Question

If the determinant of the 3 × 3 matrix \( A = \begin{pmatrix} a & 1 & 2 \\ b & 0 & -2 \\ 1 & -3 & 1 \end{pmatrix} \) is zero, then the values of \( a \) and \( b \) are:

• A. a = 0, b = 0
• B. a = $\frac{3}{2}$, b = 1
• C. a = $\frac{1}{3}$, b = 0
• D. a = $-\frac{1}{6}$, b = $-\frac{1}{7}$

Hint:

When solving for determinants, simplify the expression carefully and make sure to check the signs for each minor matrix.

Answer 1

The Correct Option is D

To solve this, we first find the determinant of the given matrix \( A \).
The determinant of a 3x3 matrix is calculated as: \[ {det}(A) = a \begin{vmatrix} 0 & -2 \\ -3 & 1 \end{vmatrix} - 1 \begin{vmatrix} b & -2 \\ 1 & 1 \end{vmatrix} + 2 \begin{vmatrix} b & 0 \\ 1 & -3 \end{vmatrix} \] Now calculate the minors: \[ \begin{vmatrix} 0 & -2 \\ -3 & 1 \end{vmatrix} = (0)(1) - (-2)(-3) = -6 \] \[ \begin{vmatrix} b & -2 \\ 1 & 1 \end{vmatrix} = (b)(1) - (-2)(1) = b + 2 \] \[ \begin{vmatrix} b & 0 \\ 1 & -3 \end{vmatrix} = (b)(-3) - (0)(1) = -3b \] Substitute these values into the determinant formula: \[ {det}(A) = a(-6) - 1(b + 2) + 2(-3b) = -6a - b - 2 - 6b = -6a - 7b - 2 \] Set the determinant equal to zero to find the values of a and b: \[ -6a - 7b - 2 = 0 ⇒ 6a + 7b = -2 \] Solving this equation gives \( a = -\frac{1}{6} \) and \( b = -\frac{1}{7} \).