If $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix}, A^{-1} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & c \\ 5/2 & -3/2 & 1/2 \end{bmatrix}$, then -
Step-by-Step Solution
Key Concept: Use the property AA^-1 = I, where I is the identity matrix. Multiply the second row of A with the third column of A^-1 to find c, and the third row of A with the second column of A^-1 to find a.
We know that $AA^{-1} = I$. The element at (2,3) of the product $AA^{-1}$ must be 0. Multiplying the second row of A [1, 2, 3] with the third column of $A^{-1}$ [1/2, c, 1/2]^T gives: 1(1/2) + 2(c) + 3(1/2) = 0 => 1/2 + 2c + 3/2 = 0 => 2 + 2c = 0 => c = -1. Similarly, the element at (3,2) of the product $AA^{-1}$ must be 0. Multiplying the third row of A [3, a, 1] with the second column of $A^{-1}$ [-1/2, 3, -3/2]^T gives: 3(-1/2) + a(3) + 1(-3/2) = 0 => -3/2 + 3a - 3/2 = 0 => 3a - 3 = 0 => a = 1. Thus, a = 1 and c = -1.
Correct Answer: 1