Elementary Row Operations: Definition, Properties, Formula and Examples

Elementary row operations

Edited By Komal Miglani | Updated on Sep 19, 2024 10:46 AM IST

An elementary matrix is a matrix that differs from the identity matrix by one single elementary row operation. Elementary row operations are used in Gaussian Elimination to reduce a matrix to row echelon form. In real life, we can use elementary row operations to quickly solve a system of equations, determine a matrix's rank, and more. The inverse of a matrix A can also be found using the basic row operations.

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Elementary Row Operations
Algorithm for finding the inverse of a singular 3 x 3 Matrix by Elementary Row Transformations
Solved Examples Based On Elementary Row Transformation
Summary

In this article, we will cover the concept of Elementary row transformation. This category falls under the broader category of Matrices, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main Exam (from 2013 to 2023), a total of 4 questions have been asked on this concept, including one in 2020, one in 2021, and three in 2022.

Elementary row transformation

In Elementary row transformation, the rows of the matrix are the only ones that are altered. The columns remain unchanged. A set of guidelines is followed when performing these row operations to ensure that the transformed matrix is identical to the original matrix.

Elementary Row Operations

Row transformation: The following three types of operation (transformation) on the rows of a given matrix are known as elementary row operation (transformation).
i) Interchange of $\mathrm{i}^{\text {th }}$ row with $\mathrm{j}^{\text {th }}$ row, this operation is denoted by
$
R_{\mathrm{i}} \leftrightarrow R_{\mathrm{j}}
$

During this operation, all the elements of $\mathrm{i}^{\text {th }}$ row get replaced by all the elements of $\mathrm{j}^{\text {th }}$ row.
ii) The multiplication of $\mathrm{i}^{\text {th }}$ row by a constant $\mathrm{k}(\mathrm{k} \neq 0)$ is denoted by
$
\mathrm{R}_{\mathrm{i}} \leftrightarrow \mathrm{kR}_{\mathrm{i}}
$

During this operation, all the elements $\mathrm{i}^{\text {th }}$ row are replaced by multiplication of elements $\mathrm{i}^{\text {th }}$ row by the constant $\mathrm{k}$.
iii) Adding of $\mathrm{i}^{\text {th }}$ row elements with of $\mathrm{j}^{\text {th }}$ row multiplied by constant $k(k \neq 0)$ is denoted by
$
R_i \leftrightarrow R_i+k R_j
$

During this operation, $\mathrm{i}^{\text {th }}$ row elements are replaced by adding the previous value of the $\mathrm{i}^{\text {th }}$ row and elements of $j^{\text {th }}$ row multiplied by a constant ( $k$ ).

In the same way, three-column operations can also be defined.

Steps for finding the inverse of a matrix of order 2 by elementary row operations

Step i: Write $A=I_n A$
Step II: Perform a sequence of elementary row operations successively on A on the LHS and the prefactor $I_n$ on the RHS till we obtain the result $I_n=B A$
Step III: Write $A^{-1}=B$

Algorithm for finding the inverse of a singular 3 x 3 Matrix by Elementary Row Transformations

Introduce unity at the intersection of the first row and first column either by interchanging two rows or by adding a constant multiple of elements of some other row to the first row.
After introducing unity at (1,1) place introduce zeros at all other places in the first column.
Introduce unity at the intersection of the 2nd row and 2nd column with the help of the 2nd and 3rd row.
Introduce zeros at all other places in the second column except at the intersection of 2nd row and 2nd column.
Introduce unity at the intersection of 3rd row and third column.
Finally, introduce zeros at all other places in the third column except at the intersection of third row and third column.

Elementary Column Transformation

In the Elementary column transformation, the matrices' columns are the only ones that are altered. The row remains unchanged. A predetermined set of guidelines is followed when performing these column operations to ensure that the transformed matrix is identical to the original matrix.

