \sin 4 \theta=\frac{-1}{2}=\sin \left(\pi+\frac{\pi}{6}\right) \\ & 4 \theta=\frac{7 \pi}{6} \Rightarrow \theta=\frac{7 \pi}{24} \end{aligned} \end{aligned} $

Given, $2 x+p y+6 z=8$

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

For infinite solution

$ \begin{aligned} & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Delta_z=0 \\ & \Rightarrow\left|\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}\right|=0 \\ & \Rightarrow \quad 2(8-5)-p(4-5)+8(1-2)=0 \\ & \Rightarrow \quad 6+p-8=0 \Rightarrow p=2 \end{aligned} $

$ \begin{aligned} &\text { We have, } x^a y^b=e^m\\ &x^a=\frac{e^m}{y^b} \Rightarrow x=\frac{e^{\frac{m}{a}}}{y^{\frac{b}{a}}} \Rightarrow x^c y^d=e^n \end{aligned} $

$ \Rightarrow \quad \frac{e^{\frac{m c}{a}}}{y^{\frac{b c}{a}}} \cdot y^d=e^n $

$ \begin{aligned} \Rightarrow & & y^{d-\frac{b c}{a}}=e^{n-\frac{m c}{a}} \\ \Rightarrow & & y^{a d-k c}=e^{a n-m c} \\ \Rightarrow & & y=e^{\frac{a n-m c}{a d-k c}}=e^{\frac{\Delta_2}{\Delta_3}} \end{aligned} $

$ \begin{aligned} &\text { and } &x=e^{\frac{\Delta_1}{\Delta_3}} \end{aligned} $

$ \quad \begin{aligned}\because & \Delta_1=m d-b n \\ & \Delta_2=a n-m c \\ & \Delta_3=a d-b c \end{aligned} $