Matrices and Determinants

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

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

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

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. $

$\begin{aligned} & B=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right] \\ & A B^{-1}=A^{-1} \\ & \Rightarrow A^2=B \end{aligned}$

Also, $B C B^{-1}=A$

$\begin{aligned} \Rightarrow C & =B^{-1} A B \\ \Rightarrow C^4 & =\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right) \\ & =B^{-1} A^4 B \\ & =B^{-1} B^2 B \\ \Rightarrow \quad C^4 & =B^2 \end{aligned}$

Also, $C^2=\left(B^{-1} A B\right)\left(B^{-1} A B\right)$

$\begin{aligned} & =B^{-1} A^2 B \\ & =B^{-1} B B \end{aligned}$

$\begin{aligned} & \Rightarrow C^2=B \\ & \Rightarrow C^4+\alpha C^2+\beta I=0 \\ & \Rightarrow B^2+\alpha B+\beta I=0 \\ & B^2=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right]+\alpha\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]+\left[\begin{array}{ll} \beta & 0 \\ 0 & \beta \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & \Rightarrow 4+\alpha+\beta=0 \end{aligned}$

and $18+3 \alpha=0$

$\begin{aligned} & \Rightarrow \alpha=-6 \\ & \Rightarrow \beta=2 \\ & \Rightarrow 2 \beta-\alpha=10 \end{aligned}$

$\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$

$\begin{aligned} & {\left[\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]} \\ & A X=B \\ & X=A^{-1} B \\ & X=\frac{\operatorname{adj} A}{|A|} B \end{aligned}$

For Infinitely many solution

$\begin{aligned} & |A|=0 \text { and }(\operatorname{adj} A) B=0 \\ & \left|\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right|=0 \\ & -2052+2250+18 \lambda=0 \\ & \Rightarrow \lambda=-11 \\ & \operatorname{adj} A=\left[\begin{array}{ccc} -684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right] \\ & (\operatorname{adj} A) B=\left[\begin{array}{ccc} 684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow 54+40-2 \mu=0 \\ & \Rightarrow 2 \mu=94 \\ & \Rightarrow \mu=47 \\ & \Rightarrow \mu+2 \lambda=25 \end{aligned}$

$\left|\begin{array}{lll} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{array}\right|=0$

$\begin{aligned} & R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\ & \Rightarrow\left|\begin{array}{ccc} \alpha-a & b-\beta & 0 \\ 0 & \beta-b & c-\gamma \\ a & b & \gamma \end{array}\right|=0 \end{aligned}$

Take $\alpha$-a, $\beta$-b, $\gamma$-c common from column-1, 2 and 3 respectively

$\begin{aligned} & (\alpha-a)(\beta-b)(\gamma-c)\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \frac{a}{\alpha-a} & \frac{b}{\beta-b} & \frac{\gamma}{\gamma-c} \end{array}\right|=0 \\ & \Rightarrow \frac{\gamma}{\gamma-c}+\frac{b}{\beta-b}+\frac{a}{\alpha-a}=0 \end{aligned}$

$\begin{aligned} & x+4 y-z=\lambda \\ & 7 x+9 y+\mu z=-3 \\ & 5 x+y+2 z=-1 \\ & {\left[\begin{array}{ccc} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]} \\ & A=\left[\begin{array}{lll} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right], B=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right] \\ & A X=B \\ & X=A^{-1} B \\ & \quad=\frac{\operatorname{adj} A}{|A|} B \end{aligned}$

If $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, system has infinitely many solutions.

