Question

The determinant of the matrix $\begin{bmatrix}3 & -1& -2\\0& 0& -1\\3 & -5& 0\end{bmatrix}$ is:

• A. 12
• B. -12
• C. 0
• D. -15

Hint:

Expansion along the row with the most zeros is the fastest way to solve determinants. Always keep track of the sign checkerboard $(\pm)$.

Answer 1

The Correct Option is B

Concept: The determinant of a $3 \times 3$ matrix can be calculated by expanding along any row or column. To simplify the calculation, choose the row or column with the most zeros.

Step 1: Choose the expansion row.
Row 2 $(R_2)$ is $[0, 0, -1]$. This is the best choice because two of the terms are zero. The signs for expansion are: $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

Step 2: Expand along Row 2.
$Det(A) = - (0) \cdot M_{21} + (0) \cdot M_{22} - (-1) \cdot M_{23}$ $Det(A) = 1 \cdot M_{23}$ Where $M_{23}$ is the minor obtained by removing Row 2 and Column 3.

Step 3: Calculate the $2 \times 2$ determinant.
$M_{23} = \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}$ $M_{23} = (3)(-5) - (-1)(3) = -15 + 3 = -12$. Therefore, $Det(A) = 1 \cdot (-12) = -12$.