Mathematics

$\begin{aligned} &\begin{aligned} & \sin x+\sin ^2 x=1 \\ & \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x \end{aligned}\\ &\therefore \text { Given expression }\\ &\begin{aligned} & =2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x \\ & =2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right] \\ & =2 \sin ^3 x\left[(\sin x+1)^3\right] \\ & =2\left[\sin ^2 x+\sin x\right]^3 \\ & =2 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)-\left(\frac{\pi}{4}\right)-(\mathrm{r}-1) \frac{\pi}{6}\right]}{\sin \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)} \\ & \frac{1}{\sin \frac{\pi}{6}} \sum_{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)\right) \\ & =2 \sqrt{3}-2=\alpha \sqrt{3}+\mathrm{b} \end{aligned}\\ &\text { So } \mathrm{a}^2+\mathrm{b}^2=8 \end{aligned}$

$\begin{aligned} & \sin 70^{\circ}\left(\cot 10^{\circ} \cot 70^{\circ}-1\right) \\ & \Rightarrow \frac{\cos \left(80^{\circ}\right)}{\sin 10}=1 \end{aligned}$

$\begin{aligned} &\begin{aligned} f(x) & =6+16\left(\frac{1}{4} \cos 3 x\right) \sin 3 x \cdot \cos 6 x \\ & =6+4 \cos 3 x \sin 3 x \cos 6 x \\ & =6+\sin 12 x \end{aligned}\\ &\text { Range of } f(x) \text { is [5, 7] }\\ &\begin{aligned} & (\alpha, \beta) \equiv(5,7) \\ & \text { distance }=\left|\frac{15+28+12}{5}\right|=11 \end{aligned} \end{aligned}$

$\begin{aligned} &\text { Let } \phi=\theta+\frac{\pi}{3} \Rightarrow \theta=\phi-\frac{\pi}{6}\\ &\begin{aligned} & x=3 \tan \left(\theta+\frac{\pi}{3}\right)=3 \tan \left(\theta+\frac{\pi}{6}\right) \\ & y=2 \tan \phi \\ & \tan \left(\phi+\frac{\pi}{6}\right)=\frac{\tan \phi+\frac{1}{\sqrt{3}}}{1-\tan \phi \cdot \frac{1}{\sqrt{3}}} \\ & \frac{x}{3}=\frac{\frac{y}{2}+\frac{1}{\sqrt{3}}}{1-\frac{y}{2} \cdot \frac{1}{\sqrt{3}}} \end{aligned} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad x=\frac{3(y \sqrt{3}+2)}{2 \sqrt{3}-y} \\ & x y+\alpha x+\beta y+r=0 \\ & 3\left(\frac{y \sqrt{3}+2}{2 \sqrt{3}-y}\right)+\alpha\left(3 \frac{(y \sqrt{3}+2)}{(2 \sqrt{3}-y)}\right)+\beta y+r=0 \\ & =(3 \sqrt{3}-\beta) y^2+(6+3 \sqrt{3} \alpha+2 \sqrt{3} \beta-y) y \\ & \quad+(6 \alpha+2 \sqrt{3} y)=0 \end{aligned}$

For this identity to hold for all $\theta$, coefficients must be 0

$\begin{aligned} & \therefore \quad \beta=3 \sqrt{3} \\ & \gamma=-\alpha \sqrt{3} \\ & 6+3 \sqrt{3} \alpha+(2 \sqrt{3})(3 \sqrt{3})+\alpha \sqrt{3}=0 \\ & \Rightarrow \alpha=-2 \sqrt{3} \\ & \Rightarrow \beta=6 \\ & \alpha^2+\beta^2+\gamma^2=75 \end{aligned}$

$\begin{aligned} &\begin{aligned} & 10 \sin ^4 \theta+15 \cos ^4 \theta=6 \\ \Rightarrow & 10 \sin ^4 \theta+10 \cos ^4 \theta+5 \cos ^4 \theta=6 \\ \Rightarrow & 10\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6 \\ \Rightarrow & 10-20\left(1-\cos ^2 \theta\right) \cos ^2 \theta+5 \cos ^4 \theta=6 \end{aligned}\\ &\text { Let } \cos ^2 \theta=x\\ &10-20\left(x-x^2\right)+5 x^2=6 \end{aligned}$

