Applications of matrices II

**Joker37** · December 18th 2008, 08:59 AM

3x - 2y = 6
6x - 4y = 12

Show that the simultaneous equations have an infinite number of solutions.

**Soroban** · December 18th 2008, 09:22 AM

Hello, Joker37!

$\begin{array}{cccc}3x - 2y &=& 6 & {\color{blue}[1]} \\ 6x - 4y &=& 12 & {\color{blue}[2]} \end{array}$

Show that the system has an infinite number of solutions.

$\begin{array}{cccc}\text{Multiply {\color{blue}[1]} by -2:} & \text{-}6x + 2y &=& \text{-}12 \\ \text{Add {\color{blue}[2]}:} & 6x - 4y &=& 12 \end{array}$

And we get: . . . . . . . . $0 \:=\:0$

This is true, regardless of the values of $x$ and $y$ .

Therefore, there are an infinite number of solutions.

**masters** · December 18th 2008, 10:12 AM

Originally Posted by Joker37

3x - 2y = 6
6x - 4y = 12

Show that the simultaneous equations have an infinite number of solutions.

Using an augmented matrix?

Perform some elementary row operations on the augmented matrix to get to reduced row echelon form.

$\left[\begin {array}{cc|c}3 & -2 & 6 \\ 6 & -4 & 12 \end{array}\right]$

Divide R1 by 3

$\left[\begin {array}{cc|c}1 & -\frac{2}{3} & 2 \\ 6 & -4 & 12 \end{array}\right]$

Multiply R1 by -6 and add to R2

$\left[\begin {array}{cc|c}1 & -\frac{2}{3} & 2 \\ 0 & 0 & 0 \end{array}\right]$

If there is a row in the augmented matrix containing all zeros (to the left and right of the line), then it is as if that row could be completely ignored, as it represents 0 = 0, which by itself is inconclusive. Whether or not there are solutions depends on the remaining row. The remaining row consists of an equation in 2 non-zero variables. We have more variables than we have equations, thus indicating an infinite number of solutions.

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