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

Step 1: Understanding the Question:

The question asks to evaluate the determinant of a given 3×3 matrix.

Step 2: Key Formula or Approach:

For a 3×3 matrix \[ M = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}, \] the determinant is calculated as: \[ \det(M) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Another method is expanding along any row or column. It is often easier to expand along a row or column containing zeros.

Step 3: Detailed Explanation:

Given matrix: \[ M = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \]

Expanding along the first row: \[ \det(M) = 1 \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2 \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3 \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} \]

Calculate each 2×2 determinant:

First minor: \[ \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = (1)(0) - (4)(6) = -24 \]

Second minor: \[ \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = (0)(0) - (4)(5) = -20 \]

Third minor: \[ \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = (0)(6) - (1)(5) = -5 \]

Substituting these values: \[ \det(M) = 1(-24) - 2(-20) + 3(-5) \] \[ = -24 + 40 - 15 \] \[ = 1 \]

Step 4: Final Answer:

Therefore, the determinant of the matrix is: \[ \boxed{1} \]