Question:

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

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Always use cofactor expansion for 3x3 matrices. It simplifies the calculation by breaking it into smaller 2x2 determinants.
  • \( 0 \)
  • \( 46 \)
  • \( -46 \)
  • \( 1 \)
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The Correct Option is C

Solution and Explanation

We are given the matrix: \[ \begin{vmatrix} 3 & 1 & 2
-4 & -2 & 3
5 & 1 & 1 \end{vmatrix} \] To evaluate the determinant, we expand along the first row: \[ \text{Determinant} = 3 \times \begin{vmatrix} -2 & 3
1 & 1 \end{vmatrix} - 1 \times \begin{vmatrix} -4 & 3
5 & 1 \end{vmatrix} + 2 \times \begin{vmatrix} -4 & -2
5 & 1 \end{vmatrix} \] Now, calculate each 2x2 determinant: \[ \begin{vmatrix} -2 & 3
1 & 1 \end{vmatrix} = (-2)(1) - (3)(1) = -2 - 3 = -5 \] \[ \begin{vmatrix} -4 & 3
5 & 1 \end{vmatrix} = (-4)(1) - (3)(5) = -4 - 15 = -19 \] \[ \begin{vmatrix} -4 & -2
5 & 1 \end{vmatrix} = (-4)(1) - (-2)(5) = -4 + 10 = 6 \] Substitute these values back: \[ \text{Determinant} = 3 \times (-5) - 1 \times (-19) + 2 \times 6 \] \[ = -15 + 19 + 12 = 16 \] Thus, the determinant is \(16\).
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