Inverse matrix

An inverse matrix is defined for square non-singular matrices. When we multiply a matrix by its inverse, we get the identity matrix.

Definition

An inverse matrix can only be worked out for a square matrix; for a rectangular matrix the inverse is not defined. The inverse of a matrix A (written A⁻¹) exists only if the matrix is non-singular (it has no linearly dependent rows). This inverse is then uniquely determined. The two main properties of the inverse matrix are:

$$(A^{-1}{)}^{-1}=A$$

And, most importantly, the key property of the inverse matrix (which is also its definition):

$$A\cdot A^{-1}=I$$

I is the identity matrix.

How to work out an inverse matrix

Now for the method of working out an inverse matrix. The simplest algorithm is Gaussian elimination. You transform the matrix A into the identity matrix, write an identity matrix next to it, and carry out the same operations on that matrix as on A. At the end of the process you have, on the left, the matrix A transformed into the identity matrix and, on the right, the inverse matrix. Let’s try it on a simple example:

$$\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$$

A quick glance shows that the matrix is non-singular, so it makes sense to work out the inverse. When working out an inverse matrix, we usually write it in this form:

$$\left(\begin{array}{cccc}1& 2& 1& 0\\ 3& 4& 0& 1\end{array}\right)$$

Now we transform the left-hand matrix into the identity matrix and carry out identical operations on the right-hand matrix. We add −3 times the first row to the second row. First the left-hand matrix:

$$\left(\begin{array}{cccc}1& 2& 1& 0\\ 0& -2& 0& 1\end{array}\right)$$

Now we add −3 times the first row of the right-hand matrix to the second row of the right-hand matrix:

$$\left(\begin{array}{cccc}1& 2& 1& 0\\ 0& -2& -3& 1\end{array}\right)$$

Now we need to get a zero in the position a₁₂. It is enough to add the second row to the first row. This time I will carry out both operations in one step:

$$\left(\begin{array}{cccc}1& 0& -2& 1\\ 0& -2& -3& 1\end{array}\right)$$

And now we get rid of the minus two by dividing the whole row by minus two:

$$\left(\begin{array}{cccc}1& 0& -2& 1\\ 0& 1& \frac{3}{2}& -\frac{1}{2}\end{array}\right)$$

The resulting inverse matrix is:

$$\left(\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right)$$

If you want to check that the inverse you have worked out is correct, multiply it by the original matrix:

$$\left(\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right)\cdot \left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

To check, you can also look at WolframAlpha.

Working it out using the determinant

We can also work out the inverse matrix using the determinant. If A is a non-singular matrix, then the inverse matrix A⁻¹ has entries $a_{ij}^{-1}$ equal to:

$$a_{ij}^{-1}=\frac{(-1{)}^{i+j}\cdot \left|A_{j,i}\right|}{\left|A\right|}$$

On the left-hand side we have i,j, but in the numerator we have j,i. Watch out: it is not a mistake that they are swapped. In the numerator we have the matrix A_j,i, which is the submatrix of A that we get by removing the jth row and the ith column from A. So if

$$A=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right),$$

then:

$$A_{1,1}=\left(\begin{array}{ccc}\text{⧸}1& \text{⧸}2& \text{⧸}3\\ \text{⧸}4& 5& 6\\ \text{⧸}7& 8& 9\end{array}\right)=\left(\begin{array}{cc}5& 6\\ 8& 9\end{array}\right),A_{1,3}=\left(\begin{array}{cc}4& 5\\ 7& 8\end{array}\right),A_{2,2}=\left(\begin{array}{cc}1& 3\\ 7& 9\end{array}\right)$$

If you do not like the subscripts being the other way round, we can write it like this instead:

$$(a_{ij}^{-1}{)}^T=\frac{(-1{)}^{i+j}\cdot \left|A_{i,j}\right|}{\left|A\right|}$$

In other words, if you do not swap the subscripts, you work out the transpose of the inverse matrix. If you transpose this matrix again, you get the inverse matrix.

Working it out this way is usually long-winded, and it is mainly suited to computers, because it is very straightforward. In the example we will only try to work out two entries. So let’s take this matrix:

$$A=\left(\begin{array}{ccc}3& -4& 5\\ 2& -3& 1\\ 3& -5& -1\end{array}\right)$$

First we work out the determinant of the whole matrix, which is used in the denominator of the fraction:

$$det(A)=-1$$

Now we work out the first entry of the inverse matrix. We denote it $a_{11}^{-1}$. It will be equal to:

$$a_{11}^{-1}=\frac{(-1{)}^{1+1}\cdot \left|A_{1,1}\right|}{\left|A\right|}$$

We will need the submatrix A_1,1. Leaving out the first row and the first column, we get:

$$A_{1,1}=\left(\begin{array}{cc}-3& 1\\ -5& -1\end{array}\right)$$

Now we need to work out the determinant of this matrix:

$$det(A_{1,1})=8$$

And now we can substitute into the formula:

$$a_{11}^{-1}=\frac{(-1{)}^2\cdot 8}{-1}=\frac{1\cdot 8}{-1}=-8$$

And we have the first entry of the inverse matrix.

$$A^{-1}=\left(\begin{array}{ccc}-8& ?& ?\\ ?& ?& ?\\ ?& ?& ?\end{array}\right)$$

We get the second number like this:

$$a_{12}^{-1}=\frac{(-1{)}^{1+2}\cdot \left|A_{2,1}\right|}{\left|A\right|}$$

The determinant of the submatrix will be

$$det(A_{2,1})=\left|\begin{array}{cc}-4& 5\\ -5& -1\end{array}\right|=29$$

And substituting everything, we get:

$$a_{12}^{-1}=\frac{(-1{)}^3\cdot 29}{-1}=\frac{-1\cdot 29}{-1}=\frac{-29}{-1}=29$$

Another piece of the jigsaw:

$$A^{-1}=\left(\begin{array}{ccc}-8& 29& ?\\ ?& ?& ?\\ ?& ?& ?\end{array}\right)$$

And so on, and so on.