Solve the system by using the inverse of the coefficient matrix. $$ \left.\qquad \begin{array}{ccc}\frac{53}{288} & \frac{2}{9} & \frac{65}{288} \ -\frac{7}{144} & \frac{1}{9} & \frac{5}{144} \ \frac{1}{32} & 0 & -\frac{3}{32}\end{array} ight] $$ $$ \begin{aligned} 3 x-6 y+5 z & =75 \ x+7 y+5 z & =-56 \ x-2 y-9 z & =25\end{aligned} $$

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Solve the system by using the inverse of the coefficient matrix. $$ \left.\qquad \begin{array}{ccc}\frac{53}{288} & \frac{2}{9} & \frac{65}{288} \ -\frac{7}{144} & \frac{1}{9} & \frac{5}{144} \ \frac{1}{32} & 0 & -\frac{3}{32}\end{array}ight] $$ $$ \begin{aligned} 3 x-6 y+5 z & =75 \ x+7 y+5 z & =-56 \ x-2 y-9 z & =25\end{aligned} $$

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To solve the system of linear equations using the inverse of the coefficient matrix, we first need to write the system in matrix form as $$ AX = B $$, where $$ A $$ is the coefficient matrix, $$ X $$ is the column matrix of variables, and $$ B $$ is the column matrix of constants.

The system of equations is:
$$ 3x - 6y + 5z = 75 $$
$$ x + 7y + 5z = -56 $$
$$ x - 2y - 9z = 25 $$

The coefficient matrix $$ A $$ is:
$$ A = \left[ \begin{array}{ccc}
3 & -6 & 5 \
1 & 7 & 5 \
1 & -2 & -9
\end{array} \right] $$

The column matrix of variables $$ X $$ is:
$$ X = \left[ \begin{array}{c}
x \
y \
z
\end{array} \right] $$

The column matrix of constants $$ B $$ is:
$$ B = \left[ \begin{array}{c}
75 \
-56 \
25
\end{array} \right] $$

Now we need to find the inverse of matrix $$ A $$, denoted as $$ A^{-1} $$, and then multiply it by matrix $$ B $$ to find the solution $$ X $$.

First, we calculate the determinant of matrix $$ A $$:
$$ \det(A) = 3(7 \cdot (-9) - 5 \cdot (-2)) - (-6)(1 \cdot (-9) - 5 \cdot 1) + 5(1 \cdot 7 - 7 \cdot 1) $$
$$ \det(A) = 3(-63 + 10) - (-6)(-9 - 5) + 5(7 - 7) $$
$$ \det(A) = 3(-53) - (-6)(-14) + 5(0) $$
$$ \det(A) = -159 - 84 + 0 $$
$$ \det(A) = -243 $$

Since the determinant is not zero, the matrix is invertible.

Next, we find the inverse of matrix $$ A $$ using the formula:
$$ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) $$

The adjoint of matrix $$ A $$ (also known as the adjugate) is the transpose of the cofactor matrix. The cofactor matrix is found by taking the determinant of each minor matrix and multiplying it by $$ (-1)^{i+j} $$, where $$ i $$ and $$ j $$ are the row and column indices of the element in the original matrix.

For simplicity, let's calculate the cofactor matrix and then its transpose to get the adjoint matrix.

The cofactor matrix $$ C $$ is:
$$ C = \left[ \begin{array}{ccc}
\det(M_{11}) & \det(M_{12}) & \det(M_{13}) \
\det(M_{21}) & \det(M_{22}) & \det(M_{23}) \
\det(M_{31}) & \det(M_{32}) & \det(M_{33})
\end{array} \right] $$

Where $$ M_{ij} $$ is the minor matrix obtained by deleting the $$ i $$-th row and $$ j $$-th column of $$ A $$.

For example, $$ M_{11} $$ is the minor matrix obtained by deleting the first row and first column of $$ A $$, and so on.

After calculating the minors and their determinants, we get the cofactor matrix $$ C $$, and then we take its transpose to get the adjoint matrix $$ \text{adj}(A) $$.

Let's calculate the minors and the adjoint matrix $$ \text{adj}(A) $$:

$$ M_{11} = \left[ \begin{array}{cc}
7 & 5 \
-2 & -9
\end{array} \right] $$
$$ \det(M_{11}) = 7 \cdot (-9) - 5 \cdot (-2) = -63 + 10 = -53 $$

$$ M_{12} = \left[ \begin{array}{cc}
1 & 5 \
1 & -9
\end{array} \right] $$
$$ \det(M_{12}) = 1 \cdot (-9) - 5 \cdot 1 = -9 - 5 = -14 $$

$$ M_{13} = \left[ \begin{array}{cc}
1 & 7 \
1 & -2
\end{array} \right] $$
\[ \det(M_{13}) = 1 \cdot (-2) - 
The solution to the system is $$ x = 5 $$, $$ y = -2 $$, and $$ z = 3 $$.

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