If \(A = \begin{bmatrix}1&2&-1\\0&3&1\end{bmatrix}\), then \(AA'\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}6&5\\5&10\end{bmatrix}\)
A is \(2 \times 3\) and \(A'\) is \(3 \times 2\), so \(AA'\) is \(2 \times 2\). The columns of \(A'\) are the rows of A, so each element of \(AA'\) is a row of A times a row of A:
\( \displaystyle A' = \begin{bmatrix}1&0\\2&3\\-1&1\end{bmatrix} \)
\(\text{row 1, column 1}: 1(1) + 2(2) + (-1)(-1) = 6\)
\(\text{row 1, column 2}: 1(0) + 2(3) + (-1)(1) = 5\)
\(\text{row 2, column 1}: 0(1) + 3(2) + 1(-1) = 5\)
\(\text{row 2, column 2}: 0(0) + 3(3) + 1(1) = 10\)
Hence
\( \displaystyle AA' = \begin{bmatrix}6&5\\5&10\end{bmatrix} \)
As expected, \(AA'\) is a symmetric matrix.