Question

Let $ A = [a_{ij}] $ be a $3 \times 3 $ matrix such that: $$ A = \begin{bmatrix} 0 & 0 & 4 \\ 1 & 0 & 0 \\ 0 & 1 & 3 \\ \end{bmatrix}, A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 0 \end{bmatrix}. $$ Then $ a_{23} $ equals:

• A. \( -1 \)
• B. \( 0 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

When multiplying matrices, each element of the resulting matrix is obtained by the dot product of the corresponding row and column. Ensure you multiply each element and sum them correctly.

Answer 1

The Correct Option is A

We are given the matrix \( A = [a_{ij}] \): \[ A = \begin{bmatrix} 0 & 0 & 4
1 & 0 & 0
0 & 1 & 3 \end{bmatrix}. \] and the matrix \( A^{-1} = \begin{bmatrix} 0 & 1 & 0
1 & 3 & 0
2 & 1 & 0 \end{bmatrix} \). We need to find \( a_{23} \), the element in the second row and third column of matrix \( A \). First, recall that if \( A^{-1} \) exists, the product of \( A \) and \( A^{-1} \) must yield the identity matrix: \[ A \cdot A^{-1} = I. \] Multiplying matrix \( A \) with \( A^{-1} \), we get the identity matrix: \[ \begin{bmatrix} 0 & 0 & 4
1 & 0 & 0
0 & 1 & 3 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 & 0
1 & 3 & 0
2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix}. \] Multiplying the first row of matrix \( A \) by the first column of matrix \( A^{-1} \): \[ 0(0) + 0(1) + 4(2) = 8. \] Thus, \( a_{23} = -1 \).