Evaluate the determinant of the matrix M = $\begin{bmatrix}3 & 4 & 6\\2 & 1 & -1\\-1 & 3 & 5\end{bmatrix}$

Analyze the question: We need the determinant of a 3×3 matrix. The usual method is cofactor expansion along the first row.

Concept: For a matrix with rows (a b c), (d e f), (g h i):

det = a(ei − fh) − b(di − fg) + c(dh − eg)

Solution:

Here a = 3, b = 4, c = 6, d = 2, e = 1, f = −1, g = −1, h = 3, i = 5.

1. First term: 3[(1)(5) − (−1)(3)] = 3(5 + 3) = 3(8) = 24

2. Second term: −4[(2)(5) − (−1)(−1)] = −4(10 − 1) = −4(9) = −36

3. Third term: 6[(2)(3) − (1)(−1)] = 6(6 + 1) = 6(7) = 42

Adding them: 24 − 36 + 42 = 30

Check (expanding along a different row, using column 1): det = 3(1·5 − (−1)(3)) − 2(4·5 − 6·3) + (−1)(4·(−1) − 6·1) = 3(8) − 2(20 − 18) − 1(−4 − 6) = 24 − 4 + 10 = 30 ✓

Answer: C (30)

Tip: Watch the sign pattern (+ − +) when expanding along a row, and be careful with negative numbers inside the brackets, since most mistakes happen there. Always verify by expanding along a second row or column if you have time.