Augmented Matrix

krclark

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Feb 14, 2016
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I'm having trouble figuring out how to get solutions or really even get an augmented matrix correct. I'm working in xyz homework and I don't even know if there's a correct way you're supposed to input the information when you get the solution for a computer program rather than x=7 etc, maybe it's supposed to be (7,0) or something (hypothetical answer). Anyway, here's the equation:

Solve the following system of equations by using matrices.

[FONT=Times New Roman, Times, serif]{x+1 = 1
{3x-y= -13

(there's only one bracket, not two and I assume it means make an augmented matrix)

Here's what I did, except there's a line directly to the left of 1
-13 to make it augmented
[/FONT]
111
3-1-13

[FONT=Times New Roman, Times, serif]Then I added R2 to R1, new matrix

[/FONT]
40-12
31-13

[FONT=Times New Roman, Times, serif]Then, to get the top left number to be zero in R1, I multiplied R1 by 1/4, new matrix

[/FONT]
10-3
3-1-13

[FONT=Times New Roman, Times, serif]Then, I multiplied R2, columns 2 and 3 by -1, trying to get R2, column 2 to 1, new matrix

[/FONT]
10-3
3113

[FONT=Times New Roman, Times, serif]But now I have no idea how to get the 3 to a zero and I don't even know if what I did was correct. Can someone point me in the right direction??


[/FONT]
 
Well, first, I'm assuming there's a typo in your first equation. If x + y = 1, then x=0 and you don't need a matrix at all. If, however, the equation is meant to be x + y = 1, then most of the work you've done so far is fine. The last step is where you encounter an error. You say:

Then, I multiplied R2, columns 2 and 3 by -1, trying to get R2, column 2 to 1, new matrix

This step is an error, because you're not allowed to do that. You can multiply an entire row by any number, positive or negative. But you can't multiply only some of the entries in a row. You can see that what you did won't work because the solution changes. Namely, in your third matrix, you've established that x=-3. If we plug that in to the remaining equation, we get: 3(-3) - y = -13, or y = 4. Allowing for your (incorrect) final step, we still have x = -3. But now the other equation is 3(-3) + y = -13 or y = -4. Performing the correct row multiplication, you'd arrive at this matrix:

\(\displaystyle \begin{pmatrix}1&0&-3\\-3&1&13\end{pmatrix}\)

From here, you're just one step away from the solution. I bet you'll feel silly when you realize what you need to do. You know that you can add any row to any other row or subtract any row from any other row and end up with an equivalent matrix. So, what if you subtract Row 1 from Row 2? In other words, let R2 = R2 - R1. What does the final row look like now? What if you repeated that subtraction? Can you see how to proceed now?
 
Still a little confused

Ok, so yes, there was a typo it is x+y=1.

That helps to know that I can't multiply only certain columns, but that I have to do the whole row. But, if I go back a step before I did that then my matrix is:

1
0
-3
3
-1
-13

Right?

I can see if I keep subtracting R2 from R1 how I can get the 3 to a 0, but then I get 0, -1, -22 for R2, then I have to multiply the whole row by (-1), right? So my final row becomes (0, 1, 22)? Are the solutions then supposed to work in the original equations, because if so, then I think I have something wrong still...
 
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