Inverse of a 3 by 3 Matrix

To find the inverse of a 3 by 3 Matrix is a little critical job but can be evaluated by following a few steps. A 3 x 3 matrix has 3 rows and 3 columns. Elements of the matrix are the numbers that make up the matrix. A singular matrix is the one in which the determinant is not equal to zero. For every m×m square matrix there exist an inverse of it. It is represented by M-1. The inverse of a matrix cannot be evaluated by calculators and using shortcuts will be inappropriate. We should practise problems to understand the concept. Let’s understand what is the inverse matrix? If M is a non-singular square matrix, there is an existence of n x n matrix M-1, which is called the inverse matrix of M such that it has a property as follows.

MM-1 = M-1 M = I

In the above property, I represents the m x m matrix. Suppose, take an example of a 2 x 2 unit matrix, i.e. I2.

\(\begin{array}{l}I_{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1  \end{bmatrix}\end{array} \)

Any m x m square matrix M, which has zero determinant always has an inverse M-1. It is mostly true for all the square matrices and is given by MM-1 = M-1M =Im

In this article, you learn how to find the inverse of a matrix of order 3 along with the formula, steps and example.

How to Find the Inverse of 3 x 3 Matrix?

The steps to find the inverse of the 3 by 3 matrix are given below.

Step 1: The first step while finding the inverse matrix is to check whether the given matrix is invertible. For this, we need to calculate the determinant of the given matrix.

If the determinant is not equal to 0, then it is an invertible matrix otherwise not. If it is invertible, proceed to the next step.

Step 2: Calculate the determinant of 2 × 2 minor matrices.

Step 3: Formulate the cofactor matrix.

Step 4: Find the Adjugate or Adjoint of the matrix by taking the transpose of the cofactor matrix.

Step 5: Finally, divide each term of the adjugate matrix by the determinant value of the given matrix.

Click here to learn how to find minors and cofactors as well as adjoint of a matrix.

Also, check: Transpose of a matrix

Inverse Matrix Formula

First, find the determinant of 3 × 3 matrices and then find its minor, cofactors and adjoint and insert the results in the Inverse Matrix formula given below:

\(\begin{array}{l}A^{-1}=\frac{1}{|A|}Adj(A)\end{array} \)

Where |A| ≠ 0

Learn how to find the inverse of 2 x 2 matrix here.

Inverse of a 3 x 3 Matrix Example

Example:Let’s see how 3 x 3 matrix looks :

M =

\(\begin{array}{l}\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}\end{array} \)

Consider the given 3×3 matrix:

\(\begin{array}{l}A =\begin{bmatrix} 1 & 2 &3 \\ 0 & 1 & 4\\ 5 & 6 & 0 \end{bmatrix}\end{array} \)

Let’s see what are the steps to find Inverse.

Check the Given Matrix is Invertible

Step 1: We can verify whether the given matrix is invertible using the value of determinant. If the determinant of the given matrix is zero, then there is no inverse for the given matrix

det (A) = 1(0 – 24) – 2(0 – 20) + 3(0 – 5)

det(A) = -24 + 40 – 15

det (A) = 1

Thus, we can say that the given matrix has an inverse matrix.

Finding the Determinants of the 2×2 Minor Matrices

Step 2: Now, we have to find the determinants of each and every 2×2 minor matrices

For first row elements:

\(\begin{array}{l}\begin{bmatrix} 1 & 4\\ 6& 0 \end{bmatrix} = -24\end{array} \)

 

\(\begin{array}{l}\begin{bmatrix} 0 & 4\\ 5 & 0 \end{bmatrix} = -20\end{array} \)

 

\(\begin{array}{l}\begin{bmatrix} 0 & 1\\ 5 & 6 \end{bmatrix} = -5\end{array} \)

 

For second-row elements:

\(\begin{array}{l}\begin{bmatrix} 2 & 3\\ 6 & 0 \end{bmatrix} =-18\end{array} \)

 

\(\begin{array}{l}\begin{bmatrix} 1 & 3\\ 5 & 0 \end{bmatrix} =-15\end{array} \)

 

\(\begin{array}{l}\begin{bmatrix} 1 & 2\\ 5 & 6 \end{bmatrix} = -4\end{array} \)

 

For third-row elements:

\(\begin{array}{l}\begin{bmatrix} 2 & 3\\ 1 & 4 \end{bmatrix} = 5\end{array} \)

 

\(\begin{array}{l}\begin{bmatrix} 1 & 3\\ 0 & 4 \end{bmatrix} = 4\end{array} \)

 

\(\begin{array}{l}\begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix} = 1\end{array} \)

Now, the new matrix formed is:

\(\begin{array}{l}A = \begin{bmatrix} -24&-20 &-5 \\ -18& -15 &-4 \\ 5 & 4 & 1 \end{bmatrix}\end{array} \)

Formulating the Matrix of Cofactors

Step 3: Now, to create the adjoint or the adjugated matrix, reverse the sign of the alternating terms as shown below:

The obtained matrix is

\(\begin{array}{l}A = \begin{bmatrix} -24&-20 &-5 \\ -18& -15 &-4 \\ 5 & 4 & 1 \end{bmatrix}\end{array} \)

\(\begin{array}{l}\begin{bmatrix} -24&-20 &-5 \\ -18& -15 &-4 \\ 5 & 4 & 1 \end{bmatrix}\times \begin{bmatrix}+ &- &+ \\ -& + & -\\ +&- & + \end{bmatrix}\end{array} \)

=

\(\begin{array}{l} \begin{bmatrix} -24&20 &-5 \\ 18& -15 &4 \\ 5 & -4 & 1 \end{bmatrix}\end{array} \)

Take the Transpose of the Cofactor Matrix to get the Adjugate Matrix

Step 4: Now take the transpose of the obtained cofactor matrix.

Thus, 

\(\begin{array}{l}Adj (A) =\begin{bmatrix} -24 & 18 &5 \\ 20 & -15 & -4\\ -5 & 4 & 1 \end{bmatrix}\end{array} \)

Finding the Inverse of the 3×3 Matrix

Step 5: Now, substitute the value of det (A) and the adj (A) in the formula:

A-1 = [1/det(A)]Adj(A)

A-1 = (1/1)

\(\begin{array}{l} \begin{bmatrix} -24&18 &5 \\ 20& -15 &-4 \\ -5 & 4 & 1 \end{bmatrix}\end{array} \)

Thus, the inverse of the given matrix is:

A-1 = (1/1)

\(\begin{array}{l} \begin{bmatrix} -24&18 &5 \\ 20& -15 &-4 \\ -5 & 4 & 1 \end{bmatrix}\end{array} \)

Note: We can also find the inverse of a 3×3 matrix using elementary row operations and elementary column operations.

Practice Problems

Try solving the practice questions given below to get a thorough understanding of the method of calculating the inverse of a 3×3 matrix using minors and cofactors.

  1. Find the inverse of a matrix 
    \(\begin{array}{l}A = \begin{bmatrix} 4 & 6 & 5\\ 1& 2 &3 \\ 0& 4 &6 \end{bmatrix}\end{array} \)
    if exists.
  2. Is the matrix 
    \(\begin{array}{l}\begin{bmatrix} 2 & 1 &2 \\ -1& -3 &-2 \\ 0& 1 &2 \end{bmatrix}\end{array} \)
    invertible? If yes, find its inverse.
  3. What is the inverse of matrix 
    \(\begin{array}{l}\begin{bmatrix} 0 & 1 &2\\ 1& 2 &3 \\ 3& 1 &1 \end{bmatrix}\end{array} \)
    ?

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