If D = $\begin{bmatrix}2& -1&3\\4&1&2\\1&-3&1\\\end{bmatrix}$ Find |D|

Analyze the question: We need the determinant of a 3×3 matrix. The options are A. 16, B. 14, C. -23, D. -37.

Concept: Expand along the first row:

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

for the matrix with rows (a, b, c), (d, e, f), (g, h, i).

Solution:

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

1. First term: 2 × (1·1 − 2·(−3)) = 2 × (1 + 6) = 2 × 7 = 14
2. Second term: −(−1) × (4·1 − 2·1) = +1 × (4 − 2) = 2
3. Third term: 3 × (4·(−3) − 1·1) = 3 × (−12 − 1) = 3 × (−13) = −39

Sum: 14 + 2 − 39 = −23

Check by expanding along a different row (the second row, signs + − + would be − + −):

|D| = −4(( −1)(1) − 3(−3)) + 1(2·1 − 3·1) − 2(2·(−3) − (−1)(1)) = −4(−1 + 9) + 1(2 − 3) − 2(−6 + 1) = −4(8) + (−1) − 2(−5) = −32 − 1 + 10 = −23 ✓

Answer: The determinant is −23, which is option C.

Tip: Watch the sign pattern (+ − +) when expanding along the first row, and the minus sign in front of the middle term. Most mistakes come from forgetting it when b is negative. Always verify by expanding along another row if time allows.