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Solving a system of linear equations with one unknown value

sparrow

New member
Sep 21, 2013
1
Hi guys, I'd be grateful for any help with this type of question. It's been driving me crazy. I have an online assignment for my linear algebra course and the question is always like this:

"Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values.

ax1+2x2+4x3 = 4
−2x1−x2−5x3 = −1
−2x1−5x2−x3 = −12"

I know the answer to this one as i got it wrong and the website told me what it was, but not how to get it. What I've been doing is putting it into a matrix (obviously) and then trying to get it into reduced row echelon form, and then trying to get the bottom row into either "0 0 0 I #" form (for infinite solutions), "0 0 0 I 0" (for no solution) and "0 0 1 I #" (for unique solution) where I just represents a line. But that isn't working so I'm clearly doing something wrong.

I can't use a calculator for this class or anything so please only solutions that can be done by hand :) Thank you very much.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
"Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values.

ax1+2x2+4x3 = 4
−2x1−x2−5x3 = −1
−2x1−5x2−x3 = −12"

I know the answer to this one as i got it wrong and the website told me what it was, but not how to get it.
What immediately strikes you about these equations is (1) that coefficient $a$ comes in a very tiresome place for getting the system in echelon form, and (2) the other two equations have minus signs everywhere.

You can make life easier for yourself by (1) swapping the first and third equations, so that the $a$ goes down to the bottom row of the matrix, (2) changing the signs on both sides of the other two equations. You will then have the system $$2x_1 + 5x_2 + x_3 = 12$$ $$2x_1 + x_2 + 5x_3 = 1$$ $$ax_1 + 2x_2 + 4x_3 = 4,$$ with matrix $$\begin{bmatrix} 2&5&1&12 \\2&1&5&1 \\ a&2&4&4\end{bmatrix}.$$

Now reduce it to echelon form in the usual way, starting by subtracting row 1 from row 2 and subtracting $a/2$ times row 1 from row 3.

What I've been doing is putting it into a matrix (obviously) and then trying to get it into reduced row echelon form, and then trying to get the bottom row into either "0 0 0 I #" form (for infinite solutions), "0 0 0 I 0" (for no solution) and "0 0 1 I #" (for unique solution) where I just represents a line. But that isn't working so I'm clearly doing something wrong.
You have the first two of those the wrong way round: $0\ 0\ 0\ |\ \#$ means no solutions (if $\#$ is a nonzero number), and $0\ 0\ 0\ |\ 0$ means infinitely many solutions.