Question

Evaluate: \( \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix} \)

• A. 1
• B. -1
• C. 5
• D. 10

Hint:

When calculating determinants, choose to expand along the row or column that has the most zeros. This minimizes the number of 2x2 determinants you need to calculate, reducing potential errors. Always remember the alternating signs for the cofactors (+ - +).

Answer 1

The Correct Option is A

To evaluate the determinant of the given 3x3 matrix, we use the formula for the determinant of a 3x3 matrix:

\[\text{If } A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}, \text{ then } \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg).\]

For the given matrix:

1	2	3
0	1	4
5	6	0

Identify the elements:

• \( a = 1 \)
• \( b = 2 \)
• \( c = 3 \)
• \( d = 0 \)
• \( e = 1 \)
• \( f = 4 \)
• \( g = 5 \)
• \( h = 6 \)
• \( i = 0 \)

Substituting these into the determinant formula, we get:

\[\text{det}(A) = 1(1 \cdot 0 - 4 \cdot 6) - 2(0 \cdot 0 - 4 \cdot 5) + 3(0 \cdot 6 - 1 \cdot 5)\]

Simplify each term:

• First term: \( 1(0 - 24) = 1 \times (-24) = -24 \)
• Second term: \( -2(0 - 20) = -2 \times (-20) = 40 \)
• Third term: \( 3(0 - 5) = 3 \times (-5) = -15 \)

Add these results:

\[-24 + 40 - 15 = 1\]

Therefore, the determinant of the given matrix is \( \boxed{1} \).

Hence, the correct answer is 1.