Question

For an invertible matrix $A$, if $A(\text{adj } A) = \begin{bmatrix} 20 & 0 \\ 0 & 20 \end{bmatrix}$, then $|A| =$

• A. $-200$
• B. $200$
• C. $-2$
• D. $20$

Hint:

For any $n \times n$ scalar matrix where $A(\text{adj } A) = k \cdot I$, the value of the determinant $|A|$ is simply the scalar constant $k$. You do not need to take any determinants or square roots of the matrix block itself!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an equation involving an invertible matrix $A$ multiplied by its adjoint matrix, which results in a diagonal scalar matrix $\begin{bmatrix} 20 & 0 \\ 0 & 20 \end{bmatrix}$. We need to find the determinant of matrix $A$, denoted as $|A|$.

Step 2: Key Formula or Approach:
We use the fundamental theorem of matrix algebra relating a matrix, its adjoint, and its determinant: $$A(\text{adj } A) = |A| \cdot I$$ where $I$ is the identity matrix of the matching dimensions ($2 \times 2$ in this scenario). By factoring out the scalar value from the given matrix, we can directly identify $|A|$.

Step 3: Detailed Explanation:
Let's analyze the given matrix on the right-hand side of our equation: $$\begin{bmatrix} 20 & 0 \\ 0 & 20 \end{bmatrix}$$ We can factor out the scalar value 20 from each element of this matrix: $$\begin{bmatrix} 20 & 0 \\ 0 & 20 \end{bmatrix} = 20 \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 20 \cdot I$$ Now substitute this back into our original matrix equation: $$A(\text{adj } A) = 20 \cdot I$$ By directly comparing this resulting expression with our standard mathematical identity $A(\text{adj } A) = |A| \cdot I$, we can conclude that: $$|A| = 20$$

Step 4: Final Answer:
The determinant $|A|$ is equal to $20$, which corresponds to option (D).