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 determine the element \( a_{23} \) of matrix \( A = [a_{ij}] \), given its inverse \( A^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 0 \end{bmatrix} \), we utilize the property \( A \times A^{-1} = I \), where \( I \) is the identity matrix.

For a \(3 \times 3\) matrix, the identity matrix is:

1	0	0
0	1	0
0	0	1

The matrix \( A \) is given as \( A = \begin{bmatrix} 0 & 0 & 4 \\ 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \). We calculate the product \( A \times A^{-1} \).

The element at position \((2, 3)\) of the product \( A \times A^{-1} \) is computed as follows:

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

We examine the elements of the product \( A \times A^{-1} \), specifically focusing on the second row and third column entry.

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

This result matches the corresponding entry in the identity matrix, which is 0.

For \( a_{23} \) to be consistent with this multiplication, it must satisfy the row-column product yielding the correct identity matrix element. The calculation \( 3 \cdot 3\,+ \, 4 \cdot 1\, = 0\) implies \( a_{23} \) should be \(-1\).

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

Therefore, the final answer is:

\(-1\)