Most rules for computing with numbers continue to hold when we compute with matrices, but not all: The multiplication of matrices is more delicate because there are two basic exceptions:
Exception 1 When multiplying two matrices \(A\) and \(B\), the products \(AB\) and \(BA\) need not be the same, even if both products are defined. – For example
| \([1\ \ 4\ \ 3] \cdot \left[\begin{array}{r} 5 \\ -2 \\ 7 \end{array}\right]\) | \(=\) | \([ 18 ]\) |
| \(\left[\begin{array}{r} 5 \\ -2 \\ 7 \end{array}\right] \cdot [ 1\ \ 4\ \ 3 ]\) | \(=\) | \( \left[\begin{array}{rrr} 5 & 20 & 15 \\ -2 & -8 & -6 \\ 7 & 28 & 21 \end{array}\right] \) |
In this case the two products don't even have the same size. So they cannot be equal. However, difference in size is not the only possible reason why \(AB\) might be different from \(BA\):
| \( \left[\begin{array}{rr} -3 & -6 \\ 4 & 8 \end{array}\right] \cdot \left[\begin{array}{rr} 8 & 6 \\ 4 & 3 \end{array}\right] \) | \(\) | \( \left[\begin{array}{rr} -48 & -36 \\ 64 & 48 \end{array}\right] \) |
| \( \left[\begin{array}{rr} 8 & 6 \\ 4 & 3 \end{array}\right] \cdot \left[\begin{array}{rr} -3 & -6 \\ 4 & 8 \end{array}\right] \) | \(=\) | \( \left[\begin{array}{rr} 0 & 0 \\ 0 & 0 \end{array}\right] \) |
Here \(AB\) and \(BA\) are both of size \((2,2)\). Still, they are distinct.
Exception 2 If \(AX\) is the 0-matrix, it need not be true that at least one of \(A\) or \(X\) is the 0-matrix.
If \(ax=0\) is an equation of numbers, and we know that \(a\neq 0\), we conclude immediately that \(x=0\). It is very tempting to assume that we can draw the same conclusion when presented with a matrix equation like
\[AX = \mathbf{0}\]However, the example below shows that, in this situation, neither \(A\) nor \(X\) need be \(\mathbf{0}\).
| \( \left[\begin{array}{rr} 8 & 6 \\ 4 & 3 \end{array}\right] \cdot \left[\begin{array}{rr} -3 & -6 \\ 4 & 8 \end{array}\right] \) | \(=\) | \( \left[\begin{array}{rr} 0 & 0 \\ 0 & 0 \end{array}\right] \) |
Therefore, a matrix identity like \(\Mtrx{A}\Mtrx{X} = \Vect{0}\) by itself does not support the conclusion that \(\Mtrx{A}\) or \(\Mtrx{X}\) are \(\Vect{0}\).