Question

For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj}A) = \begin{bmatrix} -10 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & -10 \end{bmatrix}$, then the value of the determinant of $A$ is

• A. $100$
• B. $-1000$
• C. $-10$
• D. $20$

Hint:

Whenever a matrix product $A(\operatorname{adj}A)$ forms a scalar matrix where the diagonal entries are all identical numbers $k$, you can immediately conclude that the determinant of that matrix is simply equal to $k$ ($|A| = k$). No determinant expansions are necessary!

Answer 1

The Correct Option is C

Step 1: Recall the adjugate identity.
For any square matrix, $A(\operatorname{adj}A) = |A|\,I$, where $|A|$ is the determinant and $I$ the identity.
Step 2: Compare with the given matrix.
We are told $A(\operatorname{adj}A)$ is the diagonal matrix with $-10$ on each diagonal entry.
Step 3: Factor out the scalar.
That diagonal matrix is $-10\,I$.
Step 4: Equate to the identity formula.
So $|A|\,I = -10\,I$.
Step 5: Read off the determinant.
Matching the scalar multiples gives $|A| = -10$.
Step 6: Note the shortcut.
For a $3\times 3$ matrix we could also use $|A(\operatorname{adj}A)| = |A|^3$, giving $(-10)^3 = |A|^3$, again $|A| = -10$. Both routes agree.
\[ \boxed{|A| = -10} \]