Question

If the determinant of a matrix \(A=\begin{bmatrix}1& 3& 0\\2& 6& 4-1& 0& 2\end{bmatrix}\) is \(-12\), then the determinant of the matrix \(\begin{bmatrix}2& 6& 0\\4& 12& 8-2& 0& 4\end{bmatrix}\) is

• A. \(-24\)
• B. \(24\)
• C. \(-96\)
• D. \(96\)

Hint:

For a \(3\times 3\) matrix, \(\det(kA)=k^3\det(A)\).

Answer 1

The Correct Option is C

Concept:
If every element of an \(n\times n\) matrix is multiplied by \(k\), then its determinant is multiplied by \[ k^n \]

Step 1: Compare the matrices. Given, \[ A=\begin{bmatrix}1& 3& 0\\2& 6& 4-1& 0& 2\end{bmatrix} \] The second matrix is \[ \begin{bmatrix}2& 6& 0\\4& 12& 8-2& 0& 4\end{bmatrix} \] Clearly, \[ \begin{bmatrix}2& 6& 0\\4& 12& 8-2& 0& 4\end{bmatrix}=2A \]

Step 2: Use determinant property. Since \(A\) is a \(3\times 3\) matrix, \[ \det(2A)=2^3\det(A) \] \[ =8(-12) \] \[ =-96 \]

Step 3: Final answer. \[ \boxed{-96} \]