# Basic Linear Algebra question

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• May 22nd 2009, 05:45 AM
vbiter
Basic Linear Algebra question
find a solution to the system:

5a1 + 2a2 - 3a3 = 4
a1 - a2 + 2a3 = -1

Solution:

we can eliminate a1 from the second row by using R1 - 5R2 -> R2, which
gives:

5a1 - 2a2 - 3a3 = 4
7a2 - 13a3 = 9

from the second equation we obtain:

a2 = 1/7 (9+13a3)

We substitute this expression in the first equation and solve for a1 in

terms of a3. We find that

a1 = 2/7 - 1/7 a3
....................

I can understand everything up to this point. However I cannot see how this has been substituted or how one can do this. Can anyone illustrate precisely what has happened in the last step?

Many thanks

Biter
• May 22nd 2009, 06:03 AM
scorpion007
Given $a_2=\frac{1}{7}(9+13a_3)$,

We substitute it into R1:

$5a_1 + 2a_2 - 3a_3 = 4$ (R1)
$\implies 5a_1 + 2({\color{red}\frac{1}{7}(9+13a_3)}) - 3a_3 = 4$

EDIT: You made a typo the second time you wrote R1 and reversed a sign.
• May 22nd 2009, 02:30 PM
vbiter
Hi thanks for helping me with this. The following step (or outcome) of the process is:

a1 = 2/7 - 1/7 a3

*************

I cannot follow the solution resulting from the substitution. Does the answer need to be factorised?

Thanks
• May 22nd 2009, 06:09 PM
scorpion007
It's just basic algebraic manipulation from there:

$5a_1 + 2(\frac{1}{7}(9+13a_3)) - 3a_3 = 4$

$\implies 5a_1 + \frac{18}{7}+\frac{26}{7}a_3 - 3a_3 = 4$

$\implies 5a_1 +\frac{5}{7}a_3 = 4-\frac{18}{7}$

$\implies 5a_1 = \frac{10}{7}-\frac{5}{7}a_3$

$\implies a_1 = \frac{2}{7}-\frac{1}{7}a_3$

Hope that is clear.