Question

Use the Gauss Jordan elimination method to find all solutions of the system of equations:
WILL GIVE A MEDAL!!

Answer 1

\[2x _{1}+3x _{2}=12\]

Answer 2

\[-4x _{1} + 4x _{2} = -4\]

Answer 3

\[8x _{1}-23x _{2} = -22\]

Answer 4

Find all solutions of the system of equations.

Answer 5

@genius12

Answer 6

Original System\[2x_1 + 3x_2 = 12
\\-4x_1 + 4x_2 = -4
\\8x_1 - 23x_2 = -22\]

Divide R2 by -4:
\[2x_1 + 3x_2 = 12
\\x_1 - x_2 = 1
\\8x_1 - 23x_2 = -22\]

Swap R2 and R1:
\[x_1 - x_2 = 1
\\2x_1 + 3x_2 = 12
\\8x_1 - 23x_2 = -22\]

Add -2R1 to R2:
\[x_1 - x_2 = 1
\\5x_2 = 10
\\8x_1 - 23x_2 = -22\]

Divide R2 to 5:
\[x_1 - x_2 = 1
\\x_2 = 2
\\8x_1 - 23x_2 = -22\]

Add R2 to R1:
\[x_1 = 3
\\x_2 = 2
\\8x_1 - 23x_2 = -22\]

Add 23*R2 to R3:
\[x_1 = 3
\\x_2 = 2
\\8x_1 = 24\]

Add -8*R1 to R3:
\[x_1 = 3
\\x_2 = 2\]

Answer 7

We still start off by making an Augmented matrix then apply elemental row operations to solve for A. We will call our matrix.

NOTE: In this first matrix, the reason for why you don't see the third row is because I multiplied row 3 by 0 so that it would be eliminated. The reason I eliminated it by multiplying it by 0 is because we only have 2 variables to solve for; x1 and x2. The first two rows representing the 2 equations are enough to solve for x1 and x2 so I thought it would be best to eliminate row 3 first.

\[A=\left[\begin{matrix}2 & 3 & 12 \\ -4 & 4 & -4\end{matrix}\right]\]Now we apply two elemental row operations to this matrix: row2 + (2 * row1) and row1 - (row2 * 3/4). This gives us the following matrix:\[A=\left[\begin{matrix}5 & 0 & 15 \\ 0 & 10 & 20\end{matrix}\right]\]In Gauss-Jordan Elimination, we need to have a diagonal of 1s and the rest be 0s. Thus I apply 2 more operations: row1 / 5 and row / 10. This gives us our final 'Identity Matrix':\[A=\left[\begin{matrix}1 & 0 & 3 \\ 0 & 1 & 2\end{matrix}\right]\]From this matrix, the values in the third column are basically our solutions to x1 and x2. \[\therefore x_{1}=3,x _{2}=2\]@Anita505 Sorry for the wait. I know Hero has already answered but I solved the matrix in a much faster way with only 2 steps, so I thought this might help you out.

Answer 8

@Anita505

Answer 9

I know lol. But I eliminated the third row from the matrix by multiplying it by 0 in order to get the same solution. And yes I know I can't ignore the third one, but I don't know how to add a third row with the equation box so I just eliminated it lol. And this is case, you should get the same solution with or without the third row which is why I ignored it.

Answer 10

And btw, I didn't start this question till much later on =.=

Answer 11

Multiplying the third row by zero is an illegal elementary operation.

Answer 12

So you did not solve it properly. Sorry bro.

Answer 13

If you were a student and I were the teacher, I would mark it wrong.

Answer 14

I know that multiplying by 0 is illegal, but the point of saying that was just that you don't need the third to solve for this system. It's the same thing without it. That's what I meant by 'multiplying by 0'.

Answer 15

I understand what you're trying to say, but your steps (no matter how you try to justify them) are incorrect according to the rules for elementary operations.

Answer 16

The justification for why you can eliminate the third row in this case is that if one can solve for the variables of a system without involving a certain row in the elemental row operations, then that row can be eliminated from the system as it isn't required to solve for the variables.

Answer 17

You said you multiplied the third row by 0 to eliminate it. That is not correct.

Answer 18

I did. But I have provided a justification which should be enough.

Answer 19

Well, the problem is, what you did was not Gauss-Jordan Elimination

Answer 20

Forget it....=.=