Too many equations (collinear?)

**salohcin** · Mar 1st 2011, 03:17 PM

Hello all,

I just need to clarify this. If I have too many equations (lets say 5 equations and 4 unknowns), when we say there is collinearity that means that 2 of the equations are telling us the same thing correct? And to solve this we need to try and rewrite these collinear equations as one equation?

Thanks very much for the help,

Nick

**skeeter** · Mar 1st 2011, 03:35 PM

two linear equations that represent the same line will be scalar multiples of one another ... just eliminate one of them.

**Soroban** · Mar 1st 2011, 08:05 PM

Hello, salohcin!

There is another situtation that arises . . . and it's harder to see.

Example: . $\begin{Bmatrix}3x + 2y &=& 8 & [1] \\ 2x -\; y &=& 3 & [2] \\ 4x + 5y &=& 13 & [3] \end{Bmatrix}$

If one equation is a linear combination of the other two,
. . the system is dependent; one equation can be dropped.

To test for dependence:

Multiply equation [1] by $\,a$ , multiply equation [2] by $\,b$ , and add.

If this sum can equal equation [3] for some values of $\,a$ and $\,b$ ,
. . the system is dependent.

$\begin{array}{cccccccc}<br /> \text{Multiply [1] by }a\!: & 3ax + 2ay &=& 8a \\<br /> \text{Multiply [2] by }b\!: & 2bx - by &=& 3b \end{array}$

$\text{Add: }\;(3a+2b)x + (2a-b)y \;=\;(8a+3b)$

$\text{If this equals }4x + 5y \;=\;13,\;\text{then: }\;\begin{Bmatrix}3a+2b &=& 4 \\ 2a - \;b &=& 5 \\ 8a+3b &=& 13 \end{Bmatrix}$

$\text{We find that the system has a solution: }\;a = 2,\;b = \text{-}1$

$\text{Therefore, the system is dependent; one equation may be dropped.}$

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