Question

Find the value of $x$, if \[ \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ x \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Hint:

For a matrix equation to have a solution, the system of linear equations must be consistent. If there is no single value for \( x \) that satisfies all equations, the system has no solution.

Answer 1

We are given the matrix equation: \[ A \cdot B = 0 \] where \[ A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ x \\ 2 \end{bmatrix} \] Perform the matrix multiplication: \[ A \cdot B = \begin{bmatrix} 1 \times 1 + 3 \times x + 2 \times 2 \\ 2 \times 1 + 5 \times x + 1 \times 2 \\ 15 \times 1 + 3 \times x + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 1 + 3x + 4 \\ 2 + 5x + 2 \\ 15 + 3x + 4 \end{bmatrix} = \begin{bmatrix} 3x + 5 \\ 5x + 4 \\ 3x + 19 \end{bmatrix} \] For this to equal the zero matrix: \[ \begin{bmatrix} 3x + 5 \\ 5x + 4 \\ 3x + 19 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] We now solve the system of equations: 1. \( 3x + 5 = 0 \) \[ x = -\frac{5}{3} \] 2. \( 5x + 4 = 0 \) \[ x = -\frac{4}{5} \] 3. \( 3x + 19 = 0 \) \[ x = -\frac{19}{3} \] We observe that no consistent value for \( x \) satisfies all three equations simultaneously. Therefore, there is no solution for \( x \).