Evaluate $\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}$

Analyze the question: We need the determinant of a 3×3 matrix. (Note: option A looks like it belongs to a different question, so ignore it.)

Concept: Expand along the first row:

det = 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=0, c=5, d=4, e=6, f=3, g=8, h=9, i=1.

• First term: 2(6×1 − 3×9) = 2(6 − 27) = 2(−21) = −42
• Second term: −0 × (4×1 − 3×8) = 0
• Third term: +5(4×9 − 6×8) = 5(36 − 48) = 5(−12) = −60

Total = −42 + 0 − 60 = −102

Check using the second column (where the 0 helps): det = −0(…) + 6(2×1 − 5×8) − 9(2×3 − 5×4) = 6(2 − 40) − 9(6 − 20) = 6(−38) − 9(−14) = −228 + 126 = −102 ✓

Answer: −102, which is option C.

Tip: Expand along the row or column containing a zero, since it removes one term and reduces mistakes. Remember the sign pattern + − + for the first row, and double-check by expanding along a different row or column.