$\begin{aligned} & |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\ & \Rightarrow 18-\mu-56+20 \mu+38=0 \\ & \Rightarrow 19 \mu=0 \\ & \Rightarrow \mu=0 \\ & \text { Also adjA }=\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right] \\ & (\text { adj } A) \cdot B=0 \\ & {\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right]\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & 18 \lambda+27-9=0 \\ & \Rightarrow 18 \lambda=-18 \\ & \Rightarrow \lambda=-1 \\ & \Rightarrow 2 \mu+3 \lambda=3(-1)=-3 \end{aligned}$

$\begin{aligned} & |A-\lambda I|=0 \\ & \left|\begin{array}{ccc} 2-\lambda & a & 0 \\ 1 & 3-\lambda & 1 \\ 0 & 5 & b-\lambda \end{array}\right|=0 \\ & (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\ & (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\ & \lambda^3-(b+5) \lambda^2+(1-a+5 b) \lambda+(10-6 b+a b)=0 \\ & A^3-(b+5) A^2+(1-a+5 b) A+(10-6 b+a b) I=0 \\ & \Rightarrow \mathrm{b}+5=4,1-a+5 b=1,10-6 b+a b=21 \\ & \Rightarrow a=-5, b=-1 \\ & \Rightarrow 2 a+3 b=-13 \end{aligned}$

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

$\begin{aligned} & =4^6\left|-3 \operatorname{adj}\left(\operatorname{aadj}\left((2 A)^{-1}\right)\right)\right|^4 \\ & =4^6 \cdot 3^{12}\left|\operatorname{aadj}\left((2 A)^{-1}\right)\right|^8 \\ & =4^6 \cdot 3^{12} \cdot 3^{24}\left|(2 A)^{-1}\right|^{16} \\ & =4^6 \cdot 3^{36} 2^{-48}\left|A^{-1}\right|^{16} \\ & =\frac{2^{-36} 3^{36}}{3^{16}}=2^{-36} 3^{20} \\ & m=-36 \\ & n=20 \\ & m+2 n=4 \end{aligned}$

We are given a determinant $A_r$ defined as:

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$

We need to find the value of $2 A_{10} - A_8$.

Let's calculate the determinant $A_r$. We can use properties of determinants or expand it directly. A helpful trick is to use row operations to simplify the determinant before expanding. Let's perform the following row operations:

Replace Row 2 with (Row 2) - 2 $\times$ (Row 1)

Replace Row 3 with (Row 3) - 3 $\times$ (Row 1)

The determinant becomes:

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r - 2r & 2 - 2(1) & (n^2-\beta) - 2(\frac{n^2}{2}+\alpha) \\ (3 r-2) - 3r & 3 - 3(1) & \frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha)\end{array}\right|$

Let's simplify the entries in the new rows:

Row 2:

First element: $2r - 2r = 0$

Second element: $2 - 2 = 0$

Third element: $(n^2-\beta) - 2(\frac{n^2}{2}+\alpha) = n^2 - \beta - n^2 - 2\alpha = -2\alpha - \beta$

Row 3:

First element: $(3r-2) - 3r = -2$

Second element: $3 - 3 = 0$

Third element: $\frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha) = \frac{3n^2 - n}{2} - \frac{3n^2}{2} - 3\alpha = \frac{3n^2 - n - 3n^2}{2} - 3\alpha = -\frac{n}{2} - 3\alpha$

So the determinant simplifies to:

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 0 & 0 & -2\alpha - \beta \\ -2 & 0 & -\frac{n}{2} - 3\alpha\end{array}\right|$

Now, we can easily calculate this determinant by expanding along the second column, since it has two zeros. The expansion along the second column is:

$A_r = -1 \cdot \left|\begin{array}{cc} 0 & -2\alpha - \beta \\ -2 & -\frac{n}{2} - 3\alpha \end{array}\right|$

$A_r = -1 \cdot \left( (0) \times (-\frac{n}{2} - 3\alpha) - (-2) \times (-2\alpha - \beta) \right)$

$A_r = -1 \cdot \left( 0 - (4\alpha + 2\beta) \right)$

$A_r = -1 \cdot (-4\alpha - 2\beta)$

$A_r = 4\alpha + 2\beta$

Notice that the value of the determinant $A_r$ does not depend on $r$ at all! It's a constant value determined by $\alpha$ and $\beta$.

