Find the Inverse of a 3 by 3 Matrix Online

The inverse of a 3 by 3 matrix is a bit complicated task but can be estimated by following the steps given below. A 3 by 3 matrix includes 3 rows and 3 columns.  Elements of the matrix are the numbers that form the matrix. A single matrix is one whose determinant is not equivalent to zero. For each x x x square matrix, there exists an inverse of each matrix. The inverse of matrix x * x is represented by X. The inverse of a matrix cannot be easily calculated using a calculator and shortcut method.

 XX-1 = X-1 X = I2

In the above property I2 indicates x * x matrix. For example, let us take 2 * 2 matrix as

\[\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}\]

Any x * x  square matrix X, which has zero determinant always includes an inverse X-1 .     

It is appropriate for almost all the square matrices and is given by  XX-1 = X-1 X = I2

How do you Find the Inverse of 3 by 3 Matrix?

Here, you can see the inverse of 3 by 3 matrix steps to find the inverse of 3 by 3 matrix online

1. Estimate the determinant of the given matrix

2. Find the transpose of the given matrix

3. Calculate the determinant of 2 x 2 matrix.

4. Prepare the matrix of cofactors

5. At the last, divide each term of the adjugate matrix by the determinant 

Inverse Matrix Formula

The first step is to calculate the determinant of 3 * 3 matrix and then find its cofactors, minors, and adjoint and then include the results in the below- given inverse matrix formula.

A-1 = 1/ |A | Adj (A)

Inverse of 3 X3 Matrix Example

Let us solve the 3 X 3 matrix

\[\begin{bmatrix} a & b &c \\ d & e & f\\ g & h & i\end{bmatrix}\]

Examine the given 3 X 3 matrix

A = \[\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4\\ 5 & 6 & 0\end{bmatrix}\]

Let’s learn the steps to find the inverse of 3 X 3 matrices online

Examine whether the given matrix is invertible

This can be proved if its determinant is non zero. If the determinant of the given matrix is zero, then there will be no inverse of the given matrix.

det(A) = 1(0-24) – 2(0-20) + 3(0-5)

det(A) = -24 + 40 – 15

det (A) = 1

As the value determinant is 1, we can say that a given matrix has an inverse matrix.

Finding the Transpose of the Given Matrix

To find the transpose of the given 3 X 3 matrix

Hence,A-1 = \[\begin{bmatrix} 1 & 0 & 5 \\ 2 & 1 & 6\\ 3 & 4 & 0\end{bmatrix}\]

Determining the determinants of the 2 X 2 minor matrices.

Now, we will determine the determinant of each and every 2 X 2 minor matrices.

For first row elements

\[\begin{bmatrix} 1 & 6\\ 4 & 0\end{bmatrix}\] = -24

\[\begin{bmatrix} 12 & 6\\ 3 & 0\end{bmatrix}\] = -18

\[\begin{bmatrix} 2 & 1\\ 3 & 4\end{bmatrix}\] = 5

For second row elements 

\[\begin{bmatrix} 0 & 5\\ 4 & 0\end{bmatrix}\] = -20

\[\begin{bmatrix} 1 & 5\\ 3 & 0\end{bmatrix}\] = -15

\[\begin{bmatrix} 1 & 0\\ 3 & 4\end{bmatrix}\] = 4

For third row element

\[\begin{bmatrix} 10 & 65\\ 14 & 60\end{bmatrix}\] = -5

\[\begin{bmatrix} 1 & 5\\ 2 & 6\end{bmatrix}\] = -4

\[\begin{bmatrix} 1 & 0\\ 2 & 1\end{bmatrix}\] = 1

Now, the new matrix is 

\[\begin{bmatrix} -24 & -18 & 5 \\ -20 & -15 & 4\\ -5 & 14 & 1\end{bmatrix}\]

Creating the Matrix of Cofactors

To formulate the adjoint or the adjugate matrix, reverse the sign of the alternating terms as shown below:

We get the new matrix as

A = \[\begin{bmatrix} -24 & -18 & 5 \\ -20 & -15 & 4\\ -5 & 14 & 1\end{bmatrix}\]

Adj (A) = the new matrix is 

A = \[\begin{bmatrix} -24 & -18 & 5 \\ -20 & -15 & 4\\ -5 & 14 & 1\end{bmatrix}\] X \[\begin{bmatrix} + & – & + \\ – & + & +\\ + & – & +\end{bmatrix}\]

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

Finding the inverse of 3 x 3 matrix

Now, if we will substitute the value of det (A) and the adj (A) in the formula:

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

A-1 = (1/1) =  \[\begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4\\ -5 & 4 & 1\end{bmatrix}\]

Hence, the inverse of the given matrix is

A-1 = (1/1) =  \[\begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4\\ -5 & 4 & 1\end{bmatrix}\]

Solved Examples

1. Find the Inverse of the Following Matrix

1   \[\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 1\\ 1 & -1 & 2\end{bmatrix}\]

Solution:  

-2\[\begin{bmatrix} 1 & 1\\ -1 & 2\end{bmatrix}\]  -1\[\begin{bmatrix} 1 & 1\\ 1 & 2\end{bmatrix}\] +1\[\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\] 

|A | = 2 [2-(-1)] -1 [2-1] +1[-1-1]

= 2 [2+1] -1 [1] +1 [-2]

= 2 [3] -1-2

= 6 – 3

= 3

|A | = 3  

As, A is a non-singular matrix. A -1 exists

Minor and cofactors of row 1

Minor of 2 

\[\begin{bmatrix} 1 & 1\\ -1 & 2\end{bmatrix}\] 

= [2-(-1)]

= [2 + 1]

= (3)

