Matrix inverse

As you already know, multiplying matrices is a much more difficult task than multiplying numbers. But in this topic, you will learn about the concept of inverse matrix, which can help you drastically simplify these calculations in some cases.

Since now, you'll know its most important application: find the unique solution of a square system of linear equations. Actually, when the solution is unique, the inverse is the perfect tool to calculate it explicitly.

Developing the definition

When you multiply any number by $1$, you get the same number. And nearly every number $x$ (except zero) has a partner $y$ such that $xy=1$ which is known as the multiplicative inverse of $x$. Thanks to this fact, you can solve a lot of equations. But how can you do this with matrices? To have the same flexibility, you should be able to "cancel" matrices in equations. In other words, you need to multiply inverses for them.

As a first step, the identity matrix $I$ will play the role of the number $1$ because you already know that $AI=IA = A$. The next step is clear, because to "neutralize" a square matrix $A$ you need another matrix $B$ such that $AB=I$. But as you know, the product is not commutative, so $B$ should also satisfy that $BA=I$. With this, how to define the inverse of a matrix is clear:

Just like in numbers, matrix $0$ doesn't have an inverse since $0A=0$ for any matrix $A$. But is it the only matrix that doesn't have an inverse? No! A lot of matrices don't have an inverse, and it's quite easy to see why. Just think of a matrix $C$ that has a row full of zeros, say the $i$-th row. No matter which matrix $B$ you take, the $i$-th row of $CB$ will always be $0$. Therefore $CB$ can never be $I$!

Another natural question would be if only one matrix can be the inverse of $A$. That is, if $AB=BA=I$, can there exist another matrix $C$ such that $AC=CA=I$? Fortunately, the answer is no. The inverse, when it exists, is unique. And you can easily check it:

$$B=IB=(CA)B=C(AB)=C(I)=C$$As a final note, even if there exists a matrix such that $AB=I$ you should also verify that $BA=I$. However, later you will discover that it is not necessary to do the second revision. It is enough that any of the two is fulfilled for both to do so.
Let's get our hands dirty with some matrices.

Checking out the inverse

The definition doesn't tell you much about how to find the inverse of a matrix $A$. If you had a candidate $A$, you'd have to check both that $AB=I$ and $BA=I$. Let's look at an example with some matrices $A = \begin{pmatrix} 1 & 6 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} -3 & 0 \\ 4 & -1 \end{pmatrix}$. In this case:

$$AB = \begin{pmatrix} 1 & 6 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} -3 & 0 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} 21 & -6 \\ 12 & -3 \end{pmatrix}$$

Since $AB \neq I$, you can be sure that $B \neq A^{-1}$. Now consider a new candidate $C = \begin{pmatrix} 1 & -2 \\ 0 & 1/3 \end{pmatrix}$:

$$AC = \begin{pmatrix} 1 & 6 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1/3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$Although $AC=I$, you also must check that $CA=I$, so let's get straight:

$$CA = \begin{pmatrix} 1 & -2 \\ 0 & 1/3 \end{pmatrix} \begin{pmatrix} 1 & 6 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$With this, you can be completely sure that $C=A^{-1}$.

A quick trick to rule out that two matrices are inverses of each other is the following. Calculate any entry on the diagonal of its product. If it is not $1$, then they are not inverses of each other. Alternatively, it is also true when any of the entries off the diagonal of the product is different from $0$.

After all, what is the benefit of knowing the inverse of a matrix? Time to have a look into a useful application.

Systems of linear equations

Systems of equations are very common. Sometimes they are very large and it is difficult to deal with so many equations. But thanks to the inverse, we can solve them in the blink of an eye! So the initial problem is reduced to calculating the inverse, which, as you will see in the future, is not that simple.

Take a system of linear equations $Ax=b$ where $A$ is a square matrix of size $n$ and both $x$ and $b$ are vectors in $\mathbb{R}^n$. If $A$ were invertible, then you could premultiply both sides by $A^{-1}$ to get:

$$A^{-1}Ax=A^{-1}b$$Therefore, the unique solution of the system is really simple, it's just $$x=A^{-1}b$$

The most surprising thing about this is that the value of $b$ doesn't matter. Notice also that this technique works for any square system. regardless of its size!

Getting more and more bulky

Time to see an example! A $1 \times 1$ linear system is just an equation involving numbers, like $5x=4$. You can easily solve it getting $x=4/5$. But the really interesting systems are those with greater dimensions.

Let's use the matrix from the example in the previous section $A = \begin{pmatrix} 1 & 6 \\ 0 & 3 \end{pmatrix}$. Take, for example $b = \begin{pmatrix} -4 \\ 3 \end{pmatrix}$.

Since you already know that $A^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1/3 \end{pmatrix}$, the unique solution of the system is:$$x=A^{-1}b = \begin{pmatrix}1 & -2 \\ 0 & 1/3 \end{pmatrix} \begin{pmatrix} -4 \\ 3 \end{pmatrix} = \begin{pmatrix} -10 \\ 1 \end{pmatrix}$$But what if $b = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$? You guessed it! The method works exactly the same:

$$x=A^{-1}b = \begin{pmatrix}1 & -2 \\ 0 & 1/3 \end{pmatrix} \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ -2/3 \end{pmatrix}$$

Now, you can see a more bulky example. The inverse of the $3 \times3$ matrix $A = \begin{pmatrix} 2 & 1 & -3 \\ 0 & 1 & -1 \\ 1 & -2 & 0 \end{pmatrix}$ is $A^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -6 & -2 \\ 1 & -3 & -2 \\ 1 & -5 & -2 \end{pmatrix}$. If you take $b = \begin{pmatrix} 2 \\ 8 \\-4 \end{pmatrix}$, then the solution of the system $Ax=b$ is just:

$$x=A^{-1}b = \frac{1}{2} \begin{pmatrix} 2 & -6 & -2 \\ 1 & -3 & -2 \\ 1 & -5 & -2 \end{pmatrix} \begin{pmatrix} 2 \\ 8 \\-4 \end{pmatrix} = \begin{pmatrix} -18 \\ -7 \\ -15 \end{pmatrix}$$

But how to calculate the inverse of some matrix $A$? This can be really difficult and require some new tools. You'll learn more about it in the next topic!

Conclusion

It is better that you summarize what you just learned. Take a square matrix $A$ of size $n$.

• $A$ is invertible if there is a square matrix $B$ of size $n$ such that $AB=BA=I$.
• When $A$ is invertible, its inverse is unique, and it's denoted by $A^{-1}$.
• The system $Ax=b$ has a unique solution when $A$ is invertible, and, in this case, the unique solution is $x=A^{-1}b$.