Inverse of a Matrix by Elementary Operations

Inverse of a matrix by elementary operations. A matrix is the arrangement of elements in a rectangular array. A matrix of order m x n, where m is the number of rows and n is the number of columns. We can perform the basic arithmetic operations on the matrices, such as addition, subtraction, multiplication. Elementary operations is a different type of operation that is performed on rows and columns of the matrices.

By the definition of inverse of a matrix, we know that, if A is a matrix (2×2 or 3×3) then inverse of A, is given by A-1, such that:

A.A-1 = I, where I is the identity matrix.

The basic method of finding the inverse of a matrix we have already learned. Let us learn here to find the inverse of a matrix using elementary operations.

Elementary Operations or Transformations

There are six transformations or operations performed on a matrix, that include the three transformations due to rows and three due to columns. These operations are known as elementary operations. These operations are performed on square matrices only.

These elementary operations are:

  • Interchanging any two rows or two columns
  • Multiplication of the elements of any row or column by a positive integer
  • Addition or subtraction of multiples of one row to another
  • Ri ↔ Rj or Ci ↔ Cj
  • Ri → kRi or Ci → kCi
  • Ri → Ri + kRj or Ci → Ci + kCj

Finding Inverse of Matrix Using Elementary Operations

If X = AB, is an equation of matrices, such that all the three matrices, X, A and B are of the same order. The elementary row operations will be applied on matrix X and A (on the RHS side of the matrix equation) of the product of AB, simultaneously, for the given matrix equation.

Similarly, the elementary column operations will be applied on the matrices X and B (on the RHS side of the matrix equation) of the product of AB, simultaneously.

Thus, in general if we want to evaluate the inverse of matrix A, by using elementary row operations on A = IA, in a sequence, until we get: I = BA.

Also, by using elementary column operations, on A = AI, in a sequence, till we get I = AB, we can get the value of the inverse of matrix A.

Fact: If A and B are two square matrices such that AB = BA = I, then B is the inverse matrix of A and is denoted by A–1 and A is the inverse of B.

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Solved Example

Problem: Find the inverse of the matrix using elementary operations.

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

Solution: Given,

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

Using the elementary row operation, we know;

A = IA

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

Applying R1 ←→ R2

\(\begin{array}{l}\begin{array}{l} \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \cdot A\end{array}\end{array} \)
\(\begin{array}{l}\text { Applying } R_{3} \rightarrow R_{3}-3 R_{1}\\ \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] . A\\\end{array} \)
\(\begin{array}{l}\text { Applying } R_{1} \rightarrow R_{1}-2 R_{2}, R_{3} \rightarrow R_{3}+5 R_{2}\\ \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right] \cdot A\end{array} \)
\(\begin{array}{l}\text { Applying } R_{3} \rightarrow \frac{R_{3}}{2} \\ {\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right] \cdot A} \\\end{array} \)

Applying R1 → R1+R3, R2 → R2 – 2R3

\(\begin{array}{l}{\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right] . A}\end{array} \)

I = B.A

Therefore,

\(\begin{array}{l}A^{-1}=\left[\begin{array}{ccc} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right]\end{array} \)

Practice Problem

Find the inverse of matrix A using elementary column operation.

A =

\(\begin{array}{l}A =\left[\begin{array}{ccc} 3 & 5 & 1 \\ 4 & 3 & 1 \\ 9 & 8 & 1 \end{array}\right]\end{array} \)
.

Frequently Asked Questions

Q1

What are elementary operations?

An elementary operation is applied on the matrices to find the inverse of the given matrix, by transforming rows or columns.

Q2

How many elementary operations are there?

There are six elementary operations on a matrix, which includes three transformations due to rows and three due to columns.

Q3

What is an elementary matrix?

An elementary matrix is a square matrix obtained by the swapping operation of rows or columns.

Q4

What is the inverse of a matrix A?

The inverse of a matrix A is A-1.

Q5

What is the inverse of the square matrix?

The inverse of the square matrix is unique, if it exists.

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