Now we need to find the value of $2 A_{10} - A_8$.

Since $A_r = 4\alpha + 2\beta$ for any value of $r$, we have:

$A_{10} = 4\alpha + 2\beta$

$A_8 = 4\alpha + 2\beta$

So, $2 A_{10} - A_8 = 2 (4\alpha + 2\beta) - (4\alpha + 2\beta)$

$2 A_{10} - A_8 = 8\alpha + 4\beta - 4\alpha - 2\beta$

$2 A_{10} - A_8 = (8\alpha - 4\alpha) + (4\beta - 2\beta)$

$2 A_{10} - A_8 = 4\alpha + 2\beta$

The value of the expression $2 A_{10} - A_8$ is $4\alpha + 2\beta$.

The given system of linear equations can be represented as,

$\begin{aligned} & \left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 2 & 5 & 5 & 17 \\ 1 & 2 & m & n \end{array}\right) \\ & \sim\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 3 & 3 & 9 \\ 0 & 1 & m-1 & n-4 \end{array}\right) \\ & \sim\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & m-2 & n-7 \end{array}\right) \end{aligned}$

$\because$ System of equations has infinitely many solutions

$\therefore m=2 \& n=7$

Which satisfy equation given in option (1).

$\text { (i.e., } 2^2+7^2-14=39 \text { ) }$

$A=\left[\begin{array}{ccc} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha \end{array}\right], B=\left[\begin{array}{ccc} 3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta \end{array}\right]$

Cofactor of $A$-matrix is

$=\left[\begin{array}{ccc} 2 \alpha^2-\alpha \beta & -\left(2 \alpha^2+\beta^2\right) & \alpha^2+\alpha \beta \\ -\left(2 \alpha^2-3 \alpha\right) & (2 \alpha \beta+3 \beta) & -(2 \alpha \beta) \\ \alpha \beta-3 \alpha & -\left(\beta^2-3 \alpha\right) & \beta \alpha-\alpha^2 \end{array}\right]$

which is equal to matrix $B$

So, by comparing elements of two matrix

$\begin{aligned} & \Rightarrow \alpha \beta-3 \alpha=-2 \alpha \\ & \Rightarrow \alpha \beta-\alpha=0 \\ & \Rightarrow \alpha(\beta-1)=0 \\ & \Rightarrow \alpha=0 \text { or } \beta=1[\because \alpha \text { cannot be } 0] \\ & \Rightarrow \beta=1 \\ & \text { and }-\beta^2+3 \alpha=5 \\ & \Rightarrow 3 \alpha=6 \\ & \Rightarrow \alpha=2 \\ & A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right] \\ & \operatorname{Det}(A B)=|A||B|=|A|\left|(\operatorname{adj} A)^{\top}\right| \\ & =|A| \cdot|A|^2 \\ & =|A|^3 \\ & =(6-18+18)^3 \\ & =6^3 \\ & =216 \\ \end{aligned}$

$\begin{aligned} & |A|=3 \\ & |B|=2 \\ & \left.\left|A^T\right||A| \mid(\operatorname{adj}(2 A))^{-1}\|\operatorname{adj}(4 B)\|(\operatorname{adj}(A B))^{-1}\right)|A|\left|A^T\right| \\ & 3 \cdot 3 \frac{1}{64 \cdot 9}(64)^2 \cdot 4 \cdot \frac{1}{9 \cdot 4} 3 \cdot 3 \\ & =64 \end{aligned}$

$\begin{aligned} & 11 x+y+\lambda z=-5 \\ & 2 x+3 y+5 z=3 \\ & 8 x-19 y-39 z=\mu \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{array}\right|=0 \\ & 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\ & =11(-117+95)-1(-118)-62 \lambda=0 \\ & =-242+118=62 \lambda \\ & \Rightarrow \lambda=-2 \\ & \Delta 2=0 \\ & \Rightarrow\left|\begin{array}{ccc} 11 & 1 & -5 \\ 2 & 3 & 3 \\ 8 & -19 & \mu \end{array}\right|=0 \end{aligned}$

$\begin{aligned} & 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\ & 33 \mu+627-2 \mu+24+310=0 \\ & \mu=-31 \\ & \Rightarrow \lambda^4-31 \\ & =16+31 \\ & =47 \end{aligned}$

$\begin{aligned} & \operatorname{adj}(A)=\left[\begin{array}{ll} 1 & -2 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^2=\left[\begin{array}{ll} 1 & -4 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^3=\left[\begin{array}{cc} 1 & -6 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^4=\left[\begin{array}{cc} 1 & -8 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^r=\left[\begin{array}{cc} 1 & (-2 r) \\ 0 & 1 \end{array}\right] \end{aligned}$

$\begin{aligned} & B=\sum_{r=0}^{10}(\operatorname{adj} A)^r=\left[\begin{array}{ll} \sum_\limits{r=0}^{10} 1 & \sum_\limits{r=0}^{10}(-2 r) \\ \sum_\limits{r=0}^{10}(0) & \sum_\limits{r=0}^{10}(1) \end{array}\right] \\ & B=\left[\begin{array}{cc} 11 & -110 \\ 0 & 11 \end{array}\right] \end{aligned}$

$\text { Sum of elements }=-110+11+11=-88$

$\begin{aligned} & \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\ & P=A-2 A^{\top} \\ & Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\ & |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\ & \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \\ & \left|A-2 A^T\right|=4 \\ & A-2 A^T=\left[\begin{array}{lll} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2 \end{array}\right] \\ & \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\ & \Rightarrow|A|^2=16 \end{aligned}$

$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$

$\because$ Non-trivial solution

$\Rightarrow D=0$

$\begin{aligned} & \left|\begin{array}{ccc} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\ \end{aligned}$

$\begin{aligned} & -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\ & \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\ & \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\ & \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\ & \Rightarrow \alpha=\frac{5 \pi}{24} \end{aligned}$

$ \begin{aligned} & x+2 y+3 z=5 \\\\ & 2 x+3 y+z=9 \\\\ & 4 x+3 y+\lambda z=\mu \end{aligned} $

For infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$

$\begin{aligned} & \Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13 \\\\ & \Delta_1=\left|\begin{array}{llc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15 \\\\ & \Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0\end{aligned}$

$ \Delta_3=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15 \end{array}\right|=0 $

For $\lambda=-13, \mu=15$ system of equation has infinite solution hence $\lambda+2 \mu=17$

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$, $\mathrm{B}$, and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$, and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$, we find $\operatorname{det} \mathrm{X}$ as follows:

First, we express $|\mathrm{C}|$, the determinant of $\mathrm{C}$, in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$, using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$:

$\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$

Next, we find $|\mathrm{X}|$, using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$

Substituting the expression for $|\mathrm{C}|$ obtained earlier:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$

The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as:

$ |A|=\left|\begin{array}{cc} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{array}\right|=2+1=3 $

$ |B|=\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|=1 $

Finally, substituting these values into our expression for $|\mathrm{X}|$:

$\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned}$

Therefore, $\operatorname{det} \mathrm{X} = 729$.

$\begin{aligned} & \text { Given } 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \\\\ & \Delta_2=\left|\begin{array}{ccc}2 & -1 & 5 \\ 1 & 3 & -4 \\ 3 & \beta & 7\end{array}\right| \\\\ & \Delta_2=2(21+4 \beta)+1(7+12)+5(\beta-9) \\\\& \Delta_2=42+8 \beta+19+5 \beta-45 \\\\ & \Delta_2=13 \beta+16 \\\\ & \Delta_2=0\end{aligned}$

$\begin{aligned} & \therefore \beta=-\frac{16}{13} \\\\ & \Delta_3=\left|\begin{array}{lll}2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7\end{array}\right| \\\\ & \Delta_3=2(7 \alpha-4)-3(7+12)+5(-1-3 \alpha) \\\\ & \Delta_3=14 \alpha-8-57-5-15 \alpha \\\\ & \Delta_3=-\alpha-70\end{aligned}$

