Elementary Row Operations

Table of Contents

Earlier we learnt that we can solve for unknown variables by substituting arbitrary values for the equation. You can $a_{1}x_{1} + a_{2}x_{2} = b_{1}$ by solving for $x$ or $y$ one at a time.

Now, a system of equations have $m$ equations which we change into a new system having same solution set but simplified and easy to solve. This is achieved by performing row operations on the augmented matrix.

Row Operations

A row is a linear equation with $n$ unknown variables. You can perform following operations on any row in a system of equations.

Multiply a row with a scalar i.e., non-zero constant.
Add one row to another row. You are adding matching terms from one row to another.
Interchange two rows

Multiply A Row With A Non-Zero Scalar

\begin{aligned} &A = \begin{bmatrix}1 & 3 & 8 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix}\\\\ &Now, row 1 ( R_{1} ) \hspace{5px} is \hspace{5px}\begin{bmatrix}1 & 3 & 8\end{bmatrix} which \hspace{5px}we \hspace{5px}multiply \hspace{5px}by \hspace{5px}2.\\\\ &R_{1} \times 2 = \begin{bmatrix}1 & 3 & 8\end{bmatrix} \times 2\\\\ &A = \begin{bmatrix}2 & 6 & 16 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix} \end{aligned}

Adding One Row To Another Row

\begin{aligned} &A = \begin{bmatrix}1 & 3 & 8 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix}\\\\ &Row 1 \hspace{5px}is \hspace{5px} \begin{bmatrix}1 & 3 & 8\end{bmatrix}\\\\ &Row 2 \hspace{5px}is \hspace{5px} \begin{bmatrix}2 & 1 & 1\end{bmatrix}\\\\ &Therefore,\hspace{5px} we \hspace{5px}can \hspace{5px} add \hspace{5px}row 1 \hspace{5px}to\hspace{5px} row2 \hspace{5px}as \\\\ &R_{2} = R_{1} + R_{2} = \begin{bmatrix}2 + 1 & 1 + 3 & 1 + 8\end{bmatrix}\\\\ &R_{2} = R_{1} + R_{2} = \begin{bmatrix}3 & 4 & 9\end{bmatrix}\\\\ &\begin{bmatrix}1 & 3 & 8 \\ 3 & 4 & 9 \\ 7 & 4 & 1\end{bmatrix} \end{aligned}

Interchanging Rows

\begin{aligned} &A = \begin{bmatrix}1 & 3 & 8 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix}\\\\ &Row 1 \hspace{5px} is \hspace{5px} \begin{bmatrix}1 & 3 & 8\end{bmatrix}\\\\ &Row 2 \hspace{5px} is \hspace{5px} \begin{bmatrix}2 & 1 & 1\end{bmatrix}\\\\ &R_{1} \longleftrightarrow R_{2} = \begin{bmatrix}1 & 3 & 8\end{bmatrix} \longleftrightarrow \begin{bmatrix}2 & 1 & 1\end{bmatrix}\\\\ &A = \begin{bmatrix}2 & 1 & 1 \\ 1 & 3 & 8 \\ 7 & 4 & 1\end{bmatrix} \end{aligned}

Solving System Of Linear Equations Using Row Operations

The elementary row operations can be used to solve system of linear equations. For example,

Problem #1 : Solve the following system of equations.

\begin{aligned} &2x_1 + 3x_2 + x_3 = 5\\ &5x_1 - 2x_2 + 3x_3 = 6\\ &x_1 + 2x_2 - x_3 = 2 \end{aligned}

Solution:

First we need to convert the system of equation into augmented matrix.

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

-> INTERCHANGE ROW 1 WITH ROW 3

\begin{aligned} &R_3 \longleftrightarrow R_1 = \begin{bmatrix} 1 & 2 & -1 & 2\\ 2 & -2 & 3 & 3\\ 2 & 1 & 1 & 4\end{bmatrix} \end{aligned}

-> SUBTRACT ROW 2 FROM ROW 3

\begin{aligned} &R_3 = R_3 - R_2 = \begin{bmatrix}2-2 & 1+2 & 1-3 & 4-3\end{bmatrix}\\\\ &A = \begin{bmatrix} 2 & 4 & -2 & 4\\ 2 & -2 & 3 & 3\\ 0 & 3 & -2 & 1\end{bmatrix} \end{aligned}

-> MULTIPLY R_1 BY 2

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

-> SUBTRACT ROW 1 FROM ROW 2

\begin{aligned} &R_2 = R_2 - R_1 = \begin{bmatrix}2-2 & -2-4 & 3+2 & 3-4 \end{bmatrix}\\ \\ &A = \begin{bmatrix} 2 & 4 & -2 & 4\\ 0 & -6 & 5 & -1\\ 0 & 3 & -2 & 1\end{bmatrix} \end{aligned}

-> MULTIPLY ROW 3 BY 2

\begin{aligned} &A = \begin{bmatrix} 2 & 4 & -2 & 4\\ 0 & -6 & 5 & -1\\ 0 & 6 & -4 & 2\end{bmatrix} \end{aligned}

-> SUBTRACT ROW 2 FROM ROW 3

\begin{aligned} R_3 = R_2 - R_2 = \begin{bmatrix}2 & 4 & -2 & 4\\ 0 & -6 & 5 & -1\\ 0 & 0 & 1 & 1\end{bmatrix} \end{aligned}

Convert the matrix back to system of equations.

\begin{aligned} &2x_1 + 4x_2 + -2x_3 = 4\\ &\hspace{18px}-6x_2 + 5x_3 = -1\\ &\hspace{7em} x_3 = 1 \end{aligned}

Now we will use something called back-substitution and find $x_1$ and $x_2$ .

Given the following.

\begin{aligned} &2x_1 + 4x_2 + -2x_3 = 4\\ &\hspace{1.52em}-6x_2 + 5x_3 = -1\\ &\hspace{6.8em}x_3 = 1\\\\ &Solve \hspace{3px} for \hspace{3px}x_2\\\\ & -6x_2 + 5(1) = -1\\ &-6x_2 + 5 - 5 = -1 - 5\\ &-6x_2 = -6\\\\ &Therefore, \\ &x_2 = 1\\\\ &Solve \hspace{3px}for \hspace{3px}x_1\\ &2x_1 + 4(1) - 2(1) = 4\\ &2x_1 + 4 - 2 = 4\\ &2x_1 + 2 = 4\\ &2x_1 + 2 - 2 = 4 - 2\\ & 2x_1 = 2\\\\ &Therefore, \\ &x_1 = 1 \end{aligned}

In the next post, we will discuss more about how to solve the system of linear equations using row operations.