Question

Using Cramer's rule, solve the following system of equations: \[ 2x_1 + 3x_2 = 5 \] \[ 11x_1 - 5x_2 = 6 \]

Hint:

Cramer's rule is applicable only when the determinant of the coefficient matrix is non-zero. Always check \( {det}(A) \neq 0 \) before proceeding.

Answer 1

Step 1: Write the coefficient matrix \( A \): \[ A = \begin{bmatrix} 2 & 3 \\ 11 & -5 \end{bmatrix}. \] Step 2: Compute the determinant \( \det(A) \): \[ \det(A) = (2)(-5) - (11)(3) = -10 - 33 = -43. \] Step 3: Solve for \( x_1 \) and \( x_2 \) using determinants: \[ x_1 = \frac{\det(A_1)}{\det(A)}, \quad x_2 = \frac{\det(A_2)}{\det(A)}. \] Replace the first column of \( A \) with the constants \([5, 6]\) to get \( A_1 \): \[ A_1 = \begin{bmatrix} 5 & 3 \\ 6 & -5 \end{bmatrix}, \quad \det(A_1) = (5)(-5) - (6)(3) = -25 - 18 = -43. \] Replace the second column of \( A \) with the constants \([5, 6]\) to get \( A_2 \): \[ A_2 = \begin{bmatrix} 2 & 5 \\ 11 & 6 \end{bmatrix}, \quad \det(A_2) = (2)(6) - (11)(5) = 12 - 55 = -43. \] Step 4: Calculate \( x_1 \) and \( x_2 \): \[ x_1 = \frac{\det(A_1)}{\det(A)} = \frac{-43}{-43} = 1, \quad x_2 = \frac{\det(A_2)}{\det(A)} = \frac{-43}{-43} = 1. \] Thus, the solution is: \[ x_1 = 1, \quad x_2 = 1. \]