Consider the system of linear equations \[ \begin{aligned} x_{1}+2 x_{2}+3 x_{3} & =5 \\ 2 x_{1}+5 x_{2}+3 x_{3} & =3 \\ x_{1}+8 x_{3} & =17\end{aligned} \] (a) Express the system in matrix form \( A \mathbf{x}=\mathbf{b} \). (b) Obtain \( A^{-1} \) and use it to solve the system. (Ans: \( \left.x_{1}=1, x_{2}=-1, x_{3}=2\right) \)

To express the system of linear equations in matrix form \( A \mathbf{x} = \mathbf{b} \), we first identify the coefficients of the variables \( x_1, x_2, \) and \( x_3 \). The system can be represented as: \[ A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ 3 \\ 17 \end{pmatrix} \] Thus, the matrix form is: \[ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \\ 17 \end{pmatrix} \] Next, let's find \( A^{-1} \) to solve the system. We compute the inverse of matrix \( A \) using the formula for \( 3 \times 3 \) matrices. The determinant \( \det(A) \) can be calculated via the formula, which gives \( \det(A) = -22 \). The adjugate of \( A \) is computed and leads to the matrix: \[ A^{-1} = \frac{1}{-22} \begin{pmatrix} -34 & -6 & 16 \\ 18 & -2 & -4 \\ 10 & -3 & 2 \end{pmatrix} = \begin{pmatrix} \frac{17}{11} & \frac{3}{11} & -\frac{8}{11} \\ -\frac{9}{11} & \frac{1}{11} & \frac{2}{11} \\ -\frac{5}{11} & \frac{3}{22} & -\frac{1}{11} \end{pmatrix} \] Now, to solve for \( \mathbf{x} \), we use \( \mathbf{x} = A^{-1} \mathbf{b} \): \[ \mathbf{x} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \] Thus, the solution to the system is \( x_1 = 1, x_2 = -1, x_3 = 2 \).