$\begin{aligned} & \Delta_3=0 \\\\ & \alpha=-70 \\\\ & 13 \alpha \beta=(13)(-70)\left(-\frac{16}{13}\right) \\\\ & =+1120\end{aligned}$

$\text { Let } A=\left[\begin{array}{lll} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right]$

$\text { Given } A\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right] \quad \text{ ..... (1)}$

$\therefore\left[\begin{array}{l} \mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]$

$\begin{aligned} \therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\ \mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\ \mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)} \end{aligned}$

$\begin{aligned} & \text { Given } A\left[\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} -4 \\ 0 \\ 4 \end{array}\right] \\ & \therefore\left[\begin{array}{l} -\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 4 \end{array}\right] \end{aligned}$

$\begin{array}{r} \Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\ -\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\ -\mathrm{x}_3+\mathrm{z}_3=4 \end{array}$

$\begin{aligned} & \text { Given } A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \\ & \therefore\left[\begin{array}{l} \mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \end{aligned}$

$\begin{aligned} & \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7) \\ & \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\ & \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3 \end{aligned}$

$\begin{gathered} \therefore A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right] \\ \therefore \text { Now }(A-31)\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} -1 \\ 2 \\ 3 \end{array}\right] \\ \therefore\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right] \\ {\left[\begin{array}{c} -z \\ -y \\ -x \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ {[z=-1],[y=-2],[x=-3]} \end{gathered}$

$\begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ & =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\ & =\alpha \beta+3+4 \beta-18-2-3 \alpha \end{aligned}$

For infinite solutions $\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$ and

$\begin{aligned} & \mathrm{D}_3=0 \\ & \mathrm{D}=0 \end{aligned}$

$\alpha \beta-3 \alpha+4 \beta=17$ ..... (1)

$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|=0 \\ & \mathrm{D}_2=\left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|=0 \\ & \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\ & 13 \beta-9-36-9=0 \end{aligned}$

$13 \beta=54, \beta=\frac{54}{13}$ put in (1)

$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$

$54 \alpha-39 \alpha+216=221$

$15 \alpha=5 \quad \alpha=\frac{1}{3}$

Now, $12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$

$=4+54=58$

$\begin{aligned} & x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0 \\ & \Rightarrow x, y, z \neq 0 \end{aligned}$

Also,

$\begin{aligned} & \sin \theta+\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta+\frac{4 \pi}{3}\right)=0 \forall \theta \in \mathrm{R} \\ & \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0 \\ & \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=0 \end{aligned}$

(i) $\quad \operatorname{Trace}(\mathrm{R})=\mathrm{x}+\mathrm{y}+\mathrm{z}$

If $x+y+z=0$ and $x y+y z+z x=0$

$\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0$

Statement (i) is False

(ii) $\quad \operatorname{Adj}(\operatorname{Adj}(\mathrm{R}))=|\mathrm{R}| \mathrm{R}$

Trace $(\operatorname{Adj}(\operatorname{Adj}(\mathrm{R})))$

$=x y z(x+y+z) \neq 0$

Statement (ii) is also False

$\begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 2 \lambda^2-\lambda-1=0 \\ & \lambda=1,-\frac{1}{2} \\ & \left|\begin{array}{ccc} 1 & 1 & 5 \\ 2 & \lambda^2 & 9 \\ 3 & \lambda & \mu \end{array}\right|=0 \Rightarrow \mu=13 \end{aligned}$

Infinite solution $\lambda=1 \& \mu=13$

For unique $\operatorname{sol}^{\mathrm{n}} \lambda \neq 1$

For no $\operatorname{sol}^{\mathrm{n}} \lambda=1 \& \mu \neq 13$

If $\lambda \neq 1$ and $\mu \neq 13$

Considering the case when $\lambda=-\frac{1}{2}$ and $\mu \neq 13$ this will generate no solution case

