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**

```
Now, is which we multiply by 2.
```

**Adding One Row To Another Row**

```
Row 1 is
Row 2 is
Therefore, we can add row 1 to row2 as
```

**Interchanging Rows**

```
Row 1 is
Row 2 is
```

### 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.
------------------------------------------------------------
Solution:
-------------
First we need to convert the system of equation into augmented matrix.
-> INTERCHANGE ROW 1 WITH ROW 3
-> SUBTRACT ROW 2 FROM ROW 3
-> MULTIPLY R_1 BY 2
-> SUBTRACT ROW 1 FROM ROW 2
-> MULTIPLY ROW 3 BY 2
-> SUBTRACT ROW 2 FROM ROW 3
Convert the matrix back to system of equations.
```

Now we will use something called **back-substitution** and find and .

```
Given the following
Solve for
Therefore,
Solve for
```

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