Consider the \(3 \times 3\) matrix \(\boldsymbol{M} = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 3 \\ 4 & 3 & 6 \end{bmatrix}\).
The determinant of \((\boldsymbol{M}^{2} + 12\boldsymbol{M})\) is ______.
Answer
0
Explanation
Factor first: \(\boldsymbol{M}^{2} + 12\boldsymbol{M} = \boldsymbol{M}(\boldsymbol{M} + 12I)\), so its determinant is \(\det(\boldsymbol{M}) \cdot \det(\boldsymbol{M} + 12I)\).
Expanding along the first row,
\[\det(\boldsymbol{M}) = 1(6 - 9) - 2(18 - 12) + 3(9 - 4) = -3 - 12 + 15 = 0.\]
This is no accident: row 3 is the sum of rows 1 and 2, so \(\boldsymbol{M}\) is singular. A zero factor makes the whole product zero, so \(\det(\boldsymbol{M}^{2} + 12\boldsymbol{M}) = 0\) without computing \(\boldsymbol{M}^{2}\) at all.
Answer: 0