= 3

Cofactor of 2 = + (3)

= 3

Minor of 1

\[\begin{bmatrix} 1 & 1\\ 1 & 2\end{bmatrix}\] 

= [2-1]

= (2)

= 2

Cofactor of 1 = (-1)

= -1

Minor of 1

\[\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\] 

=[-1-1]

= (-2)

= -2

Cofactor of 1 = + (-2)

= -2

Minor and cofactors of row 2

Minor of 1

\[\begin{bmatrix} 1 & 1\\ 1 & -2\end{bmatrix}\] 

= [-2-1]

= -3

Cofactor of 1 = – (-3)

= -3

Minor of 1

\[\begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix}\] 

= [4 -1]

= (3)

= 3

Cofactor of 1 = – (-3)

= -3

Minor of 1

\[\begin{bmatrix} 2 & 1\\ 1 & -1\end{bmatrix}\] 

[-2-1]

= (-3)

= -3

Cofactor of 1 = – (-3)

= 3

Minor and Cofactors of Row 3

Minor of 1

\[\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}\] 

= [1-1]

= 0

Cofactor of 1 = + (0)

= 0

Minor of -1

\[\begin{bmatrix} 2 & 1\\ 1 & 1\end{bmatrix}\] 

= [2-1]

= 1

Cofactor of -1 = – (1)

= -1

Minor of 2

\[\begin{bmatrix} 2 & 1\\ 1 & 1\end{bmatrix}\] 

= [2-1]

= 1

Cofactor of 2 = + (1)

= 1 

Cofactor Matrix

\[\begin{bmatrix} 3 & -1 & -2 \\ -3 & 3 & 3\\ 0 & -1 & 1\end{bmatrix}\]

Adjoint Matrix

\[\begin{bmatrix} 3 & -3 & 0 \\ -1 & 3 & -1\\ -2 & 3 & 1\end{bmatrix}\]

Hence, the inverse of the given matrix is 

A-1 = ⅓ \[\begin{bmatrix} -3 & 3 & 0 \\ 1 & -3 & 1\\ 2 & -3 & -1\end{bmatrix}\]

1. Find the Inverse of the Following Matrix

\[\begin{bmatrix} 6 & 2 & 3 \\ 3 & 1 & 1\\ 10 & 3 & 4\end{bmatrix}\]

Solution:

-6\[\begin{bmatrix} 1 & 1\\ 3 & 4\end{bmatrix}\] -2\[\begin{bmatrix} 3 & 1\\ 10 & 4\end{bmatrix}\] +3\[\begin{bmatrix} 13 & 1\\ 10 & 3\end{bmatrix}\] 

|A | = 2 [2-(-1)] -1 [2-1] +1[-1-1]

= 2 [2+1] -1 [1] +1 [-2]

= 2 [3] -1-2

= 6 – 3

= 3

|A | = 3  

As, A is a non-singular matrix. A -1 exists

|A| = 6 [4-3] – 2 [12-10] + 3 [9-10]

   = 6 [1] – 2 [2] + 3 [-1]

   = 6 – 4 – 3

   = 6 – 7

   = -1

|A| = -1 ≠ 0

Minor and Cofactors of Row 1

Minor of 6

\[\begin{bmatrix} 1 & 1\\ 3 & 4\end{bmatrix}\] 

= [4-3]

= (1)

= 1

Cofactor of 6 = + (1)

= 1

Minor of 2

\[\begin{bmatrix} 3 & 1\\ 10 & 4\end{bmatrix}\] 

= [12-10]

= (2)

= 2

Cofactor of 2 = (-2)

= -2

Minor of 3

\[\begin{bmatrix} 3 & 1\\ 10 & 3\end{bmatrix}\] 

[9-10]

= (-1)

= -1

Cofactor of 3 = + (-1)

= -1

Minor and Cofactors of Row 2

Minor of 3 

\[\begin{bmatrix} 2 & 3\\ 3 & 4\end{bmatrix}\] 

= [8-9]

= (-1)

= -1

Cofactor of 3 = – (-1)

= 1

Minor of 1

\[\begin{bmatrix} 6 & 3\\ 10 & 4\end{bmatrix}\] 

= [24-30]

= (-6)

= -6

Cofactor of 1 = + (-6)

= -6

Minor of 1

\[\begin{bmatrix} 6 & 2\\ 10 & 3\end{bmatrix}\] 

[18-20]

= (-2)

-2

Cofactor of 1 = – (-2)

= 2

Minor and Cofactors of Row 3

Minor of 10

\[\begin{bmatrix} 2 & 3\\ 1 & 1\end{bmatrix}\] 

= [2-3]

= -1

Cofactor of 1 = + (-1)

= (-1)

= -1

Minor of 3

\[\begin{bmatrix} 6 & 3\\ 3 & 1\end{bmatrix}\] 

= [6-9]

= (-3)

= -3

Cofactor of 3 = – (-3)

= 3

Minor of 4

\[\begin{bmatrix} 6 & 2\\ 3 & 1\end{bmatrix}\] 

= [6-6]

= (0)

= 0

Cofactor of 4 = + (0)

= 0

Cofactor Matrix

\[\begin{bmatrix} 1 & -2 & -1 \\ -1 & -6 & 2\\ -1 & 3 & 0\end{bmatrix}\]

Adjoint of Matrix

= \[\begin{bmatrix} 1 & 1 & -1 \\ -2 & -6 & 3\\ -1 & 2 & 0\end{bmatrix}\]

A-1 = 1/1 = \[\begin{bmatrix} -1 & -1 & 1 \\ 2 & 6 & -3\\ 1 & -2 & 0\end{bmatrix}\]