Elementary Row Operations

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

A = \begin{bmatrix}1 & 3 & 8 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix}

Now, row 1 ( R_{1} ) is \begin{bmatrix}1 & 3 & 8\end{bmatrix} which we multiply by 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}

Adding One Row To Another Row

A = \begin{bmatrix}1 & 3 & 8 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix}

Row 1 is \begin{bmatrix}1 & 3 & 8\end{bmatrix}
Row 2 is \begin{bmatrix}2 & 1 & 1\end{bmatrix}

Therefore, we can add row 1 to row2 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}

A = \begin{bmatrix}1 & 3 & 8 \\ 3 & 4 & 9 \\ 7 & 4 & 1\end{bmatrix}

Interchanging Rows

A = \begin{bmatrix}1 & 3 & 8 \\ 2 & 1 & 1 \\ 7 & 4 & 1\end{bmatrix}

Row 1 is \begin{bmatrix}1 & 3 & 8\end{bmatrix}
Row 2 is \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}

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.
------------------------------------------------------------
2x_1 + 3x_2 + x_3 = 5
5x_1 - 2x_2 + 3x_3 = 6
 x_1 + 2x_2 - x_3 = 2

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

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

-> INTERCHANGE ROW 1 WITH ROW 3

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

-> SUBTRACT ROW 2 FROM ROW 3

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

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

-> MULTIPLY R_1 BY 2

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

-> SUBTRACT ROW 1 FROM ROW 2

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}

-> MULTIPLY ROW 3 BY 2

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

-> SUBTRACT ROW 2 FROM ROW 3

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

Convert the matrix back to system of equations.

2x_1 + 4x_2 + -2x_3 = 4
      -6x_2 + 5x_3 = -1
               x_3 = 1

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

Given the following
2x_1 + 4x_2 + -2x_3 = 4
      -6x_2 + 5x_3 = -1
               x_3 = 1

Solve for x_2
      -6x_2 + 5(1) = -1
      -6x_2 +  5 - 5 = -1 - 5
      -6x_2  = -6
Therefore, x_2 = 1

Solve for 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
x_1 = 1

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

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