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

• Nov 30th 2009, 11:03 PM
somanyquestions
[SOLVED] using matrices to solve for a system of equations using gaussian elimination
my system of equations is

x-4y+3z-2w=9
3x-2y+z-4w= -13
-4x+3y-2z+w= -4
-2x+y-4z+3w=-10

my augmented matrix is
1 -4 3 -2 9
3 -2 1 -4 -13
-4 3 -2 1 -4
-2 1 -4 3 -10

i don't know how to plug this into my calculator to get an answer, but i got these answers using the row operations:
x= -19.5
y= -4.5
z= 0.5
w= 4.5

these answers don't seem correct, so i was wondering if anyone could tell me the actual answers when plugged into the calculator, so i could figure out what i did wrong or ask for more help.
• Dec 1st 2009, 12:13 AM
Prove It
Quote:

Originally Posted by somanyquestions
my system of equations is

x-4y+3z-2w=9
3x-2y+z-4w= -13
-4x+3y-2z+w= -4
-2x+y-4z+3w=-10

my augmented matrix is
1 -4 3 -2 9
3 -2 1 -4 -13
-4 3 -2 1 -4
-2 1 -4 3 -10

i don't know how to plug this into my calculator to get an answer, but i got these answers using the row operations:
x= -19.5
y= -4.5
z= 0.5
w= 4.5

these answers don't seem correct, so i was wondering if anyone could tell me the actual answers when plugged into the calculator, so i could figure out what i did wrong or ask for more help.

Note that if you have a system of equations of the form

$A\mathbf{x} = \mathbf{b}$

then you can use Matrix Algebra to solve for $\mathbf{x}$.

$A\mathbf{x} = \mathbf{b}$

$A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$

$I\mathbf{x} = A^{-1}\mathbf{b}$

$\mathbf{x} = A^{-1}\mathbf{b}$.

$A = \left[\begin{matrix}
1 & -4 & 3 & -2\\
3 & -2 & 1 & -4\\
-4 & 3 & -2 & 1\\
-2 & 1 & -4 &3 \end{matrix}\right]$

and

$\mathbf{b} = \left[\begin{matrix}
9 \\
-13 \\
-4 \\
-10 \end{matrix} \right]$
.

So enter these matrices into your calculator and then calculate $A^{-1}\mathbf{b}$.