$\begin{aligned} \Rightarrow & 25 x^2-20 x+4=0 \\ & (5 x-2)^2=0 \Rightarrow x=\frac{2}{5} \\ \Rightarrow & \cos ^2 \theta=\frac{2}{5} \Rightarrow \sin ^2 \theta=\frac{3}{5}, \\ & \sec ^2 \theta=\frac{5}{2}, \operatorname{cosec}^2 \theta=\frac{5}{3} \end{aligned}$

$\begin{aligned} \frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} & =\frac{27\left(\frac{5}{3}\right)^3+8\left(\frac{5}{2}\right)^3}{16\left(\frac{5}{2}\right)^4} \\ & =\frac{5^3+5^3}{5^4}=\frac{2.5^3}{5^4}=\frac{2}{5} \end{aligned}$

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 $

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

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

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

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

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

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

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

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 $\mathrm{f}(\mathrm{x})=-2$

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

$\begin{aligned} &\begin{aligned} & A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right] \\ & A^2=\left[\begin{array}{ll} 3 & -2 \\ 2 & -1 \end{array}\right], A^3=\left[\begin{array}{ll} 4 & -3 \\ 3 & -2 \end{array}\right], A^4=\left[\begin{array}{ll} 5 & -4 \\ 4 & -3 \end{array}\right] \end{aligned}\\ &\text { and so on }\\ &\begin{aligned} & \mathrm{A}^6=\left[\begin{array}{ll} 7 & -6 \\ 6 & -5 \end{array}\right] \\ & \mathrm{A}^{\mathrm{m}}=\left[\begin{array}{cc} \mathrm{m}+1 & -\mathrm{m} \\ \mathrm{~m} & -\mathrm{m}+1 \end{array}\right], \\ & \mathrm{A}^{\mathrm{m}^2}=\left[\begin{array}{cc} \mathrm{m}^2+1 & -\mathrm{m}^2 \\ \mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right) \end{array}\right] \\ & \mathrm{A}^{\mathrm{m}^2}+\mathrm{A}^{\mathrm{m}}=3 \mathrm{I}-\mathrm{A}^{-6} \\ & {\left[\begin{array}{cc} \mathrm{m}^2+1 & -\mathrm{m}^2 \\ \mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right) \end{array}\right]+\left[\begin{array}{cc} \mathrm{m}+1 & -\mathrm{m} \\ \mathrm{~m} & -\mathrm{m}+1 \end{array}\right]} \end{aligned} \end{aligned}$

$\begin{aligned} & =3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} -5 & 6 \\ -6 & 7 \end{array}\right] \\ & =\left[\begin{array}{ll} 8 & -6 \\ 6 & -4 \end{array}\right] \\ & =\mathrm{m}^2+1+\mathrm{m}+1=8 \\ & =\mathrm{m}^2+\mathrm{m}-6=0 \Rightarrow \mathrm{~m}=-3,2 \\ & \mathrm{n}(\mathrm{~s})=2 \end{aligned}$

$\begin{aligned} & \text { As } A \operatorname{adj} A=|A| I, \operatorname{det}(\lambda A)=\lambda^n \operatorname{det} A \\\\ & \operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=3^3 \operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3 A))) \\\\ & =3^3(-6 \operatorname{adj}(3 A))^2 \\\\ & =3^3(-6)^6|3 A|^4 \\\\ & =3^9 2^6 \cdot 3^{12} \cdot(-2)^4 \\\\ & =3^{21} \cdot 2^{10}\end{aligned}$

Now comparing with given condition

$ \begin{aligned} & 2^{m+n} 3^{m n}=2^{10} \cdot 3^{21} \\\\ & m+n=10, m n=21 \\\\ & \Rightarrow m=7, n=3(m>n) \\\\ & \therefore 4 m+2 n=28+6=34 \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. $

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

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 $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$

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

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

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

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

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

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