[SOLVED] using matrices to solve for a system of equations using gauss-jordan elimina

**somanyquestions** · November 28th 2009, 07:06 PM

i have to use matrices to solve the system of equations using gauss-jordan elimination

my system of equations is:
2x -y +3z =24
2y -z =14
7x -5y =6

my matrix would be
2 -1 3 24
0 2 -1 14
7 -5 0 6

can i have a walk through of this equations and an explanation on what to do? this is what i have so far:

i want to get this matrix into reduced row-echelon form, so first i put it into row-echelon form then reduce.
-i times Row 1 by 1/2 so i would end up with:
1 -1/2 3/2 12
0 2 -1 14
7 -5 0 6

-then i times Row 2 by 1/2 and end up with:
1 -1/2 3/2 12
0 1 -1/2 7
7 -5 0 6

-then i times Row 1 by -7 and add it to Row 3 and end up with:
1 -1/2 3/2 12
0 2 -1 14
0 -1.5 -10.5 -78

what would i do to get the leading 1 of row 3? am i doing this right? are my answers right so far? can anyone give me a walkthrough through the rest of the problem? sorry, my spacing isn't working.

**somanyquestions** · November 28th 2009, 07:40 PM

calculator answer=........ x = 8, y = 10, and z = 6

**Prove It** · November 28th 2009, 07:54 PM

Originally Posted by somanyquestions

i have to use matrices to solve the system of equations using gauss-jordan elimination

my system of equations is:
2x -y +3z =24
2y -z =14
7x -5y =6

my matrix would be
2 -1 3 24
0 2 -1 14
7 -5 0 6

can i have a walk through of this equations and an explanation on what to do? this is what i have so far:

i want to get this matrix into reduced row-echelon form, so first i put it into row-echelon form then reduce.
-i times Row 1 by 1/2 so i would end up with:
1 -1/2 3/2 12
0 2 -1 14
7 -5 0 6

-then i times Row 2 by 1/2 and end up with:
1 -1/2 3/2 12
0 1 -1/2 7
7 -5 0 6

-then i times Row 1 by -7 and add it to Row 3 and end up with:
1 -1/2 3/2 12
0 2 -1 14
0 -1.5 -10.5 -78

what would i do to get the leading 1 of row 3? am i doing this right? are my answers right so far? can anyone give me a walkthrough through the rest of the problem? sorry, my spacing isn't working.

$\left[\begin{matrix} 2 & -1 & 3 &|& 24\\ 0 & 2 & -1 &|& 14\\ 7 & -5 & 0 &|& 6 \end{matrix}\right]$

$\frac{1}{2}R_1 \to R_1$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 2 & -1 &|& 14\\ 7 & -5 & 0 &|& 6 \end{matrix}\right]$

$R_3 - 7R_1 \to R_3$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 2 & -1 &|& 14\\ 0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]$

$\frac{1}{2}R_2 \to R_2$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 1 & -\frac{1}{2} &|& 7\\ 0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]$

$R_3 + \frac{3}{2}R_2 \to R_3$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 1 & -\frac{1}{2} &|& 7\\ 0 & 0 & -\frac{45}{4} &|& -\frac{135}{2} \end{matrix}\right]$

$-\frac{4}{45}R_3 \to R_3$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 1 & -\frac{1}{2} &|& 7\\ 0 & 0 & 1 &|& 6 \end{matrix}\right]$

Now it is in row-echelon form and you can solve the system of equations.

**somanyquestions** · November 28th 2009, 08:01 PM

Originally Posted by Prove It

$\left[\begin{matrix} 2 & -1 & 3 &|& 24\\ 0 & 2 & -1 &|& 14\\ 7 & -5 & 0 &|& 6 \end{matrix}\right]$

$\frac{1}{2}R_1 \to R_1$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 2 & -1 &|& 14\\ 7 & -5 & 0 &|& 6 \end{matrix}\right]$

$R_3 - 7R_1 \to R_3$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 2 & -1 &|& 14\\ 0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]$

$\frac{1}{2}R_2 \to R_2$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 1 & -\frac{1}{2} &|& 7\\ 0 & -\frac{3}{2} & -\frac{21}{2} &|& -78 \end{matrix}\right]$

$R_3 + \frac{3}{2}R_2 \to R_3$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 1 & -\frac{1}{2} &|& 7\\ 0 & 0 & -\frac{45}{4} &|& -\frac{135}{2} \end{matrix}\right]$

$-\frac{4}{45}R_3 \to R_3$

$\left[\begin{matrix} 1 & -\frac{1}{2} & \frac{3}{2} &|& 12\\ 0 & 1 & -\frac{1}{2} &|& 7\\ 0 & 0 & 1 &|& 6 \end{matrix}\right]$

Now it is in row-echelon form and you can solve the system of equations.

how do you type out all these matrices?

**Prove It** · November 28th 2009, 08:06 PM

Originally Posted by somanyquestions

how do you type out all these matrices?

Click on them - that gives you the code I used.

**somanyquestions** · November 28th 2009, 08:21 PM

can you please make your answer into reduced row-echelon form?

**mr fantastic** · November 28th 2009, 09:39 PM

Originally Posted by somanyquestions

can you please make your answer into reduced row-echelon form?

The complete solution has almost been handed to you. Surely you can do the last few steps to get the reduced row echelon form.

**somanyquestions** · November 29th 2009, 08:46 AM

yeah, it's fine i was just freaking out.

Thread: [SOLVED] using matrices to solve for a system of equations using gauss-jordan elimina

LinkBack

Thread Tools

Display

[SOLVED] using matrices to solve for a system of equations using gauss-jordan elimina

Similar Math Help Forum Discussions

Solve a system using Gauss-Jordan elimination.

System of equations using Gaussian elimination or Gauss-Jordan elimination method

How to solve the following systems of equation in matrices/Gauss Jordan form?

[SOLVED] using matrices to solve for a system of equations using gaussian elimination

Solve system, using either Gaussian or Gauss-Jordan elimination

Search Tags