Question

If the determinant of the matrix \[ \begin{vmatrix} 1 & 3 & 2\\ 0 & 5 & -6\\ 2 & 7 & 8 \end{vmatrix} \] is \(26\), then the determinant of the matrix \[ \begin{vmatrix} 2 & 7 & 8\\ 0 & 5 & -6\\ 1 & 3 & 2 \end{vmatrix} \] is

• A. \(-26\)
• B. \(26\)
• C. \(0\)
• D. \(52\)

Hint:

If two rows or two columns of a determinant are interchanged, the determinant changes its sign.

Answer 1

The Correct Option is A

Concept:
This question is based on the property of determinants. If two rows of a determinant are interchanged, then the value of the determinant changes its sign. That means: \[ R_i \leftrightarrow R_j \quad \Longrightarrow \quad \Delta \text{ changes to } -\Delta \]

Step 1: Write the given determinant.
Let: \[ \Delta= \begin{vmatrix} 1 & 3 & 2\\ 0 & 5 & -6\\ 2 & 7 & 8 \end{vmatrix} \] It is given that: \[ \Delta=26 \]

Step 2: Compare the second determinant with the first determinant.
The second determinant is: \[ \Delta'= \begin{vmatrix} 2 & 7 & 8\\ 0 & 5 & -6\\ 1 & 3 & 2 \end{vmatrix} \] Now compare rows. In the first determinant: \[ R_1=(1,3,2) \] \[ R_2=(0,5,-6) \] \[ R_3=(2,7,8) \] In the second determinant: \[ R_1'=(2,7,8) \] \[ R_2'=(0,5,-6) \] \[ R_3'=(1,3,2) \] So the second determinant is obtained by interchanging the first and third rows of the first determinant: \[ R_1 \leftrightarrow R_3 \]

Step 3: Apply determinant row interchange property.
When two rows are interchanged, the determinant changes sign. So: \[ \Delta'=-\Delta \] Since: \[ \Delta=26 \] Therefore: \[ \Delta'=-26 \] Hence, the correct answer is: \[ \boxed{(A)\ -26} \]