In my last post, I multiplied a 2 × 2 matrix by another 2 × 2 matrix. Now let’s multiply a 2 × 2 matrix by a 2 × 1 matrix. If you were paying attention last time, this is possible because the inside dimensions are the same (2= 2) and the resulting matrix will be a 2 × 1 matrix, that is, the outside dimensions.

This is done exactly the same way as illustrated in my last post, only there is only 1 column in the second matrix to multiply with the first matrix:

\[\left[{\begin{array}{cc}{1}&{{-}{2}}\\{3}&{{-}{4}}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{c}{5}\\{6}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{(}{5}\times{1}{)}{+}{(}{6}\times{(}{-}{2}{))}}\\{{(}{5}\times{3}{)}{+}{(}{6}\times{(}{-}{4}{))}}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{-}{7}}\\{{-}{9}}\end{array}}\right]

\]

Now let the second matrix be composed of variables. This does not change the method at all. It just means that the result is a matrix with algebraic expressions:

\[\left[{\begin{array}{cc}{1}&{{-}{2}}\\{3}&{{-}{4}}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{(}{x}\times{1}{)}{+}{(}{y}\times{(}{-}{2}{))}}\\{{(}{x}\times{3}{)}{+}{(}{y}\times{(}{-}{4}{))}}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{x}{-}{2}{y}}\\{{3}{x}{-}{4}{y}}\end{array}}\right]

\]

Please keep this example in mind for my next post when I use matrices to solve a system of equations.

The last skill I need to present is matrix division. When using matrices, you do not actually divide a matrix by another matrix. Rather, you multiply by the inverse of a matrix.

In scalar arithmetic, you can think of dividing a number say 4, by another number, say 2, as multiplying the 4 by the inverse (reciprocal) of 2:

\[\frac{4}{2}\hspace{0.33em}{=}\hspace{0.33em}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}

\]

The same thing is done with matrices. However, finding the inverse of a matrix is a little involved and I will not cover that in this set of posts. Rather I will just give you the result when needed. However I will say a few things about the properties of matrix inverses.

In scalar arithmetic, multiplying a number by it’s reciprocal (inverse) equals 1:

\[\frac{2}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}

\]

The same thing is true with matrices, only what is “1” in the matrix world?

The equivalent “1” for matrices is the *Identity Matrix*. This is a square (rows = columns) matrix with 1’s down its diagonal and 0’s everywhere else:

\left[{\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}}\right]

\]

This is the identity matrix for a 2 × 2 matrix. The inverse of a matrix is that matrix where multiplying it by the original matrix, results in the identity matrix. The inverse of a matrix **A**, is denoted as **A**^{-1}.

{\mathbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathbf{A}}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{-}{2}}&{1}\\{\frac{3}{2}}&{{-}\frac{1}{2}}\end{array}}\right]

\] \[

\begin{array}{l}

{{\mathbf{A}}\times{\mathbf{A}}^{{-}{1}}{=}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{{-}{2}}&{1}\\{\frac{3}{2}}&{{-}\frac{1}{2}}\end{array}}\right]{=}\left[{\begin{array}{cc}{{(}{-}{2}\times{1}{)}{+}{(}\frac{3}{2}\times{2}{)}}&{{(}{1}\times{1}{)}{+}{(}{-}\frac{1}{2}\times{2}{)}}\\{{(}{-}{2}\times{3}{)}{+}{(}\frac{3}{2}\times{4}{)}}&{{(}{1}\times{3}{)}{+}{(}{-}\frac{1}{2}\times{4}{)}}\end{array}}\right]}\\

{{=}\hspace{0.33em}\left[{\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}}\right]}

\end{array}

\]

And it turns out that when multiplying a matrix by its inverse, order does not matter: **A** × **A**^{-1} = **A**^{-1} × **A**.

In my next post, I will put all this talk about matrices in practice and use them to solve a system of equations.