Inverse of 3 x 3 Matrix – Explanation & Examples

The inverse of a matrix is significant in linear algebra. It helps us solve a system of linear equations. We can find the inverse of square matrices only. Some matrices do not have inverses. So, what is the inverse of a matrix?

The inverse of a matrix $A$ is $A^{–1}$, such that multiplying the matrix with its inverse results in the identity matrix, $I$.

In this lesson, we will take a brief look at what an inverse matrix is, how to find the inverse of a $3\times 3$ matrix, and the formula for the inverse of a $3\times 3$ matrix. We will look at a couple of examples and some practice problems for you to try out!

What is the Inverse of a Matrix?

In matrix algebra, matrix inverse plays the same role as a reciprocal in number systems. Inverse matrix is the matrix with which we can multiply another matrix to get the identity matrix (the matrix equivalent of the number $1$)! To know more about the identity matrix, please check here.

Consider the $3\times 3$ matrix shown below:

$B=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$

We denote the inverse of this matrix as $B^{–1}$.

The multiplicative inverse (reciprocal) in the number system and the inverse matrix in matrices play the same role. Also, the identity matrix ($I$ ) (in matrices domain) plays the same role as the number one ( $1$ ).

How to Find the Inverse of a 3 x 3 Matrix

So how do we find the inverse of a $3\times 3$ matrix?

To find the inverse of a matrix, we can use a formula that requires a few points to be satisfied before its usage.

For a matrix to have an inverse, it has to satisfy $2$ conditions:

1. The matrix needs to be a square matrix (the number of rows must be equal to the number of columns).
2. The determinant of the matrix (this is a scalar value of a matrix from a few operations done on its elements) must not be $0$.

Remember, not all matrices that are square matrices have an inverse. A matrix whose determinant is $0$ is not invertible (doesn’t have an inverse) and is known as a singular matrix.

Read more about singular matrices here!

The formula for the inverse of a $3\times 3$ matrix is quite messy! Nonetheless, let’s tackle it!!

3 x 3 Inverse Matrix Formula

Consider the $3\times 3$ matrix shown below:

$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$

The formula for the inverse of a $3\times 3$ matrix (Matrix $A$) is given as:

$A^{–1}=\frac{1}{det(A)}\left[\begin{array}{ccc}(ei–fh)& –(bi–ch)& (bf–ce)\\ –(di-fg)& (ai–cg)& –(af–cd)\\ (dh–eg)& –(ah–bg)& (ae–bd)\end{array}\right]$

Where $det(A)$ is the determinant of the $3\times 3$ matrix given as:

$det(A)=a(ei–fh)–b(di–fg)+c(dh–eg)$

Tough!
Tough!
But don’t worry, after working out several questions, it will come to you naturally!

Let’s calculate the inverse of  a $3\times 3$ matrix ( Matrix $C$ ) shown below:

$C=\left[\begin{array}{ccc}1& 2& 1\\ 3& 4& 1\\ –1& 2& –1\end{array}\right]$

Before we calculate the inverse, we have to check the $2$ conditions outlined above.

• Is it a square matrix?

Yes, it is a $3\times 3$ square matrix!

• Is the determinant equal to $0$?

Let’s calculate the determinant of Matrix $C$ by using the determinant formula for a $3\times 3$ matrix.

$|C|=a(ei–fh)–b(di–fg)+c(dh–eg)$

$=1(–4–2)–2(-3–(–1))+1(6–(–4))$

$=1(–6)–2(–2)+1(10)$

$=8$

The determinant isn’t $0$. So, we can go ahead and calculate the inverse using the formula we just learned. Shown below:

$C^{–1}=\frac{1}{det(C)}\left[\begin{array}{ccc}(ei–fh)& –(bi–ch)& (bf–ce)\\ –(di–fg)& (ai–cg)& –(af–cd)\\ (dh–eg)& –(ah–bg)& (ae–bd)\end{array}\right]$

$C^{–1}=\frac{1}{8}\left[\begin{array}{ccc}–6& 4& –2\\ 2& 0& 2\\ 10& –4& –2\end{array}\right]$

$C^{–1}=\left[\begin{array}{ccc}–\frac{6}{8}& \frac{4}{8}& –\frac{2}{8}\\ \frac{2}{8}& 0& \frac{2}{8}\\ \frac{10}{8}& –\frac{4}{8}& –\frac{2}{8}\end{array}\right]$

Note: We multiplied the scalar constant, $\frac{1}{8}$, with each element of the matrix. This is the scalar multiplication of a matrix. 

Let’s reduce the fractions and write the final answer:

$C^{–1}=\left[\begin{array}{ccc}–\frac{3}{4}& \frac{1}{2}& -\frac{1}{4}\\ \frac{1}{4}& 0& \frac{1}{4}\\ \frac{5}{4}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right]$

Let us look at some examples to enhance our understanding further!

Example 1

Given $A=\left[\begin{array}{ccc}0& 1& 4\\ –1& –1& 1\\ 4& –2& 0\end{array}\right]$, find $A^{–1}$.