Solved Examples Based On Elementary Row Transformation

Example 1: Find the inverse of matrix A, if matrix A= $\left[\begin{array}{cc}a & b \\ c & \left(\frac{1+b c}{a}\right)\end{array}\right]$

Solution:

Use $\mathrm{A A^{-1}}=I$

$
\begin{aligned}
& {\left[\begin{array}{lc}
a & b \\
c & \left(\frac{1+b c}{a}\right)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_1 \rightarrow \frac{1}{{a}} \mathrm{R}_1 \\
& {\left[\begin{array}{lc}
1 & \frac{b}{a} \\
c & \left(\frac{1+b c}{a}\right)
\end{array}\right]=\left[\begin{array}{ll}
\frac{1}{a} & 0 \\
0 & 1
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{cR_{1 }} \\
& {\left[\begin{array}{ll}
1 & \frac{b}{q} \\
0 & \frac{1}{a}
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{a} & 0 \\
\frac{c}{a} & 1
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_2 \rightarrow \mathrm{aR}_2 \\
& {\left[\begin{array}{ll}
1 & \frac{b}{a} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{a} & 0 \\
-c & a
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_1-\frac{\mathrm{b}}{\mathrm{a}} \mathrm{R}_2 \\
& {\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
\frac{1+b c}{a} & -b \\
-c & a
\end{array}\right] \mathrm{A}} \\
& \mathrm{A}^{-1}=\left[\begin{array}{cc}
\frac{1+b c}{a} & -b \\
-c & a
\end{array}\right]
\end{aligned}
$

Example 2: Find the inverse of a matrix $
\mathrm{A}=\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 2 \\
3 & 1 & 1
\end{array}\right]
$

Solution:
First write, $\mathrm{A}=\mathrm{IA}$
$
\Rightarrow\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 2 \\
3 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$

Apply, $\mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_1$
$
\Rightarrow\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & -5 & -8
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
-3 & 0 & 1
\end{array}\right] \mathrm{A}
$

Apply, $\mathrm{R}_1 \rightarrow \mathrm{R}_1-2 \mathrm{R}_2$
$
\Rightarrow\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & -5 & -8
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
-3 & 0 & 1
\end{array}\right] \mathrm{A}
$

Apply, $\mathrm{R}_3 \rightarrow \mathrm{R}_3+5 \mathrm{R}_2$
$
\Rightarrow\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
-3 & 5 & 1
\end{array}\right] \mathrm{A}
$

$
\begin{aligned}
& \text { Apply, } \mathrm{R}_3 \rightarrow \frac{1}{2} \mathrm{R}_3 \\
& \Rightarrow\left[\begin{array}{lll}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \mathrm{A} \\
& \text { Apply, } \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_3 \\
& \Rightarrow\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
0 & 1 & 0 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \mathrm{A} \\
& \text { Apply, } \mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_3 \\
& \Rightarrow\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} \\
3 & -4 & -1 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \mathrm{A}
\end{aligned}
$

Hence, $
\mathrm{A}^{-1}=\left[\begin{array}{ccc}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
3 & -4 & -1 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right]
$

Example 3 Find the inverse of the matrix $A=\left[\begin{array}{lll}1 & 2 & 0 \\ 3 & 2 & 5 \\ 1 & 2 & 3\end{array}\right]$
Solution
Use $A A^{-1}=I$
$
\begin{aligned}
& {\left[\begin{array}{lll}
1 & 2 & 0 \\
3 & 2 & 5 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] A} \\
& R_1 \leftrightarrow R_2 \\
& {\left[\begin{array}{lll}
3 & 2 & 5 \\
1 & 2 & 0 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right] A}
\end{aligned}
$
$
\begin{aligned}
& R_2 \rightarrow R_2-\frac{1}{3} \cdot R_1 \\
& {\left[\begin{array}{ccc}
3 & 2 & 5 \\
0 & \frac{4}{3} & -\frac{5}{3} \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
1 & -\frac{1}{3} & 0 \\
0 & 0 & 1
\end{array}\right] A} \\
& R_3 \rightarrow R_3-\frac{1}{3} \cdot R_1 \\
& {\left[\begin{array}{ccc}
3 & 2 & 5 \\
0 & \frac{4}{3} & -\frac{5}{3} \\
0 & \frac{4}{3} & \frac{4}{3}
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
1 & -\frac{1}{3} & 0 \\
0 & -\frac{1}{3} & 1
\end{array}\right] A} \\
& R_3 \rightarrow R_3-1 \cdot R_2 \\
&
\end{aligned}
$

