Question

If $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{2}\begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2c \\ 5 & -3 & 1 \end{bmatrix}$, then values of $a$ and $c$ are respectively

• A. $\frac{1}{2},\ \frac{1}{2}$
• B. $-1,\ 1$
• C. $2,\ -\frac{1}{2}$
• D. $1,\ -1$

Hint:

Never compute a full matrix product if you only need a couple of missing variables!
Isolate rows and columns that contain only one unknown element. Multiplying Row 1 of $A$ by Column 3 of $A^{-1}$ skips variable $a$ entirely, letting you isolate $c$ in a single mental equation step: $2c + 2 = 0 \rightarrow c = -1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem provides a matrix $A$ containing an unknown variable $a$, and its inverse matrix $A^{-1}$ containing an unknown variable $c$. We need to compute the specific values of $a$ and $c$.

Step 2: Key Formula or Approach:
By the fundamental definition of matrix inverses, the product of a matrix and its inverse yields the identity matrix $I$:
$$A A^{-1} = I$$ Instead of computing the entire matrix multiplication or deriving the full inverse manually, we can choose a couple of strategic element positions to solve for $a$ and $c$ directly.

Step 3: Detailed Explanation:
Let's look at the product element at position $(1,3)$ of the resulting identity matrix, which must equal 0:
$$(A A^{-1})_{1,3} = 0$$ Multiply the first row of $A$ by the third column of $A^{-1}$ (factoring out the $\frac{1}{2}$ constant):
$$\frac{1}{2} \big[ (0)(1) + (1)(2c) + (2)(1) \big] = 0$$ $$2c + 2 = 0 \implies 2c = -2 \implies c = -1$$ Next, let's look at the product element at position $(2,3)$ of the identity matrix, which must also equal 0:
$$(A A^{-1})_{2,3} = 0$$ Multiply the second row of $A$ by the third column of $A^{-1}$:
$$\frac{1}{2} \big[ (1)(1) + (2)(2c) + (3)(1) \big] = 0$$ Substitute $c = -1$:
$$\frac{1}{2} \big[ 1 + 4(-1) + 3 \big] = \frac{1}{2} [ 1 - 4 + 3 ] = 0 \quad \text{(This cleanly validates our value for } c\text{)}$$ Now, let's look at the product element at position $(3,1)$ of the identity matrix, which must equal 0:
$$(A A^{-1})_{3,1} = 0$$ Multiply the third row of $A$ by the first column of $A^{-1}$:
$$\frac{1}{2} \big[ (3)(1) + (a)(-8) + (1)(5) \big] = 0$$ $$3 - 8a + 5 = 0$$ $$8 - 8a = 0 \implies 8a = 8 \implies a = 1$$

Step 4: Final Answer:
The values are $a = 1$ and $c = -1$, which corresponds to option (D).