Question

Evaluate the determinant of the following matrix: \[ \begin{vmatrix} 1 & 2 & 5 \\ 1 & 1 & 4 \\ -2 & -3 & -9 \end{vmatrix} \]

• A. \( 2 \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( -1 \)

Hint:

When calculating a determinant of a 3x3 matrix, you can expand along any row or column. This helps simplify the computation.

Answer 1

The Correct Option is C

We are given the matrix: \[ \begin{vmatrix} 1 & 2 & 5
1 & 1 & 4
-2 & -3 & -9 \end{vmatrix} \] To evaluate the determinant, we can expand along the first row: \[ \text{Determinant} = 1 \times \begin{vmatrix} 1 & 4
-3 & -9 \end{vmatrix} - 2 \times \begin{vmatrix} 1 & 4
-2 & -9 \end{vmatrix} + 5 \times \begin{vmatrix} 1 & 1
-2 & -3 \end{vmatrix} \] Now calculate each 2x2 determinant: \[ \begin{vmatrix} 1 & 4
-3 & -9 \end{vmatrix} = (1)(-9) - (4)(-3) = -9 + 12 = 3 \] \[ \begin{vmatrix} 1 & 4
-2 & -9 \end{vmatrix} = (1)(-9) - (4)(-2) = -9 + 8 = -1 \] \[ \begin{vmatrix} 1 & 1
-2 & -3 \end{vmatrix} = (1)(-3) - (1)(-2) = -3 + 2 = -1 \] Substituting back: \[ \text{Determinant} = 1 \times 3 - 2 \times (-1) + 5 \times (-1) \] \[ = 3 + 2 - 5 = 0 \] Thus, the determinant of the matrix is \(0\).