Choose appropriate values for \( a \) and \( b \), PG19 a. solve the SLE by using matrix equation, where get the inverse matrix by using i. elementary row operation PG23 ii. matrix method b. verify your answer with Gauss Elimination Method. P6 10,P616 c. if the system has no solution, what is the value of \( a \) and \( b \) ?

To solve a system of linear equations (SLE) using matrix methods, we first need to define a specific system of equations. Let's choose the following system for our example: \[ \begin{align*} 2x + 3y &= 8 \quad (1) \\ 4x + 6y &= 16 \quad (2) \end{align*} \] ### Part a: Solve the SLE using matrix equations We can express this system in matrix form as \( AX = B \), where: \[ A = \begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 16 \end{pmatrix} \] #### i. Finding the inverse matrix using elementary row operations To find the inverse of matrix \( A \), we augment it with the identity matrix: \[ \left( \begin{array}{cc|cc} 2 & 3 & 1 & 0 \\ 4 & 6 & 0 & 1 \end{array} \right) \] We will perform row operations to convert the left side into the identity matrix. 1. Divide the first row by 2: \[ \left( \begin{array}{cc|cc} 1 & \frac{3}{2} & \frac{1}{2} & 0 \\ 4 & 6 & 0 & 1 \end{array} \right) \] 2. Subtract 4 times the first row from the second row: \[ \left( \begin{array}{cc|cc} 1 & \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & 0 & -2 & 1 \end{array} \right) \] 3. The second row indicates that the system is dependent (the second equation is a multiple of the first). Thus, we cannot find an inverse for \( A \) because it is singular. #### ii. Finding the inverse matrix using the matrix method The determinant of \( A \) is calculated as follows: \[ \text{det}(A) = (2)(6) - (3)(4) = 12 - 12 = 0 \] Since the determinant is zero, the matrix \( A \) does not have an inverse. ### Part b: Verify your answer with Gauss Elimination Method Using Gauss elimination on the augmented matrix: \[ \left( \begin{array}{cc|c} 2 & 3 & 8 \\ 4 & 6 & 16 \end{array} \right) \] 1. Divide the first row by 2: \[ \left( \begin{array}{cc|c} 1 & \frac{3}{2} & 4 \\ 4 & 6 & 16 \end{array} \right) \] 2. Subtract 4 times the first row from the second row: \[ \left( \begin{array}{cc|c} 1 & \frac{3}{2} & 4 \\ 0 & 0 & 0 \end{array} \right) \] The second row indicates that the system is dependent, confirming that there are infinitely many solutions along the line defined by the first equation. ### Part c: If the system has no solution, what is the value of \( a \) and \( b \)? For a system to have no solution, the equations must be inconsistent. For example, if we modify the second equation to: \[ 4x + 6y = 20 \quad (2') \] Now the augmented matrix becomes: \[ \left( \begin{array}{cc|c} 2 & 3 & 8 \\ 4 & 6 & 20 \end{array} \right) \] Performing the same row operations: 1. Divide the first row by 2: \[ \left( \begin{array}{cc|c} 1 & \frac{3}{2} & 4 \\ 4 & 6 & 20 \end{array} \right) \] 2. Subtract 4 times the first row from the second row: \[ \left( \begin{array}{cc|c} 1 & \frac{3}{2} & 4 \\ 0 & 0 & 4 \end{array} \right) \] The second row \( 0 = 4 \) is a contradiction, indicating that the system has no solution. Thus, for