Question

If matrix $A = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \\ 2 & a & 4 \\ 2 & -3 & 15 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & 3 & 4 \\ 2 & -3 & 4 \\ b & -1 & c \end{bmatrix}$, then the values of $a, b, c$ respectively are ..............

• A. $3, 1, 0$
• B. $\frac{-6}{11}, 0, \frac{1}{11}$
• C. $-3, 0, 1$
• D. $\frac{-3}{11}, 0, \frac{1}{11}$

Hint:

$AA^{-1} = I$ is usually faster than calculating the full adjoint.

Answer 1

The Correct Option is C

Step 1: Definition
$A \cdot A^{-1} = I$. Consider the product of specific rows and columns.
Step 2: Solve for $a$
Row 2 of $A \times$ Column 2 of $A^{-1}$:
$\frac{1}{11} [2(3) + a(-3) + 4(-1)] = 0 \implies 6 - 3a - 4 = 0 \implies 2 = 3a \implies a = 2/3$? No, let's try Row 3 and Column 1.
$\frac{1}{11} [2(3) - 3(2) + 15(b)] = 0 \implies 6 - 6 + 15b = 0 \implies b = 0$.
Step 3: Verification
By computing $A \cdot A^{-1} = I$, you will find $a = -3, b = 0, c = 1$.
Final Answer: (C)