Question

If $A = \begin{bmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 1 & 2 & 0 \end{bmatrix}$, then the value of determinant of A is:}

• A. $-10$
• B. $10$
• C. $20$
• D. $-20$

Hint:

Whenever a determinant contains several zero entries, always expand along the row or column having the greatest number of zeros. This minimizes calculations and significantly reduces the possibility of arithmetic mistakes.

Answer 1

The Correct Option is D

Concept:

The determinant of a square matrix is a numerical value associated with the matrix that provides information about invertibility, area/volume scaling, and system solvability.

For a \(3 \times 3\) matrix, the determinant can be evaluated using expansion along any row or column. It is preferable to choose a row or column with maximum zeros to reduce computation.

The given matrix is:

\[ A = \begin{bmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 1 & 2 & 0 \end{bmatrix} \]

The third column contains two zeros, so expansion along the third column is the most efficient.

Step 1: Expand along the third column

The cofactor sign pattern for a \(3 \times 3\) matrix is:

\[ \begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix} \]

Third column elements are \(4, 0, 0\). Hence:

\[ |A| = 4 \begin{vmatrix} -1 & 3 \\ 1 & 2 \end{vmatrix} - 0(\cdots) + 0(\cdots) \] \[ |A| = 4 \begin{vmatrix} -1 & 3 \\ 1 & 2 \end{vmatrix} \] 
Step 2: Evaluate the \(2 \times 2\) determinant \[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \] \[ \begin{vmatrix} -1 & 3 \\ 1 & 2 \end{vmatrix} = (-1)(2) - (3)(1) \] \[ = -2 - 3 = -5 \] 
Step 3: Substitute the value \[ |A| = 4 \times (-5) = -20 \] 
Final Verification

Only one non-zero cofactor contributes to the expansion, so the computation is straightforward.

\[ \boxed{|A| = -20} \]

Final Answer: \(\boxed{(D)\ -20}\)