Open In App

Inverse of a Matrix by Elementary Operations – Matrices | Class 12 Maths

Improve
Improve
Like Article
Like
Save
Share
Report

The Gaussian Elimination method is also known as the row reduction method and it is an algorithm that is used to solve a system of linear equations. It is usually understood as a sequence of operations performed on the corresponding matrix of coefficients. This algorithm is used to find :

  • The rank of a matrix.
  • The determinant of a matrix.
  • The inverse of a matrix.

The operations we can perform on the matrix to modify are:

  • Interchanging/swapping two rows.
  • Multiplying or Dividing a row by a positive integer.
  • Adding or subtracting a multiple of one row to another.

Now using these operations we can modify a matrix and find its inverse. The steps involved are:

  • Step 1: Create an identity matrix of n x n.
  • Step 2: Perform row or column operations on the original matrix(A) to make it equivalent to the identity matrix.
  • Step 3: Perform similar operations on the identity matrix too. 

Now the resultant identity matrix after all the operations is the inverse matrix.

Examples

Note: 

Here, R1: Row 1, R2: Row 2, R3: Row 3

Example 1: Find the inverse of the following matrix by elementary operations?

A =\begin{bmatrix}2&0&3\\-1&3&-4\\-3&1&-4\end{bmatrix}

Solution:

Let’s perform row or column operations on the original matrix(A) to make it equivalent to the identity matrix.

Step 1: Interchange R2 and R3 rows (to make A[2][2] = 1)

\begin{bmatrix}2&0&3\\-3&1&-4\\-1&3&-4\end{bmatrix}

Step 2: R1 = R1 + R3 (to make A[1][1] = 1)

\begin{bmatrix}1&3&-1\\-3&1&-4\\-1&3&-4\end{bmatrix}

Step 3: R2 = R2 – 3R3 (to make A[2][1] = 0)

\begin{bmatrix}1&3&-1\\0&-8&8\\-1&3&-4\end{bmatrix}

Step 4: R3 = R3 + R1 (to make A[3][1] = 0)

\begin{bmatrix}1&3&-1\\0&-8&8\\0&6&-5\end{bmatrix}

Step 5: R2 = R2/-8 (to make A[2][2] = 1)

\begin{bmatrix}1&3&-1\\0&1&-1\\0&6&-5\end{bmatrix}

Step 6: R1 = R1 – R2 (to make A[1][3] = 0)

\begin{bmatrix}1&2&0\\0&1&-1\\0&6&-5\end{bmatrix}

Step 7: R3 – 6R2 (to make A[3][2] = 0)

\begin{bmatrix}1&2&0\\0&1&-1\\0&0&1\end{bmatrix}

Step 8: R2 = R2 + R3 (to make A[2][3] = 0)

\begin{bmatrix}1&2&0\\0&1&0\\0&0&1\end{bmatrix}

Step 9: R1 = R1 – 2R2 (to make A[1][2] = 0)

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

Now perform the same operation as above on the identity matrix. Result after each similar operations as above on the identity matrix, we get:

Identity \ matrix(I) = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Step 1: Interchange R2 and R3 rows

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

Step 2: R1 = R1 + R3

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

Step 3: R2 = R2 – 3R3

\begin{bmatrix} 1 & 1 & 0\\ 0 & -3 & 1\\ 0 & 1 & 0 \end{bmatrix}

Step 4: R3 = R3 + R1

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

Step 5: R2 = R2/-8

\begin{bmatrix} 1 & 1 & 0\\ 0 & 3/8 & -1/8\\ 1 & 2 & 0 \end{bmatrix}

Step 6: R1 = R1 – R2

\begin{bmatrix} 1 & 5/8 & 1/8\\ 0 & 3/8 & -1/8\\ 1 & 2 & 0 \end{bmatrix}

Step 7: R3 – 6R2

\begin{bmatrix} 1 & 5/8 & 1/8\\ 0 & 3/8 & -1/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

Step 8: R2 = R2 + R3

\begin{bmatrix} 1 & 5/8 & 1/8\\ 0 & 1/8 & 5/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

Step 9: R1 = R1 – 2R2

\begin{bmatrix} -1 & 3/8 & -9/8\\ 1 & 1/8 & 5/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

So, the inverse of matrix A is:

A^{-1}  = \begin{bmatrix} -1 & 3/8 & -9/8\\ 1 & 1/8 & 5/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

Example 2: Find the inverse of the following matrix by elementary operations?

A = \begin{bmatrix} 1& 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Solution:

Step 1: R1 = R1 + R2

\begin{bmatrix} 1& 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Step 2: R2 = R2 x -1

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

Similar operations on the identity matrix will result in:

A^{-1} = \begin{bmatrix} 1& 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Example 3: Find the inverse of the following matrix by elementary operations?

A = \begin{bmatrix} 1& 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix}

Solution:

Step 1: Swap R2 and R3

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

Step 2: R2 = R2 – R3

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

Similar operations on the identity matrix will result in:

A^{-1} = \begin{bmatrix} 1& 0 & 0\\ 0 & -1 & 1\\ 0 & 1 & 0 \end{bmatrix}

Last Updated : 17 Nov, 2020
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads