Question

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \text{adj } A = A A^T$, then $5a + b = \dots$

• A. 7
• B. 9
• C. 13
• D. 5

Hint:

Always use the property $A(\text{adj } A) = |A|I$. Calculating the actual adjoint matrix and manually performing the multiplication is a massive and completely unnecessary waste of exam time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We know the property $A \cdot \text{adj } A = |A|I$. We are given this equals $A A^T$. Therefore, $AA^T = |A|I$, which implies $A$ is proportional to an orthogonal matrix.

Step 2: Formula Application:
$A A^T = \begin{bmatrix} 5a & -b
3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3
-b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2+b^2 & 15a-2b
15a-2b & 13 \end{bmatrix}$. $|A| = 10a + 3b$. $|A|I = \begin{bmatrix} 10a+3b & 0 \\ 0 & 10a+3b \end{bmatrix}$.

Step 3: Explanation:
Equating the off-diagonal elements: $15a - 2b = 0 \implies b = \frac{15a}{2}$. Equating the $(2,2)$ elements: $13 = 10a + 3b$. Substitute $b$: $13 = 10a + 3(\frac{15a}{2}) \implies 13 = \frac{20a + 45a}{2} \implies 26 = 65a \implies a = \frac{26}{65} = \frac{2}{5}$. Then $b = \frac{15}{2}(\frac{2}{5}) = 3$. We need $5a + b = 5(2/5) + 3 = 2 + 3 = 5$. (Wait, checking calculation: $5a+b = 2+3=5$. Let me re-verify Step 2). If $AA^T = |A|I$, then $13 = 10a + 3b$. Since we need $5a+b$, and the logic holds, the result is 5.

Step 4: Final Answer:
The value of $5a + b$ is 5 (Option D).