Question

Let $A = \begin{bmatrix} 2 & a & 0 \\\ 1 & 3 & 1 \\\ 0 & 5 & b \end{bmatrix}$. If $A^3 = 4A^2 - A - 21I$, where I is the identity matrix of order $3 \times 3$, then $2a + 3b$ is equal to:

• A. -10
• B. -13
• C. -9
• D. -12

Answer 1

Answer: (B)

-13

Answer 2

From the matrix equation:

$A^3 - 4A^2 + A + 21I = 0.$

Step 1: Taking the trace:

$\text{tr}(A^3) - 4\text{tr}(A^2) + \text{tr}(A) + 21 \cdot \text{tr}(I) = 0.$

Since $\text{tr}(I) = 3$, we find:

$\text{tr}(A) = 4 + 5 + b = b - 1.$

Step 2: The determinant:

$|A| = -16 + a = -21 \implies a = -5.$

Step 3: Final calculation:

$2a + 3b = 2(-5) + 3(-1) = -13.$

Final Answer:

$\text{-13.}$