Inverse of Matrix by Elementary Operations

An arrangement of numbers, expressions, or symbols in a rectangular array is called a matrix (plural: matrices). With an order of the number of rows × the number of columns, this layout comprises horizontal and vertical columns. In three-dimensional space, every pair of points represents a different equation with one or more solutions.

Matrix algebra involves matrix operations like addition, subtraction, multiplication, etc.

Inverse of Matrix

There exists an inverse matrix A-1 if A is a non-singular square matrix, which satisfies the below condition:

AA-1 = A-1A = I, where I’ is an Identity matrix.

Elementary operations

A matrix is subjected to six operations or transformations, three caused by the rows and three by the columns. These actions are referred to as elementary actions. Only square matrices are used for these operations.

These elementary operations are:

• Interchanging any two rows or columns ($R_i \leftrightarrow R_j$ or $C_i \leftrightarrow C_j$)

• Multiplying the elements of any row or column by a positive integer ($R_i \rightarrow kR_i$ or $C_i \rightarrow kC_i$)

• Addition or subtraction of multiples of one row to another ($R_i \rightarrow R_i + kR_j$ or $C_i$ → $C_i+ kC_j$)

How to find the inverse of the 3×3 matrix?

In order to determine a matrix's inverse, we must do the following steps:

• Finding the minors matrix is the first step.

• Change that matrix into a matrix of cofactors

• Find the adjoint of the matrix

• Multiply by \[\dfrac{1}{\text{determinant}}\]

Finding Inverse of Matrix Using Elementary Operations

If the three matrices, X, A, and B, are all of the same order, then X = AB is an equation of matrices. The basic row operations will be applied simultaneously to the matrices X and A, which is on the RHS side of the matrix equation of the product of AB for the given matrix equation.

Similarly, we can apply the basic column operations simultaneously on the matrices X and B (on the RHS side of the matrix equation) of the product of AB.

Thus, in general, if we calculate the inverse of a matrix A by doing simple row operations on A = IA in a certain order until we get: I = BA.

Also, the value of the inverse of matrix A must be obtained by performing basic column operations on the value of A = AI sequentially until we reach I = AB.

Inverse of Matrix Using Elementary Row transformation

As the names suggest, only the rows of the matrices are changed; the columns remain unchanged. A certain set of guidelines is followed while performing these row operations to ensure that the converted matrix is identical to the original matrix. Let’s understand how to find the inverse of a matrix using elementary row operations with the help of an example:

Example:

A = \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\]

Applying $R_1 \rightarrow R_2$

\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\3&1&1\end{array}} \right]\] 

Inverse of Matrix Using Elementary Column Transformation

The basic column operation is obtained by similarly applying those three-row operations to columns. 

Solved Examples:

1. Find the inverse of the matrix using elementary operations.

A = \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\] 

Solution:

A = \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\] 

Using the elementary row operation, we know;

$A = IA$

\[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\] = \[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\]. A

Applying $R_1 \rightarrow R_2$

\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\3&1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&1&0\\1&0&0\\0&0&1\end{array}} \right].A\]

Applying $R_3 \rightarrow R_3 - 3R_1$

\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&{ - 5}&{ - 8}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&1&0\\1&0&0\\0&{ - 3}&1\end{array}} \right].A\]

Applying $R_1 \rightarrow R_1-2R_2$ and $R_3\rightarrow R-3 + 5R_2$

$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2  \end{bmatrix}= \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{bmatrix} \cdot A$

Applying $ R_3 \rightarrow \frac{R_3}{2}$

\[\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}\\0&1&2\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}&1&0\\1&0&0\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{array}} \right].A\]

Applying $R_1 \rightarrow R_1+R_3, R_2 \rightarrow R_2 – 2R_3$

\[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&{\dfrac{{ - 1}}{2}}&{\dfrac{1}{2}}\\{ - 4}&3&1\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{array}} \right].A\]

Therefore,

$A^{-1}=\begin{bmatrix}{\dfrac{1}{2}}&{\dfrac{{ - 1}}{2}}&{\dfrac{1}{2}}\\{ - 4}&3&1\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{bmatrix}$

2. Find the inverse of the following matrix by elementary operations?

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

Solution: First rewrite the matrix as:

$[A|I]=\left[\begin{array}{ccc:ccc} 1 & 0 & 0 & -2 & 1 & 3 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 1 & 2 & 0 \end{array}\right]$

Applying $R_3 \rightarrow 2R_3+R_1$

$\left[\begin{array}{ccc:ccc} 1 & 0 & 0 & -2 & 1 & 3 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 1 & 0 & 2 & 0 & 5 & 3 \end{array}\right]$

Applying $R_1$  →  $R_1 + R_2$ and $R_3$ →  $R_3+ 5R_2$

$\left[\begin{array}{ccc:ccc} 1 & 1 & 0 & -2 & 0 & 4 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 1 & 5 & 2 & 0 & 0 & 8 \end{array}\right]$

Applying $R_1$  →  $2R_1 - R_3$  and  $R_2$ → $8R_2 - R_3$

$\left[\begin{array}{ccc:ccc} 1 & -3 & -2 & -4 & 0 & 0 \\ -1 & 3 & -2 & 0 & -8 & 0 \\ 1 & 5 & 2 & 0 & 0 & 8 \end{array}\right]$

Applying $R_1$ → $\frac{1}{-4}, R_2$ → $\frac{1}{-8}$, and $R_3$ → $\frac{1}{8} $

$\left[\begin{array}{ccc:ccc} -1 / 4 & +3 / 4 & +2 / 4 & 1 &0 &0 \\ +1 / 8 & -3 / 8 & 2 / 8 & 0 &1 & 0 \\ 1 / 8 & 5 / 8 & 2 / 8 & 0 & 1 & 0 \end{array}\right]$

Hence $A^{-1}=\left[\begin{array}{ccc:ccc} -1 / 4 & 3 / 4 & 2 / 4 \\ 1 / 8 & -3 / 8 & 2 / 8 \\ 1 / 8 & 5 / 8 & 2 / 8 \end{array}\right]$

Important formulas

• $AA^{-1} = A^{-1} A = I$, where I’ is an Identity matrix.

• $A^{-1} = \dfrac{(adjA)}{(\det A)}$

Interesting facts

• If A and B are 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.

• The inverse of the inverse matrix is equal to the original matrix.

• If A and B are invertible matrices, then AB is also invertible.

Summary

We learn about the Inverse of a matrix and how to find the inverse of a matrix using elementary operations. An arrangement of numbers, expressions, or symbols in a rectangular array is called a matrix and there exists an inverse matrix $A^{-1}$  if A is a non-singular square matrix, which satisfies the $AA^{-1}  = A^{-1} A = I$ condition. A matrix is subjected to six elementary operations. 

Problems for solving

1. Find the inverse of matrix A using elementary column operation.

A =\[\left[ {\begin{array}{*{20}{c}}3&5&1\\4&3&1\\9&8&1\end{array}} \right]\]

2.  Find the inverse of matrix A using elementary column operation.

A = \[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&0&1\\0&1&1\end{array}} \right]\] 

Answer:

1. \[\left[{\begin{array}{*{20}{c}}{-\frac{1}{3}}&{\frac{1}{5}}&{\frac{2}{{15}}}\\{\frac{1}{3}}&{ - \frac{2}{5}}&{\frac{1}{{15}}}\\{\frac{1}{3}}&{\frac{7}{5}}&{ - \frac{{11}}{{15}}}\end{array}} \right]\]

2. \[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&1\\0&1&0\end{array}} \right]\]