Math CentralQuandaries & Queries


Question from Lisa, a student:

I am supposed to use augmented matrices to solve this system of equations.
x - 3y = 5; 4x - 12y = 13. I think it is unsolvable but I want to be sure I am not just giving up. Thanks in advance for your help! Lisa.

We have two responses for you

Hello Lisa.

Yes in fact this matrix has no solution, which in turn means that these two lines do not intersect. The reason being is if you set up your matrix it will look like this

1   -3|  5
4 -12|13

when you try to remove the 4 below the 1 you take row two and subtract 4 times row 1. Doing this gets rid of the 4, but it also gets rid of the -12. Leaving you with a matrix that looks like this

1 -3 | 5
0  0 |-7

If you ever have row, such as the second row, where the last entry is not zero but all the other entries are zero, then there is no solution, which means the lines do not cross. However if you have the same number of equations as variables and row reduction yield a row of all zeros , then there are infinitely many solutions. So basically a long story short, you are correct there is no solution to this problem.

I hope this helped

Brennan Yaremko


Hi Lisa.

Another way to tell that this is insoluble is to remember that a solution means an ordered pair (x,y) such that both equations are true. This means the intersection of the two lines.

Remember back to when you first learned about line equations. Standard form for a line is:

Ax + By = C.

and from this, the slope is -A/B.

Since your two equations
x - 3y = 5;
4x - 12y = 13

have the same -A/B, they are parallel and hence, don't intersect, so there is no solution.

This isn't proving by augmented matrices, but it is a perfectly valid way to check your answer.

Stephen La Rocque.

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