Answer:
(a) To compute the determinant of the system of equations we need to write the system in matrix form. The matrix equation can be written as:
\[ \begin{pmatrix} 0 & -3 & 7 \\ 1 & 2 & -1 \\ 5 & -2 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} \]
The determinant of the coefficient matrix is given by:
\[ \begin{vmatrix} 0 & -3 & 7 \\ 1 & 2 & -1 \\ 5 & -2 & 0 \end{vmatrix} \]
Using cofactor expansion along the first row the determinant can be calculated as follows:
\[ \begin{vmatrix} 0 & -3 & 7 \\ 1 & 2 & -1 \\ 5 & -2 & 0 \end{vmatrix} = 0 \times \begin{vmatrix} 2 & -1 \\ -2 & 0 \end{vmatrix} - (-3) \times \begin{vmatrix} 1 & -1 \\ 5 & 0 \end{vmatrix} + 7 \times \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} \]
Simplifying the determinants:
\[ \begin{vmatrix} 2 & -1 \\ -2 & 0 \end{vmatrix} = (2 \times 0) - (-1 \times -2) = 2 \]
\[ \begin{vmatrix} 1 & -1 \\ 5 & 0 \end{vmatrix} = (1 \times 0) - (-1 \times 5) = 5 \]
\[ \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} = (1 \times -2) - (2 \times 5) = -12 \]
Substituting back into the original equation:
\[ \begin{vmatrix} 0 & -3 & 7 \\ 1 & 2 & -1 \\ 5 & -2 & 0 \end{vmatrix} = 0 - (-3 \times 5) + 7 \times (-12) = 45 + (-84) = -39 \]
Therefore the determinant of the system of equations is -39.
(b) To use Cramer's rule to solve for the \(x\)'s we can find the values of \(x_1\ \(x_2\ and \(x_3\) using the following formulas:
\[ x_1 = \frac{{D_1}}{{D}} \]
\[ x_2 = \frac{{D_2}}{{D}} \]
\[ x_3 = \frac{{D_3}}{{D}} \]
where \(D_1\ \(D_2\ and \(D_3\) are the determinants of the matrices obtained by replacing the first second and third column of the coefficient matrix with the column on the right side of the equation while \(D\) is the determinant of the coefficient matrix.
\[ D_1 = \begin{vmatrix} 2 & -3 & 7 \\ 3 & 2 & -1 \\ 2 & -2 & 0 \end{vmatrix} \]
\[ D_2 = \begin{vmatrix} 0 & 2 & 7 \\ 1 & 3 & -1 \\ 5 & 2 & 0 \end{vmatrix} \]
\[ D_3 = \begin{vmatrix} 0 & -3 & 2 \\ 1 & 2 & 3 \\ 5 & -2 & 2 \end{vmatrix} \]
Calculating the determinants:
\[ D_1 = 2 \times \begin{vmatrix} 2 & -1 \\ -2 & 0 \end{vmatrix} - (-3) \times \begin{vmatrix} 3 & -1 \\ 2 & 0 \end{vmatrix} + 7 \times \begin{vmatrix} 3 & 2 \\ 2 & -2 \end{vmatrix} = 2(2 \times 0 - (-1) \times -2) - 3(3 have -3x2 + 7(x3) = 2. Substituting x3 = 61/23 we have -3x2 + 7(61/23) = 2. Solving for x2 we get x2 = (2 - 7(61/23))/(-3) = -5/3.
From the first row we have x1 + 2(x2) - 1(x3) = 3. Substituting x2 = -5/3 and x3 = 61/23 we have x1 + 2(-5/3) - 1(61/23) = 3. Solving for x1 we get x1 = 195/69 = 65/23.
Therefore the solution to the system of equations using Gauss elimination with partial pivoting is \(x_1 = \frac{{65}}{{23}}\ \(x_2 = -\frac{{5}}{{3}}\ and \(x_3 = \frac{{61}}{{23}}\).
(d) To check the solution we substitute the values of \(x_1\ \(x_2\ and \(x_3\) back into the original equations:
For the first equation: -3(65/23) + 7(61/23) = 2. This equation holds true.
For the second equation: 2(65/23) - (5/3) + (61/23) = 3. This equation holds true.
For the third equation: 5(65/23) - 2(5/3) = 2. This equation holds true.
Therefore the solution obtained through Cramer's rule and Gauss elimination with partial pivoting is valid and satisfies the original system of equations.