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

To find \( a_{23} \) of the matrix \( A = [a_{ij}] \), given the inverse \( A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 0 \end{bmatrix} \), we need to use the property that \( A \times A^{-1} = I \), where \( I \) is the identity matrix.

The identity matrix for a \(3 \times 3\) matrix is: 

1	0	0
0	1	0
0	0	1

To calculate the product \( A \times A^{-1} \):

\( A = \begin{bmatrix} 0 & 0 & 4 \\ 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \) and \( A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 0 \end{bmatrix} \), calculate \( A \times A^{-1} \).

The element \((2, 3)\) of the product \( A \times A^{-1} \) is given by:

\(a_{23} \times (1, 0, 0)^T = 0 \times 0 + 0 \times 1 + a_{23} \times 0 = 0\)

Calculating for each element (i, j) of the product \( A \times A^{-1} \), particularly for the third row and second column:

• The second row, third column: \( (0 \cdot 0) + (1 \cdot 0) + (0 \cdot 1) = 0 + 0 + 0 = 0\).

This matches the identity matrix entry at \( (2, 3) \), which is 0.

Thus, \( a_{23} \) correctly reproduces this multiplication result. Therefore, the missing \( a_{23} \) should satisfy the row and column multiplication correctly, giving \( a_{23} = -1 \) so that: \( 3 \cdot 3\,+ \, 4 \cdot 1\, = 0\). Thus:

The value \( a_{23} \) equals \(-1\).

Hence, the correct answer is:

\(-1\)

Answer 2

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, \( a_{23} = -1 \).