Solution

We will use the formula for the inverse of a $3\times 3$ matrix to find the inverse of Matrix $A$. Shown below:

$A^{-1}=\frac{1}{a(ei–fh)–b(di–fg)+c(dh–eg)}\left[\begin{array}{ccc}(ei–fh)& –(bi–ch)& (bf–ce)\\ –(di–fg)& (ai–cg)& –(af–cd)\\ (dh–eg)& –(ah–bg)& (ae–bd)\end{array}\right]$

$A^{–1}=\frac{1}{0(2)–1(-4)+4(6)}\left[\begin{array}{ccc}2& -8& 5\\ 4& -16& -4\\ 6& 4& 1\end{array}\right]$

$A^{–1}=\frac{1}{28}\left[\begin{array}{ccc}2& -8& 5\\ 4& -16& -4\\ 6& 4& 1\end{array}\right]$

$A^{–1}=\left[\begin{array}{ccc}\frac{1}{14}& –\frac{2}{7}& \frac{5}{28}\\ \frac{1}{7}& -\frac{4}{7}& -\frac{1}{7}\\ \frac{3}{14}& \frac{1}{7}& \frac{1}{28}\end{array}\right]$

Example 2

Given $A=\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 0\\ 1& 2& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& –2& 2\end{array}\right]$, confirm if  Matrix $B$ is the inverse of Matrix $A$.

Solution

For Matrix $B$ to be the inverse of Matrix $,A$, the matrix multiplication between these two matrices should result in an identity matrix ($3\times 3$ identity matrix). If so, $B$ is the inverse of $A$.

Let’s check:

$A\times B=\left[\begin{array}{ccc}2& 2& 1\\ 0& 1& 0\\ 1& 2& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& -2& 2\end{array}\right]$

$=\left[\begin{array}{ccc}(2)(1)+(2)(0)+(1)(1)& (2)(0)+(2)(1)+(1)(-2)& (2)(1)+(2)(0)+(1)(2)\\ (0)(1)+(1)(0)+(0)(1)& (0)(0)+(1)(1)+(0)(-2)& (0)(1)+(1)(0)+(0)(2)\\ (1)(1)+(2)(0)+(1)(1)& (1)(0)+(2)(1)+(1)(-2)& (1)(1)+(2)(0)+(1)(2)\end{array}\right]$

$=\left[\begin{array}{ccc}3& 0& 4\\ 0& 1& 0\\ 2& 0& 3\end{array}\right]$

This is not the  $3\times 3$ identity matrix! 

Thus, Matrix $B$ is not the inverse of Matrix $A$.

If you want to review matrix multiplication, please check this lesson out!

Practice Questions

1. Given $K=\left[\begin{array}{ccc}0& 2& -1\\ 3& -2& 1\\ 3& 2& -1\end{array}\right]$, find $K^{–1}$.

2. Calculate $A^{–1}$ for Matrix $A$ shown below:
   $A=\left[\begin{array}{ccc}1& –9& 1\\ –3& –1& 9\end{array}\right]$
3. Calculate the inverse of the $3\times 3$ matrix shown below:
   $D=\left[\begin{array}{ccc}2& 4& 8\\ 0& 1& 0\\ 1& -4& 1\end{array}\right]$

Answers

1. This matrix does not have an inverse because this matrix’s determinant is equal to $0$!

   Recall that the determinant cannot be $0$ for a matrix to have an inverse. Let’s check the value of the determinant:

   $|K|=0(2–2)–2(–3–3)+(–1)(6+6)$ 
   $|K|=0(0)–2(–6)–1(12)$
   $|K|=12–12$
   $|K|=0$

   Since the determinant is $0$, this matrix will not have an inverse!

2. If you look at this matrix carefully, you will see that it is not a square matrix!. It is a $2\times 3$ matrix ( $2$ rows and $3$  columns). Recall that we cannot find the inverse of a non-square matrix. 
   Thus, Matrix $A$ doesn’t have an inverse!
3. We will use the formula for the inverse of a $3\times 3$ matrix to find the inverse of Matrix $D$. Shown below:

   $D^{–1}=\frac{1}{a(ei–fh)–b(di–fg)+c(dh–eg)}\left[\begin{array}{ccc}(ei–fh)& –(bi–ch)& (bf–ce)\\ –(di–fg)& (ai–cg)& –(af–cd)\\ (dh–eg)& –(ah–bg)& (ae–bd)\end{array}\right]$

   $D^{–1}=\frac{1}{2(1)–4(0)+8(–1)}\left[\begin{array}{ccc}1& –36& –8\\ 0& –6& 0\\ –1& 12& 2\end{array}\right]$

   $D^{–1}=\frac{1}{–6}\left[\begin{array}{ccc}1& –36& –8\\ 0& –6& 0\\ –1& 12& 2\end{array}\right]$

   $D^{–1}=\left[\begin{array}{ccc}–\frac{1}{6}& 6& \frac{4}{3}\\ 0& 1& 0\\ \frac{1}{6}& –2& –\frac{1}{3}\end{array}\right]$

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