Matrix form of given linear programming is AX = B,

Where A = \(\begin{bmatrix}3&2&-2\\-1&1&-4\\2&-3&4\end{bmatrix},\) x = \(\begin{bmatrix}x\\y\\z\end{bmatrix},\) B = \(\begin{bmatrix}1\\13\\8\end{bmatrix}\)

\(\therefore\) D = |A| = \(\begin{vmatrix}3&2&2\\-1&1&-4\\2&-3&4\end{vmatrix}\) = \(3\begin{vmatrix}1&-4\\-3&4\end{vmatrix}-2\begin{vmatrix}-1&-4\\2&4\end{vmatrix}\)\(-2\begin{vmatrix}-1&1\\2&-3\end{vmatrix}\)

 = 3(4 - 12) - 2(-4 + 8) -2 (3 - 2)

 = -24 - 8 - 2 = -34

D_x = \(\begin{vmatrix}1&2&-2\\13&1&-4\\8&-3&4\end{vmatrix}\) = \(1\begin{vmatrix}1&-4\\-1&4\end{vmatrix}-2\begin{vmatrix}13&-4\\8&4\end{vmatrix}\)\(-2\begin{vmatrix}13&1\\8&-3\end{vmatrix}\)

 = (4 - 12) - (52 + 32) - 2(-39 - 8)

 = -8 - 168 + 94 = -176 + 94 = -82

D_y =   \(\begin{vmatrix}3&1&-2\\-1&13&-4\\2&8&4\end{vmatrix}\) = \(3\begin{vmatrix}13&-4\\8&4\end{vmatrix}-1\begin{vmatrix}-1&-4\\2&4\end{vmatrix}\)\(-2\begin{vmatrix}-1&13\\2&8\end{vmatrix}\)

= 3(52 + 32) - 1(-4 + 8) - 2(-8 - 26)

 = 252 - 4 + 68 = 320 - 4 = 316

D_z =   \(\begin{vmatrix}3&2&1\\-1&1&13\\2&-3&8\end{vmatrix}\) = \(3\begin{vmatrix}1&13\\-3&8\end{vmatrix}-2\begin{vmatrix}-1&13\\2&8\end{vmatrix}\)\(1\begin{vmatrix}-1&1\\2&-3\end{vmatrix}\)

 = 3(8 + 39) - 2(-8 - 26) + (3 - 2)

 = 141 + 68 + 1 = 210

\(\therefore\) x = \(\frac{Dx}D=\frac{-82}{-34}=\frac{41}{17}\) 

y = \(\frac{Dy}D=\frac{316}{-34}=\frac{-158}{17}\)

z = \(\frac{Dz}D=\frac{210}{-34}=\frac{-105}{17}\) 

Hence, solution of given linear programming is 

x = 41/17, y = -158/17 and -105/17.