$\begin{aligned} & {\left[\begin{array}{ccc}3 & 2 & 5 \\ 0 & \frac{4}{3} & -\frac{5}{3} \\ 0 & 0 & 3\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ 1 & -\frac{1}{3} & 0 \\ -1 & 0 & 1\end{array}\right] A} \\ & R_3 \rightarrow \frac{1}{3} \cdot R_3 \\ & {\left[\begin{array}{ccc}3 & 2 & 5 \\ 0 & \frac{4}{3} & -\frac{5}{3} \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ 1 & -\frac{1}{3} & 0 \\ -\frac{1}{3} & 0 & \frac{1}{3}\end{array}\right] A} \\ & R_2 \rightarrow R_2+\frac{5}{3} \cdot R_3 \\ & {\left[\begin{array}{lll}3 & 2 & 5 \\ 0 & \frac{4}{3} & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ \frac{4}{9} & -\frac{1}{3} & \frac{5}{9} \\ -\frac{1}{3} & 0 & \frac{1}{3}\end{array}\right] A} \\ & R_1 \rightarrow R_1-5 \cdot R_3 \\ & {\left[\begin{array}{lll}3 & 2 & 0 \\ 0 & \frac{4}{3} & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{5}{3} & 1 & -\frac{5}{3} \\ \frac{4}{9} & -\frac{1}{3} & \frac{5}{9} \\ -\frac{1}{3} & 0 & \frac{1}{3}\end{array}\right] A} \\ & \end{aligned}$

$
\begin{aligned}
& R_2 \rightarrow \frac{3}{4} \cdot R_2 \\
& {\left[\begin{array}{lll}
3 & 2 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\frac{5}{3} & 1 & -\frac{5}{3} \\
\frac{1}{3} & -\frac{1}{4} & \frac{5}{12} \\
-\frac{1}{3} & 0 & \frac{1}{3}
\end{array}\right] A} \\
& R_1 \rightarrow R_1-2 \cdot R_2 \\
& {\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & \frac{3}{2} & -\frac{5}{2} \\
\frac{1}{3} & -\frac{1}{4} & \frac{5}{12} \\
-\frac{1}{3} & 0 & \frac{1}{3}
\end{array}\right] A} \\
& R_1 \rightarrow \frac{1}{3} \cdot R_1 \\
& {\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{1}{2} & -\frac{5}{6} \\
\frac{1}{3} & -\frac{1}{4} & \frac{5}{12} \\
-\frac{1}{3} & 0 & \frac{1}{3}
\end{array}\right] A} \\
& \text{Hence}, A^{-1}=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{1}{2} & -\frac{5}{6} \\
\frac{1}{3} & -\frac{1}{4} & \frac{5}{12} \\
-\frac{1}{3} & 0 & \frac{1}{3}
\end{array}\right] \\
&
\end{aligned}
$

Example 4: Find the inverse matrix of $
A=\left[\begin{array}{rrr}
1 & 1 & 2 \\
2 & 3 & 5 \\
-1 & 0 & 2
\end{array}\right]
$

Solution:
Use $A A^{-1}=I$
$
\left[\begin{array}{rrr}
1 & 1 & 2 \\
2 & 3 & 5 \\
-1 & 0 & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] A
$

Swap matrix rows: $R_1 \leftrightarrow R_2$
$
\begin{aligned}
& {\left[\begin{array}{ccc}
2 & 3 & 5 \\
1 & 1 & 2 \\
-1 & 0 & 2
\end{array}\right]=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right] A} \\
& R_2 \leftarrow R_2-\frac{1}{2} \cdot R_1 \\
& {\left[\begin{array}{ccc}
2 & 3 & 5 \\
0 & -\frac{1}{2} & -\frac{1}{2} \\
-1 & 0 & 2
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
1 & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{array}\right] A}
\end{aligned}
$

$
\begin{gathered}
R_3 \leftarrow R_3+\frac{1}{2} \cdot R_1 \\
{\left[\begin{array}{ccc}
2 & 3 & 5 \\
0 & -\frac{1}{2} & -\frac{1}{2} \\
0 & \frac{3}{2} & \frac{9}{2}
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
1 & -\frac{1}{2} & 0 \\
0 & \frac{1}{2} & 1
\end{array}\right]}
\end{gathered}
$

Swap matrix rows : $R_2 \leftrightarrow R_3$
$
\begin{aligned}
& {\left[\begin{array}{ccc}
2 & 3 & 5 \\
0 & \frac{3}{2} & \frac{9}{2} \\
0 & -\frac{1}{2} & -\frac{1}{2}
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
0 & \frac{1}{2} & 1 \\
1 & -\frac{1}{2} & 0
\end{array}\right] A} \\
& R_3 \rightarrow R_3+\frac{1}{3} \cdot R_2 \\
& {\left[\begin{array}{lll}
2 & 3 & 5 \\
0 & \frac{3}{2} & \frac{9}{2} \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
0 & \frac{1}{2} & 1 \\
1 & -\frac{1}{3} & \frac{1}{3}
\end{array}\right] A} \\
& R_2 \rightarrow R_2-\frac{9}{2} \cdot R_3 \\
& {\left[\begin{array}{lll}
2 & 3 & 5 \\
0 & \frac{3}{2} & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
-\frac{9}{2} & 2 & -\frac{1}{2} \\
1 & -\frac{1}{3} & \frac{1}{3}
\end{array}\right] A} \\
& R_1 \rightarrow R_1-5 \cdot R_3 \\
&
\end{aligned}
$

