# Math Help - I need help with matricies

1. ## I need help with matricies

Use Gaussian Elimination to find a solution to the given system or show that it is inconsistent?

x-3y-2z=2
3x-9y-9z=9
-x+4y+3z=3

This is a problem i have in the book and this is an (odd) question in which i have the answer too, which is x=18, y=6, z=-1. I really don't know how to do it. Can someone teach me or show me how they did it. I tried alot of times to get the answer but i still can't get the answer they got. Help Plzzz!

2. First, form your augmented matrix as follows:

$\left( \begin {array} {ccc|c}
1 & -3 & -2 & 2 \\
3 & -9 & -9 & 9 \\
-1 & 4 & 3 & 3
\end {array}\right)$

The column at the right contains the numbers to the right of the equals sign in the set of equations and the columns to the left are the coefficients of x, y and z. Each row now represents an equation.

We want only the top row to have a number other than 0 in the first column, so we will add multiples of that row to the other rows to get 0s in the other places.

$
\left( \begin {array}{ccc|c}
1 & -3 & -2 & 2 \\
0 & 0 & -3 & 3 \\
0 & 1 & 1 & 5
\end {array} \right)$

We have added -3 times the first row to the second row and 1 times the first row to the third row.

Now because we have a 0 in the second row and second column, we must switch it with a lower row.

$
\left( \begin {array}{ccc|c}
1 & -3 & -2 & 2 \\
0 & 1 & 1 & 5 \\
0 & 0 & -3 & 3
\end {array} \right)$

We have here swapped the second and 3rd rows and now have a one in the second row and 2nd column so we can proceed straight to the third row.

to get a 1 in the third row and the third column, we just need to multiply the third row by $\frac {1}{-3}$

$
\left( \begin {array}{ccc|c}
1 & -3 & -2 & 2 \\
0 & 1 & 1 & 5 \\
0 & 0 & 1 & -1
\end {array} \right)$

Now the matrix is in row echelon form. This is always what you are aiming to find in Gaussian elimination: the first non-zero number in each row is a 1 and all numbers below the first in a row are 0. The steps for getting there will vary with each matrix, but if you just work through one row and column at a time it is pretty easy to see the next step. If you have trouble remembering what you are "allowed" to do, think of this as just another way of setting out your work: you can do the same things you always could with the rows as if they were equations; you just don't write in the variable names and have a line instead of an =.

Now we have the matrix in row echelon form, we turn the rows back into equations.

x-3y-2z=2
y+z = 5
z = -1.

You can probably figure out how to solve the rest of this yourself.

3. I have a probably more confusing approach than his. But if you know how to find the inverse of a matrix then all you have to do to solve an equation for all the variables is multiply the inverse by the 'equals' matrix and out will pop out a matrix with the answers.

I don't quite remember the technique for finding the inverse...but it simple in a 2x2 but gets really large for 4x4 and more.

4. The inverse of a 2X2 matrix $\left(\begin {array} {cc} a & b \\ c & d \end {array} \right)$ is $\frac {1}{ad-bc} \left( \begin{array}{cc} d & -b \\ -c & a \end {array} \right)$. For larger matrices, you need will need to reduce the matrix to row echelon form anyway, but with more extra columns attached and some extra steps at the end. So it is never easier to use this to solve equations unless you already know the inverse of the matrix.

5. Originally Posted by TrevorP
...
I don't quite remember the technique for finding the inverse...but it simple in a 2x2 but gets really large for 4x4 and more.
the technique i use is the following.

first we note that if an nxn matrix is invertible, its reduced row echelon form is $I_n$.

we set up the following structure, the matrix on the left and $I_n$ on the right

$\begin{array}{ccc|ccc} 1 & -3 & -2 & 1 & 0 & 0 \\ 3 & -9 & -9 & 0 & 1 & 0 \\ -1 & 4 & 3 & 0 & 0 & 1 \\ \hline \end{array}$

now or objective is to bring the matrix on the left of the line to reduced row echelon form (it should look exactly like the matrix on the right when done). then the matrix on the right would end up being the inverse (of course whatever operation we do on the matrix on the left to reduce it, we do the same on the right)

as you said, 2x2 matrices are easy, there's a formula for them

EDIT: Ah, badgerigar posted the formula. i agree with him, his method is easier for solving these. there are actually other ways to find the inverse, they take too long though