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

The Correct Option is B

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:

\[ \text{-13.} \]