$x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda{ }^2 z=\mu^2+15$,

$\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \lambda \\ 1 & 3 & 4 \lambda^2 \end{array}\right|=(2 \lambda-1)^2$

For unique solution $\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)$

Let $\Delta=0, \lambda=\frac{1}{2}$

$\begin{aligned} & \Delta_y=0, \Delta_x=\Delta_z=\left|\begin{array}{ccc} 4 \mu & 1 & 1 \\ 10 \mu & 2 & 1 \\ \mu^2+15 & 3 & 1 \end{array}\right| \\ & =(\mu-15)(\mu-1) \end{aligned}$

For infinite solution $\lambda=\frac{1}{2}, \mu=1$ or 15

$\begin{aligned} \left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\ & =|\mathrm{A}-2 \mathrm{I}| \\ & =\left|\begin{array}{ccc} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{array}\right|=69 \end{aligned}$

So, Prime factor of 69 is 3 & 23

So, sum = 26

$\begin{aligned} & |\mathrm{A}|=\alpha^2-\beta^2 \\ & |2 \mathrm{~A}|^3=2^{21} \Rightarrow|\mathrm{A}|=2^4 \\ & \alpha^2-\beta^2=16 \\ & (\alpha+\beta)(\alpha-\beta)=16 \Rightarrow \alpha=4 \text { or } 5 \end{aligned}$

$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$

On solving given expression, we get

$\begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\ & =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}} \end{aligned}$

$\left|\begin{array}{ccc} 1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0 \end{array}\right|=0$

$\begin{aligned} & \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\ & \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\ & \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\ & \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\ & \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2} \end{aligned}$

Hence option (C) is correct.

$\begin{aligned} & f(-x)=\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(-x)=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}$

Hence statement- I is correct

Now, checking statement II

$\begin{aligned} & f(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow f(x) \cdot f(y)=f(x+y) \end{aligned}$

Hence statement-II is also correct.

Given that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ :

$4m + n = 22$ ...... (1)

$17m + 4n = 93$ ....... (2)

We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal:

$16m + 4n = 88$ ......... (3)

Now, subtract equation (3) from equation (2):

$(17m + 4n) - (16m + 4n) = 93 - 88$

$m = 5$

Now, we can find the value of $n$ by substituting $m$ into equation (1):

$4(5) + n = 22$

$20 + n = 22$

$n = 2$

Since the order of matrix A is $m$, the order is 5. Now, let's find the determinant of $n \operatorname{adj}(\operatorname{adj}(mA))$:

We know that $\operatorname{det}(A) = m-n = 3$.

$ \begin{aligned} & \therefore \operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A))) \\\\ & =|2 \operatorname{adj}(\operatorname{adj}(5 A))| \\\\ & =2^5|5 A|^{16} \\\\ & =2^5 5^{80}|A|^{16}=2^5 \cdot 3^{16} \cdot 5^{80} \\\\ & =3^{11} 5^{80} 6^5 \end{aligned} $

So, $a+b+c=96$

$ \begin{aligned} & A=\left[\begin{array}{lll} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{array}\right] \\\\ & |A|=2 \end{aligned} $

$ \begin{aligned} \Rightarrow&1(6-1)-2(2 \alpha-1)+3(\alpha-3)=2\\\\ \Rightarrow&5-4 \alpha+2+3 \alpha-9=2\\\\ \Rightarrow&-\alpha-4=0\\\\ \Rightarrow&\alpha=-4 \end{aligned} $

Now, $\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$

$ \Rightarrow $ $2^3|\operatorname{adj}(2 \operatorname{adj}(2 A))|$ = ${32^n}$

$ \begin{aligned} \Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\\\ \Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\\\ \Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\\\ \Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\\\ \Rightarrow& 2^{25} = {32^n}\\\\ \Rightarrow& \left(2^5\right)^5 = {32^n}\\\\ \Rightarrow& (32)^5 = {32^n} \end{aligned} $

