i have the following matrix : 
 B = ( 1, 2x, 8
 1, 2x +2, 2x+8
 1, 2x+6, 7x-15 )
x is a parameter
it is given to me that for the parameters z, y the following is correct
(B^99)\* (-392, y, z) = (0, 0, 0) 
find x
user
sure, i can help you with that.

the matrix $b$ can be written as follows:

$$b = \begin{bmatrix} 1 & 2x & 8 \\ 1 & 2x+2 & 2x+8 \\ 1 & 2x+6 & 7x-15 \end{bmatrix}$$

the determinant of $b$ can be calculated as follows:

$$\det(b) = -27x^3 - 10x^2 + 80x + 120$$

the characteristic polynomial of $b$ can be calculated as follows:

$$p(t) = \det(ti - b) = t^3 - 392t^2 - 80t + 120$$

the roots of the characteristic polynomial are $t_1 = 2, t_2 = -4\sqrt{11} - 3, t_3 = -4\sqrt{11} + 3$.

the eigenvalues of $b$ are the roots of the characteristic polynomial. therefore, the eigenvalues of $b$ are $2, -4\sqrt{11} - 3, -4\sqrt{11} + 3$.

the eigenvectors of $b$ can be calculated using the following formula:

$$v_i = \frac{1}{\sqrt{\det(b - \lambda_i i)}}(b - \lambda_i i)^{-1} e_i$$

where $e_i$ is the $i$th standard basis vector.

the eigenvectors of $b$ are as follows:

$$v_1 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$
$$v_2 = \begin{bmatrix} 2\sqrt{11} \\ -4 \\ 0 \end{bmatrix}$$
$$v_3 = \begin{bmatrix} -2\sqrt{11} \\ 4 \\ 0 \end{bmatrix}$$

the matrix $b$ can be written in the following eigenvalue-eigenvector form:

$$b = udv^{-1}$$

where $u$ is the matrix of eigenvectors, $d$ is the diagonal matrix of eigenvalues, and $v^{-1}$ is the matrix of the inverses of the eigenvectors.

in this case, $u$ is the following matrix:

$$u = \begin{bmatrix} 1 & 2\sqrt{11} & -2\sqrt{11} \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

$d$ is the following matrix:

$$d = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -4\sqrt{11} - 3 & 0 \\ 0 & 0 & -4\sqrt{11} + 3 \end{bmatrix}$$

and $v^{-1}$ is the following matrix:

$$v^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 2\sqrt{11} & 1 & 0 \\ -2\sqrt{11} & 0 & 1 \end{bmatrix}$$

the matrix $b$ can be written in the following form:

$$b = u\begin{bmatrix} 2 & 0 & 0 \\ 0 & -4\sqrt{11} - 3 & 0 \\ 0 & 0 & -4\sqrt{11} + 3 \end{bmatrix}v^{-1}$$

the matrix $b^99$ can be calculated as follows:

$$b^99 = \left(u\begin{bmatrix} 2 & 0 & 0 \\ 0 & -4\sqrt{11} - 3 & 0 \\ 0 & 0 & -4\sqrt{11} + 3 \end{bmatrix}v^{-1}\right)^99 = u\begin{bmatrix} 2^99 & 0 & 0 \\ 0 & (-4\sqrt{11} - 3)^99 & 0 \\ 0 & 0 & (-4\sqrt{11} + 3)^99 \end{bmatrix}v^{-1}$$

the vector $(-392, y
assistant
Sure, I can help you with that.

The matrix $B$ can be written as follows:

$$B = \begin{bmatrix} 1 & 2x & 8 \\ 1 & 2x+2 & 2x+8 \\ 1 & 2x+6 & 7x-15 \end{bmatrix}$$

The determinant of $B$ can be calculated as follows:

$$\det(B) = -27x^3 - 10x^2 + 80x + 120$$

The characteristic polynomial of $B$ can be calculated as follows:

$$p(t) = \det(tI - B) = t^3 - 392t^2 - 80t + 120$$

The roots of the characteristic polynomial are $t_1 = 2, t_2 = -4\sqrt{11} - 3, t_3 = -4\sqrt{11} + 3$.

The eigenvalues of $B$ are the roots of the characteristic polynomial. Therefore, the eigenvalues of $B$ are $2, -4\sqrt{11} - 3, -4\sqrt{11} + 3$.

The eigenvectors of $B$ can be calculated using the following formula:

$$v_i = \frac{1}{\sqrt{\det(B - \lambda_i I)}}(B - \lambda_i I)^{-1} e_i$$

where $e_i$ is the $i$th standard basis vector.

The eigenvectors of $B$ are as follows:

$$v_1 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$
$$v_2 = \begin{bmatrix} 2\sqrt{11} \\ -4 \\ 0 \end{bmatrix}$$
$$v_3 = \begin{bmatrix} -2\sqrt{11} \\ 4 \\ 0 \end{bmatrix}$$

The matrix $B$ can be written in the following eigenvalue-eigenvector form:

$$B = UDV^{-1}$$

where $U$ is the matrix of eigenvectors, $D$ is the diagonal matrix of eigenvalues, and $V^{-1}$ is the matrix of the inverses of the eigenvectors.

In this case, $U$ is the following matrix:

$$U = \begin{bmatrix} 1 & 2\sqrt{11} & -2\sqrt{11} \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

$D$ is the following matrix:

$$D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -4\sqrt{11} - 3 & 0 \\ 0 & 0 & -4\sqrt{11} + 3 \end{bmatrix}$$

and $V^{-1}$ is the following matrix:

$$V^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 2\sqrt{11} & 1 & 0 \\ -2\sqrt{11} & 0 & 1 \end{bmatrix}$$

The matrix $B$ can be written in the following form:

$$B = U\begin{bmatrix} 2 & 0 & 0 \\ 0 & -4\sqrt{11} - 3 & 0 \\ 0 & 0 & -4\sqrt{11} + 3 \end{bmatrix}V^{-1}$$

The matrix $B^99$ can be calculated as follows:

$$B^99 = \left(U\begin{bmatrix} 2 & 0 & 0 \\ 0 & -4\sqrt{11} - 3 & 0 \\ 0 & 0 & -4\sqrt{11} + 3 \end{bmatrix}V^{-1}\right)^99 = U\begin{bmatrix} 2^99 & 0 & 0 \\ 0 & (-4\sqrt{11} - 3)^99 & 0 \\ 0 & 0 & (-4\sqrt{11} + 3)^99 \end{bmatrix}V^{-1}$$

The vector $(-392, y
assistant