24.6 Practice Problems
Problem 24.1. Let \(A=\begin {pmatrix}1&0&3\end {pmatrix}\) and \(B=\begin {pmatrix}1\\2\\4\end {pmatrix}\). Find \(AB\) and \(BA\).
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Solution. \(A\) is \(1\times 3\) and \(B\) is \(3\times 1\).
For \(AB\) the inner dimensions match (\(3\) and \(3\)), and the result is \(1\times 1\): \[AB=\begin {pmatrix}1(1)+0(2)+3(4)\end {pmatrix}=\begin {pmatrix}13\end {pmatrix}.\]
For \(BA\) the inner dimensions also match (\(1\) and \(1\)), and the result is \(3\times 3\): \[BA=\begin {pmatrix}1\\2\\4\end {pmatrix}\begin {pmatrix}1&0&3\end {pmatrix} =\begin {pmatrix}1&0&3\\2&0&6\\4&0&12\end {pmatrix}.\]
Note 24.11. This is the sharpest possible illustration that matrix multiplication does not commute: \(AB\) and \(BA\) are not merely different numbers, they are different sizes — a single number against a \(3\times 3\) array.
The rule behind it: an \(m\times n\) matrix times an \(n\times p\) matrix gives \(m\times p\). The two inner dimensions must agree and they vanish; the outer two survive. Here \((1\times 3)(3\times 1)\) leaves \(1\times 1\), and \((3\times 1)(1\times 3)\) leaves \(3\times 3\).
Note also that \(BA\) has a column of zeros and its rows are all multiples of \(\begin {pmatrix}1&0&3\end {pmatrix}\). Its determinant is therefore zero, so it has no inverse — a product of a column and a row can never be invertible for a matrix larger than \(1\times 1\).
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