$\begin{aligned} & {\left[\begin{array}{lll}2 & 3 & 0 \\ 0 & \frac{3}{2} & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-5 & \frac{8}{3} & -\frac{5}{3} \\ -\frac{9}{2} & 2 & -\frac{1}{2} \\ 1 & -\frac{1}{3} & \frac{1}{3}\end{array}\right] A} \\ & R_2 \rightarrow \frac{2}{3} \cdot R_2 \\ & {\left[\begin{array}{lll}2 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-5 & \frac{8}{3} & -\frac{5}{3} \\ -3 & \frac{4}{3} & -\frac{1}{3} \\ 1 & -\frac{1}{3} & \frac{1}{3}\end{array}\right] A} \\ & R_1 \rightarrow R_1-3 \cdot R_2 \\ & {\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & -\frac{4}{3} & -\frac{2}{3} \\ -3 & \frac{4}{3} & -\frac{1}{3} \\ 1 & -\frac{1}{3} & \frac{1}{3}\end{array}\right] A} \\ & \end{aligned}$

$
\begin{aligned}
& \qquad R_1 \rightarrow \frac{1}{2} \cdot R_1 \\
& {\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
2 & -\frac{2}{3} & -\frac{1}{3} \\
-3 & \frac{4}{3} & -\frac{1}{3} \\
1 & -\frac{1}{3} & \frac{1}{3}
\end{array}\right] A} \\
\end{aligned}
$

Hence, $
A^{-1}=\left[\begin{array}{ccc}
2 & -\frac{2}{3} & -\frac{1}{3} \\
-3 & \frac{4}{3} & -\frac{1}{3} \\
1 & -\frac{1}{3} & \frac{1}{3}
\end{array}\right]
$

Summary

Elementary row transformations are a method for manipulating matrices while preserving essential properties and solution sets. They form the backbone of algorithms for solving systems of linear equations, finding matrix inverses, and performing various matrix operations. Elementary row operations provide a systematic approach for manipulating matrices while preserving their essential properties, making them indispensable in various mathematical applications and algorithmic implementations.

Frequently Asked Questions (FAQs)

1. What is Elementary row Transformation?

In Elementary row transformation, the matrices' rows are the only ones that are altered. The columns remain unchanged. A predetermined set of guidelines is followed when performing these row operations to ensure that the transformed matrix is identical to the original matrix.

2. What are the steps to find the inverse of a nonsingular 3 x3 matrix?

Introduce unity at the intersection of the first row and the first column either by interchanging two rows or by adding a constant multiple of elements of some other row to the first row. After introducing unity at (1,1) place introduce zeros at all other places in the first column. Introduce unity at the intersection of the 2nd row and 2nd column with the help of the 2nd and 3rd row.

3. Are row transformation and column Transformation the same?

No, Row transformation and column Transformation are different because, In Elementary row transformation, rows are the only ones that are altered. Whereas In the Elementary column transformation, the matrices' columns are the only ones that are altered.

4. While performing row operations, is the entire row multiplied by the constant?

Yes, While performing row operations entire row is multiplied by the constant. We multiply the entire row by a constant.

5. What are the different elementary row operations?

i) Interchange of $\mathrm{i}^{\text {th }}$ row with $\mathrm{j}^{\text {th }}$ row, this operation is denoted by $\mathrm{R}_{\mathrm{i}} \leftrightarrow \mathrm{R}_{\mathrm{j}}$
ii) The multiplication of $\mathrm{i}^{\text {th }}$ row by a constant $k(k \neq 0)$ is denoted by $R_i \leftrightarrow k R_i$
iii) Adding of $\mathrm{i}^{\text {th }}$ row elements with of $\mathrm{j}$ th row multiplied by constant $\mathrm{k}(\mathrm{k} \neq 0)$ is denoted by $R_i \leftrightarrow R_i+k R_j$