Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n

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Step 1: Compute AB

First, we compute the product of matrices $A$ and $B$:\r\n$$AB = \begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix} \begin{bmatrix}1&-1&1\\ 1&-2&-2\\ 2&1&3\end{bmatrix}$$\r\n$$AB = \begin{bmatrix}-4+4+8&4-8+4&-4-8+12\\ -7+1+6&7-2+3&-7-2+9\\ 5-3-2&-5+6-1&5+6-3\end{bmatrix}$$\r\n$$AB = \begin{bmatrix}8&0&0\\ 0&8&0\\ 0&0&8\end{bmatrix} = 8I$$\r\nWhere $I$ is the identity matrix.

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Step 2: Express the system of equations in matrix form

The given system of linear equations is:\r\n$x-y+z=4$\r\n$x-2y-2z=9$\r\n$2x+y+3z=1$\r\nThis can be written in matrix form as:\r\n$$\begin{bmatrix}1&-1&1\\ 1&-2&-2\\ 2&1&3\end{bmatrix} \begin{bmatrix}x\\ y\\ z\end{bmatrix} = \begin{bmatrix}4\\ 9\\ 1\end{bmatrix}$$\r\n$$BX = C$$\r\nWhere $B=\begin{bmatrix}1&-1&1\\ 1&-2&-2\\ 2&1&3\end{bmatrix}$, $X=\begin{bmatrix}x\\ y\\ z\end{bmatrix}$, and $C=\begin{bmatrix}4\\ 9\\ 1\end{bmatrix}$.

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Step 3: Use the result AB = 8I to find B^{-1}

Since $AB = 8I$, we have $A(\frac{1}{8}B) = I$. Therefore, $A = 8B^{-1}$ and $B^{-1} = \frac{1}{8}A$.\r\n$$B^{-1} = \frac{1}{8} \begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix}$$\r\n

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Step 4: Solve for X

We have $BX = C$. Multiplying both sides by $B^{-1}$, we get:\r\n$B^{-1}BX = B^{-1}C$\r\n$IX = B^{-1}C$\r\n$X = B^{-1}C$\r\nSubstituting the values of $B^{-1}$ and $C$:\r\n$$X = \frac{1}{8} \begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix} \begin{bmatrix}4\\ 9\\ 1\end{bmatrix}$$\r\n$$X = \frac{1}{8} \begin{bmatrix}-16+36+4\\ -28+9+3\\ 20-27-1\end{bmatrix}$$\r\n$$X = \frac{1}{8} \begin{bmatrix}24\\ -16\\ -8\end{bmatrix}$$\r\n$$X = \begin{bmatrix}3\\ -2\\ -1\end{bmatrix}$$\r\nTherefore, $x=3$, $y=-2$, and $z=-1$.

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\r\n Final Answer: x = 3, y = -2, z = -1<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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