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: Understanding the Question:
We are given a matrix property equation for a $3 \times 3$ square matrix $A$, where the product of the matrix and its adjugate, $A(\operatorname{adj}A)$, is a scalar diagonal matrix. We need to determine the value of the determinant of matrix $A$, denoted as $|A|$.

Step 2: Key Formula or Approach:
According to a foundational theorem of matrix algebra, the product of a matrix and its adjugate satisfies the identity: $$A(\operatorname{adj}A) = |A| \cdot I$$ Where $|A|$ is the scalar determinant of the matrix and $I$ is the matching $3 \times 3$ identity matrix.

Step 3: Detailed Explanation:
Let's factor out the common scalar value $-10$ from the given matrix expression: $$A(\operatorname{adj}A) = \begin{bmatrix} -10 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & -10 \end{bmatrix} = -10 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$A(\operatorname{adj}A) = -10 \cdot I$$ Now, compare this result directly with our standard identity theorem formula: $$A(\operatorname{adj}A) = |A| \cdot I$$ By equating the scalar coefficients from both equations, we find: $$|A| = -10$$ This matches option (C).

Step 4: Final Answer:
The value of the determinant of $A$ is $-10$, which corresponds to option (C).