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 the 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:

For matrix inverses, verify by computing \( A \cdot A^{-1} = I \) row by row and column by column for consistency.

Answer 1

The Correct Option is D

For \( A \cdot A^{-1} = I \) (the identity matrix), we verify the values of \( a \) and \( c \) such that: \[ \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \cdot \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2c \\ 5 & -3 & 1 \end{bmatrix} = I. \]
Step 1: Simplify for \( a \)
Consider the third row of \( A \) and the first column of \( A^{-1} \): \[ (3)(1) + (a)(-8) + (1)(5) = 0. \]
Simplify: \[ 3 - 8a + 5 = 0. \]
\[ 8 - 8a = 0 \quad \implies \quad a = 1. \]
Step 2: Simplify for \( c \)
Consider the second row of \( A \) and the third column of \( A^{-1} \): \[ (1)(1) + (2)(2c) + (3)(1) = 0. \]
Simplify: \[ 1 + 4c + 3 = 0. \]
\[ 4c + 4 = 0 \quad \implies \quad c = -1. \]
Final Answer: \[ \boxed{1, -1} \]