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: Understanding the Concept:
By definition of an inverse matrix, $A \cdot A^{-1} = I$, where $I$ is the identity matrix.
Equating corresponding elements of the product will give the unknown values.
Step 2: Key Formula or Approach:
The $(i, j)$ element of $I$ is $1$ if $i=j$ and $0$ otherwise.
We use the multiplication $11(AA^{-1}) = 11I$.
Step 3: Detailed Explanation:
Row 2 of A $\times$ Col 2 of $A^{-1}$:
$2(3) + a(-3) + 4(-1) = 11(1) \implies 6 - 3a - 4 = 11 \implies -3a = 9 \implies a = -3$.
Row 3 of A $\times$ Col 1 of $A^{-1}$:
$2(3) + (-3)(2) + 15(b) = 11(0) \implies 6 - 6 + 15b = 0 \implies b = 0$.
Row 3 of A $\times$ Col 3 of $A^{-1}$:
$2(4) + (-3)(4) + 15(c) = 11(1) \implies 8 - 12 + 15c = 11 \implies 15c = 15 \implies c = 1$.
Step 4: Final Answer:
The values are $a = -3, b = 0, c = 1$.