Given a system of linear equations. \[ \begin{array}{l}x-2 y+z=2 \\ b x+a y=3 \\ -2 y+z=-4\end{array} \] Choose appropriate values for \( a \) and \( b \), a. solve the SLE by using matrix equation, where get the inverse matrix by using i. elementary row operation

To solve the given system of linear equations using matrix methods, we first need to express the system in matrix form. The system of equations is: \[ \begin{array}{l} x - 2y + z = 2 \\ bx + ay = 3 \\ -2y + z = -4 \end{array} \] We can represent this system in the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix. ### Step 1: Define the matrices Let: \[ A = \begin{bmatrix} 1 & -2 & 1 \\ b & a & 0 \\ 0 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 3 \\ -4 \end{bmatrix} \] ### Step 2: Choose values for \( a \) and \( b \) To ensure that the system has a unique solution, we can choose values for \( a \) and \( b \) such that the determinant of matrix \( A \) is non-zero. Let's choose \( a = 1 \) and \( b = 1 \): \[ A = \begin{bmatrix} 1 & -2 & 1 \\ 1 & 1 & 0 \\ 0 & -2 & 1 \end{bmatrix} \] ### Step 3: Calculate the determinant of \( A \) The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] For our matrix \( A \): \[ A = \begin{bmatrix} 1 & -2 & 1 \\ 1 & 1 & 0 \\ 0 & -2 & 1 \end{bmatrix} \] The determinant is calculated as follows: \[ \text{det}(A) = 1 \cdot (1 \cdot 1 - 0 \cdot -2) - (-2) \cdot (1 \cdot 1 - 0 \cdot 0) + 1 \cdot (1 \cdot -2 - 1 \cdot 0) \] Calculating each term: 1. \( 1 \cdot (1 - 0) = 1 \) 2. \( -(-2) \cdot (1 - 0) = 2 \) 3. \( 1 \cdot (-2 - 0) = -2 \) Putting it all together: \[ \text{det}(A) = 1 + 2 - 2 = 1 \] Since the determinant is non-zero, the matrix \( A \) is invertible. ### Step 4: Find the inverse of \( A \) using elementary row operations To find the inverse of \( A \), we can augment \( A \) with the identity matrix and perform row operations: \[ \left[ A | I \right] = \left[ \begin{array}{ccc|ccc} 1 & -2 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -2 & 1 & 0 & 0 & 1 \end{array} \right] \] We will perform row operations to convert the left side into the identity matrix. 1. Subtract Row 1 from Row 2: \[ \left[ \begin{array}{ccc|ccc} 1 & -2 & 1 & 1 & 0 & 0 \\ 0 & 3 & -1 & -1 & 1 & 0 \\ 0 & -2 & 1 & 0 & 0 & 1 \end{array} \right] \] 2. Multiply Row 2 by \( \frac{1}{3} \): \[ \left[ \begin{array}{ccc|ccc} 1 & -2 & 1 & 1 & 0 & 0 \\ 0 & 1 & -\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & -2 & 1 & 0 & 0 & 1 \end{array} \right] \] 3. Add \( 2