Gaussian Elimination Method

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JMJ127

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Write the system of equations as an augmented matrix. Then solve for x and y.

−4x−2y=8

16x+5y=−20
 
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Write the system of equations as an augmented matrix. Then solve for x and y.

−4x−2y=8

16x+5y=−20
 
Write the system of equations as an augmented matrix. Then solve for x and y.

−4x−2y=8 My work: -4x-2y=8 -4 -2 8 16 5 =20
16x+5y=-20 16 5 -20 = -4 -2 = 8
16x+5y=−20 Someone please help, this is as far as I got and I have a quiz tomorrow :(
 
-4 -2 8
16 5 -20

Replace the 2nd row with: 4*1st row + the 2nd row. This will make the 16 a 0.

Post back with your work.
 
JMJ127 said:
Write the system of equations as an augmented matrix. Then solve for x and y.

−4x−2y=8 My work: -4x-2y=8 -4 -2 8 16 5 =20
16x+5y=-20 16 5 -20 = -4 -2 = 8
16x+5y=−20 Someone please help, this is as far as I got and I have a quiz tomorrow :(
Click to expand...
I think you mean [math]\begin{bmatrix}-4 & -2 & 9 \\ 16 & 5 & -20\end{bmatrix}[/math].

As Jomo suggested, add 4 times the first row to the second row.
This is NOT going to give nice, integer, numbers!
 
JMJ127 said:
−4x−2y=8 My work: -4x-2y=8 -4 -2 8 16 5 =20
16x+5y=-20 16 5 -20 = -4 -2 = 8
16x+5y=−20
Click to expand...
Hi JMJ. You may have noticed the forum software removed most of your spacing. For aligning columns, the forum guidelines suggest using [code] and [/code] tags with the Courier New font. Use the Preview button, to view and make adjustments before posting.
Code: 
My work:

-4  -2    8       16   5  = 20
16   5  -20       -4  -2  =  8
Your augmented coefficient matrix is correct.

Your first row operation swapped row1 with row2. That's a valid row operation, but it's not always necessary.

Gaussian Elimination requires a lot of arithmetic, so we need to be mindful of minor errors (like the missing negative sign on 20, above). I suggest double-checking each step as you go (check both the copying of numbers and the arithmetic). I organize my scratch paper, too.

The goal is to change the coefficient matrix to obtain all ones on the diagonal and zeros everywhere else. The augmented column will then display the solutions for x and y (shown as X and Y below).
Code: 
1  0  X
0  1  Y
We call that "reduced row echelon form". I usually choose to begin by getting a 1 in the upper-left corner (position R1,C1). Jomo and Halls chose to begin by getting a 0 in the lower-left corner (position R2,C1). That's okay! There are different ways, all of which yield that diagonal of 1s.

Following Jomo's lead, we replace (row2) with (4*row1+row2) because we can see that 4×(-4)+16 gives us the zero we want.
Code: 
-4  -2    8
16   5  -20

   4 * R1 + R2 --> R2

-4  -2    8
 0  -3   12
Next, it's easy to see how to get 1s on the diagonal. To change -4 into 1, we divide by -4. To change -3 into 1, we divide by -3.
Code: 
-4  -2    8
 0  -3   12

   -1/4 * R1 --> R1
   -1/3 * R2 --> R2

1   ½   -2
0   1   -4
The last step (to obtain reduced row echelon form) is to change 1/2 into 0. We do that using the 1 in row2.

-1/2 * R2 + R1 --> R1

If you find yourself making too many mistakes, please upload an image of your work. We can offer suggestions on how to organize.

JMJ127 said:
… I have a quiz tomorrow :(
Click to expand...
Pro Tip: Don't wait until the day before a quiz to become confident with the material.

?
 
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