# Inverse Matrices - A Primer

Orry Messer · 2018/11/18 · 7 minute read

# 1. Introduction

In this post we talk about the inverse matrix. We discuss what it means for a matrix to be invertible. We then discuss what it means for a matrix not to be invertible - singular. Finally, we briefly go through how to find a matrix’s inverse.

# 2. Invertibility

Let’s think about a square matrix, $A$. In a previous post we mentioned that a matrix acts on a vector. The matrix does something to that vector.

The inverse of a matrix undoes what the original matrix did. The inverse of $A$ is $A^{-1}$. $A^{-1}$ undoes what $A$ did. It’s kind of like dividing by a number after multiplying by it.

If we have a vector $x$, and we multiply it by our matrix $A$, we get $Ax$. If we undo this operation by applying the inverse, we get back to $x = A^{-1}Ax$. The one matrix which always leaves any vector unchanged is the identity matrix, $I$. That is: $Ix = x$ . This is why $A^{-1}A = I$.

When matrix $A$ has an inverse it is said to be invertible. However, this is not always the case. When a matrix is not invertible, it is also known as singular.

# 3. Non-Invertibility - Singular Matrices

Not all matrics are invertible. Matrices which are not invertible are known as singular. This means that there exists no matrix which can undo what the original matrix just did.

Say we are working in two dimensions - a plane. Most $2 \times 2$ matrices will move vectors around the $2 \times 2$ plane. They’ll either stretch them, or squash them, or even change their direction. But they’ll usually use up all of the plane; the output of that matrix multiplication won’t just be limited to a single line on that plane. The output of that matrix multiplication can usually fall anywhere on the plane.

But if the output of a $2 \times 2$ matrix always lands on a single line, then there is no way to undo that operation. We’ve, in effect, lost some information.

This extends to higher dimensions. If you cannot select a vector in $\mathbb{R}^n$ so that the output of an $n \times n$ matrix multiplication with that vector can fall anywhere within the $\mathbb{R}^n$ hyperplane, then that $n \times n$ matrix is singular. This means that the output of that matrix multiplication is constrained to a lower dimension (even a line, or a point).

It’s kind of like how $0$ has no inverse. If we have the number $10$, and we multiply it by $5$, we get $50$. If we want to undo this operation, we multiply by $5$’s (multiplicative) inverse: $\frac {1} {5}$ and get back $50 \times \frac {1} {5} = 10$. However, if we multiply $10$ by $0$, we get $0$. There really isn’t much we can do to $0$ to get back to $10$. This is a tenuous analogy at best, as matrices and numbers are different animals. But, hopefully this helps convey some intuition.

# 4. Testing for Singularity

There are several equivalent conditions which, if met, mean that a matrix is singular. Some of these conditions are:

• The matrix’s determinant is 0
• The matrix does not have full rank
• The $n \times n$ matrix does not have all of its pivots.

Discussions about rank and determinants are enough to warrant their own posts (and more), and are beyond the scope of this post. When I do get around to writing those posts, I’ll be sure to link them here!

For now, let’s just keep in mind that not all matrices have inverses, and that if the determinant of a matrix is zero, then it doesn’t have an inverse.

But if it does have an inverse, let’s try out finding it:

# 5. Inverse of a 2x2 Matrix

A $2 \times 2$ matrix has this form:

$\begin{bmatrix} a & b \newline c & d \end{bmatrix}$

If a $2 \times 2$ matrix is invertible, then its inverse is:

$\frac {1} {ad - bc} \times \begin{bmatrix} d & -b \newline -c & a \end{bmatrix}$

$\frac {1} {ad - bc}$ is the determinant of the matrix. If it’s zero, then we’d have $\frac {1} {0}$. In other words, the inverse wouldn’t exist - our matrix would be singular.

Consider the matrix:

$A % = % \begin{bmatrix} 1 & 3 \newline 2 & 7 \end{bmatrix}$

According to our formula above, its inverse is:

$\frac {1} {1 \times 7 - 3 \times 2} \times \begin{bmatrix} 7 & -3 \newline -2 & 1 \end{bmatrix} % = % \frac {1} {1} \times \begin{bmatrix} 7 & -3 \newline -2 & 1 \end{bmatrix} % = % \begin{bmatrix} 7 & -3 \newline -2 & 1 \end{bmatrix}$

Let’s verify that this is indeed the inverse in two ways.