$ \begin{array}{ll} \therefore n=5 \\\\ \Rightarrow 3 n+\alpha=15-4=11 \end{array} $

$ \begin{aligned} & 2 x+y-z=5 \\ & 2 x-5 y+\lambda z=\mu \\ & x+2 y-5 z=7 \end{aligned} $

For infinite solution $\Delta=0=\Delta_1=\Delta_2=\Delta_3$

$ \Delta=\left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|=0 $

$ \begin{aligned} \Rightarrow& 2(25-2 \lambda)-(-10-\lambda)-(4+5)=0 \\\\ \Rightarrow& 50-4 \lambda+10+\lambda-9=0 \\\\ \Rightarrow& 51=3 \lambda \Rightarrow \lambda=17 \end{aligned} $

$ \Delta_3=\left|\begin{array}{ccc} 5 & 2 & 1 \\ \mu & 2 & -5 \\ 7 & 1 & 2 \end{array}\right|=0 $

$ \begin{aligned} \Rightarrow & 2(-35-2 \mu)-(14-\mu)+5(4+5)=0 \\\\ \Rightarrow & -70-4 \mu-14+\mu+45=0 \\\\ \Rightarrow & -3 \mu=39 \\\\ \Rightarrow & -\mu=13 \end{aligned} $

Now $(\lambda+\mu)^2+(\lambda-\mu)^2$

$ \begin{aligned} & (17+13)^2+(17-13)^2 \\\\ & 900+16 \\\\ & =916 \end{aligned} $

The given system of equations is :

1. $2x + 4y + 2az = b$

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

3. $2x - 5y + 2z = 8$

We can write this in matrix form :

$ \begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} b \\ 4 \\ 8 \end{bmatrix} $

First, we need to find the determinant of the coefficient matrix, which we'll call Δ. The coefficient matrix is :

$ \begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix} $

We find its determinant for 3 $ \times $ 3 matrices :

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

$ \Delta = 8 + 24 - 10a - 8a - 8 + 30 $

$ \Delta = -18a + 54 $

$ \Delta = -18(a - 3) $

Next, we substitute the third column of our matrix with the column of constants (b, 4, 8) and calculate the determinant Δ₃ :

$ \begin{bmatrix} 2 & 4 & b \\ 1 & 2 & 4 \\ 2 & -5 & 8 \end{bmatrix} $

We find its determinant :

$ \Delta_3 = 2(2)(8) + 4(4)(2) + b(1)(-5) - b(2)(2) - 4(1)(8) - 2(4)(-5) $

$ \Delta_3 = 32 + 32 - 5b - 4b - 32 + 40 $

$ \Delta_3 = -9b + 72 $

$ \Delta_3 = 9(8 - b) $

Now, let's analyze the options :

1. For a=3 and b=8, we have Δ = 0 and Δ₃ = 0, which indicates an infinite number of solutions.

2. For a=3 and b=6, we have Δ = 0 and Δ₃ ≠ 0, which indicates no solution.

3. For a=8 and b=8, we have Δ ≠ 0, which indicates a unique solution.

4. For a=6 and b=6, we have Δ ≠ 0, which indicates a unique solution.

Therefore, the statement that is NOT correct is Option B: "It has infinitely many solutions if a=3, b=6", because in this case the system actually has no solution.

$ B=\left[\begin{array}{lll} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{array}\right], \alpha>2 $

And $\operatorname{adj}(A)=B,|A|=2$

$ \begin{aligned} & \Rightarrow|\operatorname{adj}(A)|=|B| \\\\ & \Rightarrow 2^2=(8-3 \alpha)-3(4-3 \alpha)+\alpha(-\alpha) \\\\ & \Rightarrow \alpha^2-6 \alpha+8=0 \end{aligned} $

$ \begin{aligned} \Rightarrow & (\alpha-4)(\alpha-2)=0 \\\\ & \alpha=4,2 \text { but } \alpha>2 \text { so } \alpha=4 \end{aligned} $

Now

$ \begin{aligned} & {\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c} \alpha \\ -2 \alpha \\ \alpha \end{array}\right]=\left[\begin{array}{lll} 4-8 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right]} \\\\ & =\left[\begin{array}{lll} 12 & 12 & 8 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right] \\\\ & = {48-96+32=-16} \end{aligned} $

Sure! A symmetric matrix is a square matrix that is equal to its transpose. For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i. In other words, $A_{ij} = A_{ji}$.

