Suppose we have two matrices:

• [latex]A=\left(\begin{array}{c}-1 & 2\\4 & 2\end{array}\right)[/latex]
• [latex]B=\left(\begin{array}{c}1 & 4 & -4\\4 & 4 & 1\end{array}\right)[/latex]

We can multiply A and B (in that order) since the two columns of A matches the two rows of B. To so, we calculate the inner product of each row of A with each column of B:

• [latex]AB=\left(\begin{array}{c}-1(1)+2(4) & -1(4)+2(4) & -1(-4)+2(1)\\4(1)+2(4) & 4(4)+2(4) & 4(-4)+2(1)\end{array}\right)=\left(\begin{array}{c}7 & 4 & 6\\12 & 24 & -14\end{array}\right)[/latex]

We cannot multiply B and A (in that order) since the three columns of B do not match the two rows of A.

The inverse of the two (row) by two (column) matrix A, denoted A^–1, is:

• [latex]A^{-1}=\frac{1}{-1(2)-4(2)}\left(\begin{array}{c}2 & -2\\-4 & -1\end{array}\right)=\left(\begin{array}{c}-1/5 & 1/5\\2/5 & 1/10\end{array}\right)[/latex]

Check that this is correct by calculating:

• [latex]A^{-1}A=\left(\begin{array}{c}(-1/5)(-1)+(1/5)(4) & (-1/5)(2)+(1/5)(2)\\(2/5)(-1)+(1/10)(4) & (2/5)(2)+(1/10)(2)\end{array}\right)=\left(\begin{array}{c}1 & 0\\0 & 1\end{array}\right)[/latex]

The transpose of a matrix, denoted A^T, reverses its rows and columns, so:

• [latex]B^TA^T=\left(\begin{array}{c}1 & 4\\4 & 4\\-4 & 1\end{array}\right)\left(\begin{array}{c}-1 & 4\\2 & 2\end{array}\right)=\left(\begin{array}{c}7 & 12\\4 & 24\\6 & -14\end{array}\right)[/latex]

This illustrates the general result, (AB)^T = B^TA^T.

If we multiply A by itself we get AA = A², a matrix power:

• [latex]A^2=\left(\begin{array}{c}-1(-1)+2(4) & -1(2)+2(2)\\4(-1)+2(4) & 4(2)+2(2)\end{array}\right)=\left(\begin{array}{c}9 & 2\\4 & 12\end{array}\right)[/latex]

Note that this has to be calculated using matrix multiplication, not simply squaring each element of A.