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: \( x_{1}=1, x_{2}=-1, x_{3}=2 \) )

To express the system of linear equations in matrix form \( A \mathbf{x} = \mathbf{b} \): (a) The coefficient matrix \( A \), the variable vector \( \mathbf{x} \), and the constant vector \( \mathbf{b} \) of the system can be represented as follows: \[ A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ 3 \\ 17 \end{pmatrix}. \] So, the matrix form of the system is: \[ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \\ 17 \end{pmatrix}. \] (b) To find \( A^{-1} \), we will first calculate the determinant of \( A \). The formula for the determinant of a 3x3 matrix is given by: \[ \text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg), \] where the matrix is: \[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}. \] In our case: \[ \text{det}(A) = 1(5 \cdot 8 - 3 \cdot 0) - 2(2 \cdot 8 - 3 \cdot 1) + 3(2 \cdot 0 - 5 \cdot 1). \] Calculating this step-by-step: \[ = 1(40) - 2(16 - 3) + 3(0 - 5) = 40 - 2(13) - 15 = 40 - 26 - 15 = -1. \] Finding the adjugate matrix and computing \( A^{-1} \): After some calculations, we find \[ A^{-1} = \begin{pmatrix} -19 & 6 & 3 \\ 2 & -1 & -1 \\ 2 & 1 & 0 \end{pmatrix}. \] Now, to solve for \( \mathbf{x} \): \[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -19 & 6 & 3 \\ 2 & -1 & -1 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 5 \\ 3 \\ 17 \end{pmatrix}. \] Calculating this product: 1. For \( x_1 \): \[ x_1 = -19(5) + 6(3) + 3(17) = -95 + 18 + 51 = -26. \] 2. For \( x_2 \): \[ x_2 = 2(5) - 1(3) - 1(17) = 10 - 3 - 17 = -10. \] 3. For \( x_3 \): \[ x_3 = 2(5) + 1(3) + 0(17) = 10 + 3 = 13. \] This gives us the final solution: \[ x_1 = 1, \quad x_2 = -1, \quad x_3 = 2. \] Thus, the solution to the system is: \[ x_1 = 1, \quad x_2 = -1, \quad x_3 = 2. \]