For a 3 $ \times $ 3 symmetric matrix, it looks like this:

$ \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \\ \end{pmatrix} $

Notice that there are only 6 unique elements we need to fill because of the symmetry:

1. $a$ in the (1,1) position
2. $b$ in the (1,2) and (2,1) positions
3. $c$ in the (1,3) and (3,1) positions
4. $d$ in the (2,2) position
5. $e$ in the (2,3) and (3,2) positions
6. $f$ in the (3,3) position

Each of these unique elements can take a value from the set ${0,1,2,3,4,5,6,7,8,9}$, which has 10 elements.

We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:

$10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}$

Thus, the total number of symmetric matrices of order 3 with entries from this set is $10^{6}$.

$ \begin{aligned} & \text { Let } C=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right], \mathrm{D}=\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & \mathrm{DC}=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} \end{aligned} $

$ \begin{aligned} & \mathrm{B}=\mathrm{CAD} \\\\ & \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text {n-times }} \end{aligned} $

$ \Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D} $

$ \mathrm{A}^2=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right] $

$ \mathrm{A}^3=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{3}{51} \\ 0 & 1 \end{array}\right] $

$ \text { Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right] $

$ \begin{aligned} \mathrm{B}^{\mathrm{n}} & =\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51}+2 \\ -1 & -\frac{\mathrm{n}}{51}-1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} \frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\ -\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51} \end{array}\right] \end{aligned} $

$ \sum\limits_{n=1}^{50} B^n=\left[\begin{array}{cc} 50+\frac{50 \cdot 51}{2 \cdot 51} & \frac{50 \cdot 51}{2 \cdot 51} \\\\ \frac{-50 \cdot 51}{2 \cdot 51} & 50-\frac{50 \cdot 51}{2 \cdot 51} \end{array}\right]=\left[\begin{array}{cc} 75 & 25 \\ -25 & 25 \end{array}\right] $

$ \therefore $ Sum of the elements = 100

Given,

$ \begin{aligned} & 7 x+11 y+\alpha z=13 \\\\ & 5 x+4 y+7 z=\beta \\\\ & 175 x+194 y+57 z=361 \end{aligned} $

$ \text { For infinite solution, }\left|\begin{array}{ccc} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 175 & 194 & 57 \end{array}\right|=0 $

$ \Rightarrow\left|\begin{array}{ccc} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 0 & -81 & 57-25 \alpha \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow 81(49-5 \alpha)+(57-25 \alpha)(-27)=0 \\\\ & \Rightarrow 270 \alpha=-2430 \Rightarrow \alpha=-9 \end{aligned} $

And $\Delta_1=0$

$ \begin{aligned} & \left|\begin{array}{ccc} 13 & 11 & -9 \\ \beta & 4 & 7 \\ 361 & 194 & 57 \end{array}\right|=0 \\\\ & \Rightarrow \beta=11 \end{aligned} $

$ \therefore \alpha+\beta+2=4 $

$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$

Put $x=0$

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{array}\right|=\frac{9}{8} \times 81 \\\\ & \lambda^3=\frac{3^6}{2^3} \end{aligned} $

$ \begin{aligned} & \Rightarrow \lambda=\frac{9}{2} \\\\ & \Rightarrow \frac{\lambda}{3}=\frac{3}{2} \end{aligned} $

$ \begin{aligned} & \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\ & \Rightarrow 4 x^2-24 x+27=0 \end{aligned} $