## 5.1 Verification #1 - Matrix times Inverse Equals Identity

We’ll multiply the original matrix by the inverse. If we get the identity matrix, then we’ve done our calculation correctly:

$\begin{bmatrix} 1 & 3 \newline 2 & 7 \end{bmatrix} \begin{bmatrix} 7 & -3 \newline -2 & 1 \end{bmatrix} % = % \begin{bmatrix} 1 \times 7 + 3 \times (-2) & 1 \times (-3) + 3 \times 1 \newline 2 \times 7 + 7 \times (-2) & 2 \times (-3) + 7 \times 1 \end{bmatrix} % = % \begin{bmatrix} 1 & 0 \newline 0 & 1 \end{bmatrix}$

## 5.2 Verification #2 - Inverse Undoes what the Original Matrix Did

Next, we’ll multiply a vector by the original matrix and then the result of that calculation, by the inverse. We should get back our original vector.

Let’s pick the vector $\begin{bmatrix} 1 & 2 \end{bmatrix} ^ {T}$:

$\begin{bmatrix} 1 & 3 \newline 2 & 7 \end{bmatrix} \begin{bmatrix} 1 \newline 2 \end{bmatrix} % = % \begin{bmatrix} 1 \times 1 + 3 \times 2 \newline 2 \times 1 + 7 \times 2 \end{bmatrix} % = % \begin{bmatrix} 7 \newline 16 \end{bmatrix}$

Let’s now see if the inverse matrix brings $\begin{bmatrix} 7 & 16 \end{bmatrix} ^ {T}$ back to $\begin{bmatrix} 1 & 2 \end{bmatrix} ^ {T}$

$\begin{bmatrix} 7 & -3 \newline -2 & 1 \end{bmatrix} \begin{bmatrix} 7 \newline 16 \end{bmatrix} % = % \begin{bmatrix} 7 \times 7 + (-3) \times 16 \newline (-2) \times 7 + 1 \times 16 \end{bmatrix} % = % \begin{bmatrix} 1 \newline 2 \end{bmatrix}$

So it looks like we’ve done our calculations correctly!

# 6. Inverse of a Larger Matrix

For larger matrices, we can’t compute the inverse in the same way that we did for $2 \times 2$ matrices. Here, we’ll employ a different strategy using Gaussian Elimination.

Say we wish to find the inverse of the 3x3 matrix:

$A % = % \begin{bmatrix} 2 & -1 & 0 \newline -1 & 2 & -1 \newline 0 & -1 & 2 \end{bmatrix}$

We’ll start off by creating the augmented matrix $[A \space \space I]$, where $I$ is the identity matrix:

$[A \space \space I] % = % \begin{bmatrix} 2 & -1 & 0 & 1 & 0 & 0 \newline -1 & 2 & -1 & 0 & 1 & 0 \newline 0 & -1 & 2 & 0 & 0 & 1 \end{bmatrix}$

We then apply Gaussian Elimination, and end up with the matrix $[I \space \space A^{-1}]$:

$[I \space \space A^{-1}] % = % \begin{bmatrix} 1 & 0 & 0 & \frac{3} {4} & \frac{1} {2} & \frac{1} {4} \newline 0 & 1 & 0 & \frac{1} {2} & 1 & \frac{1} {2} \newline 0 & 0 & 1 & \frac{1} {4} & \frac{1} {2} & \frac{3} {4} \end{bmatrix}$

Yielding the inverse matrix:

$A^{-1} % = % \begin{bmatrix} \frac{3} {4} & \frac{1} {2} & \frac{1} {4} \newline \frac{1} {2} & 1 & \frac{1} {2} \newline \frac{1} {4} & \frac{1} {2} & \frac{3} {4} \end{bmatrix}$

We can then apply the same sort of calculations that we did for the $2 \times 2$ case, and test that this is indeed the inverse matrix. If you don’t believe that this is in fact the inverse, then feel free to do so :) I’m going to stop right here, though!