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

The matrix equation \( A \cdot B = 0 \) is provided, with \[ A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \] and \[ B = \begin{bmatrix} 1 \\ x \\ 2 \end{bmatrix} \]. Performing the matrix multiplication yields: \[ A \cdot B = \begin{bmatrix} 1 \times 1 + 3 \times x + 2 \times 2 \\ 2 \times 1 + 5 \times x + 1 \times 2 \\ 15 \times 1 + 3 \times x + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 1 + 3x + 4 \\ 2 + 5x + 2 \\ 15 + 3x + 4 \end{bmatrix} = \begin{bmatrix} 3x + 5 \\ 5x + 4 \\ 3x + 19 \end{bmatrix} \] For \( A \cdot B \) to be the zero matrix, the following system of equations must be satisfied: \[ \begin{bmatrix} 3x + 5 \\ 5x + 4 \\ 3x + 19 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] This results in three separate equations: 1. \( 3x + 5 = 0 \), yielding \( x = -\frac{5}{3} \). 2. \( 5x + 4 = 0 \), yielding \( x = -\frac{4}{5} \). 3. \( 3x + 19 = 0 \), yielding \( x = -\frac{19}{3} \). Since no single value of \( x \) satisfies all three equations simultaneously, there is no solution for \( x \).