Earlier we learnt that we can solve for unknown variables by substituting arbitrary values for the equation. You can 
by solving for 
or 
one at a time.
Now, a system of equations have 
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 
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 
and 
.
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.
