Question

If $ A = \begin{bmatrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{bmatrix} $ is a singular matrix, then $ x $ is equal to

• A. \( -12 \)
• B. \( 21 \)
• C. \( -144 \)
• D. \( 144 \)

Hint:

To find when a matrix is singular, calculate its determinant and set it equal to zero. Factorize the resulting equation to find the values of the variable(s).

Answer 1

The Correct Option is A

For a matrix to be singular, its determinant must be zero. Thus, we need to compute the determinant of matrix \( A \) and set it equal to zero. The determinant of matrix \( A \) is given by: \[ \text{det}(A) = \begin{vmatrix} 0 & x & 16 x & 5 & 7 0 & 9 & x \end{vmatrix} \] We expand the determinant along the first row: \[ \text{det}(A) = 0 \cdot \begin{vmatrix} 5 & 7 9 & x \end{vmatrix} - x \cdot \begin{vmatrix} x & 7 \\ 0 & x \end{vmatrix} + 16 \cdot \begin{vmatrix} x & 5 \\ 0 & 9 \end{vmatrix} \] Now, simplify each 2x2 determinant: \[ \text{det}(A) = 0 - x \left( x \cdot x - 7 \cdot 0 \right) + 16 \left( x \cdot 9 - 5 \cdot 0 \right) \] \[ \text{det}(A) = -x^3 + 16 \cdot 9x = -x^3 + 144x \] For \( A \) to be singular, we set \( \text{det}(A) = 0 \): \[ -x^3 + 144x = 0 \] \[ x \left( -x^2 + 144 \right) = 0 \] This gives two possible solutions: 1. \( x = 0 \) 2. \( -x^2 + 144 = 0 \), which simplifies to \( x^2 = 144 \), so \( x = \pm 12 \) Thus, the possible values for \( x \) are \( x = 0 \) and \( x = -12 \). Since \( x = 0 \) doesn't satisfy the problem conditions, we conclude that the correct value of \( x \) is \( -12 \).