Question

If \(A\) is a \(3\times3\) matrix with \[ |A|=-1, \] then \[ \left|\,3(\operatorname{adj}(A^T))A^2\,\right| \] equals

• A. \(81\)
• B. \(9\)
• C. \(27\)
• D. \(3\)

Hint:

For a \(3\times3\) matrix: \[ |\operatorname{adj}(A)|=|A|^2 \] and \[ |kA|=k^3|A| \] These two formulas are frequently used in determinant problems.

Answer 1

The Correct Option is C

Step 1: List the rules we need.
For a $3\times3$ matrix: $|kA|=k^3|A|$, $|\operatorname{adj}A|=|A|^{2}$, $|A^T|=|A|$, and $|A^m|=|A|^m$. We use these one by one.

Step 2: Pull out the scalar 3.
The matrix inside is $B=\operatorname{adj}(A^T)\,A^2$, a $3\times3$ matrix. So $|3B|=3^3|B|=27\,|B|$.

Step 3: Split $|B|$ into factors.
Since determinant of a product is the product of determinants, $|B|=|\operatorname{adj}(A^T)|\cdot|A^2|$.

Step 4: Determinant of the adjoint.
Because $|A^T|=|A|=-1$, we have $|\operatorname{adj}(A^T)|=|A^T|^{2}=(-1)^2=1$.

Step 5: Determinant of $A^2$.
$|A^2|=|A|^2=(-1)^2=1$.

Step 6: Put it all together.
$|3B|=27\times1\times1=27$. \[ \boxed{27} \]