Help solving matrices

tkhunny

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Apr 12, 2005
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Avoiding fractions is an important pursuit, but it is not the goal. You will have to get used to your fractions or develop methods around most of them. Sometimes, you can simply wait until later. For example, if you don't need R2 to clear up R3, do R3 first.

-2* R1 + R3

gives

0 0 -1 -2
or
0 0 1 2 <== New R3

Then

7*R3 + R2

gives

0 -2 0 -4
or
0 2 0 4
or
0 1 0 2

How do you feel about those fractions, now?

Rule #1 - Don't panic.
Rule #2 - Think!
 
Hello, trip20!

Those fractions aren't too bad . . . deal with them.


Solve the system of equations using matrices.
Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.

. . \(\displaystyle \begin{array}{ccc}3x\,+\,y\,-\,z & = & 0\\ x\,+\,y\,+\,2z & = & 6\\ 2x\,+\,2y\,+\,3z& = & 10\end{array}\)

First I created the augmented matrix:

. . \(\displaystyle \begin{pmatrix}3 & 1 &-1 & | & 0\\ 1 & 1 & 2 & | & 6 \\ 2 & 2 & 3 & | & 10\end{pmatrix}\)


\(\displaystyle R_1\,\leftrightarrow\,R_2\;\begin{pmatrix}1& 1& 2& |& 6\\ 3 & 1 &-1 & | & 0 \\ 2& 2 & 3 & | & 10\end{pmatrix}\)


\(\displaystyle \begin{array}{cccccc}\quad \\ \quad\\ \quad\\R_2\,-\,3R_1\\ \quad \\R_3-2R_1\end{array}\;
\begin{pmatrix}1 & 1 & 2 & | & 6 \\ 0 & -2 & -7 & | & -18 \\ 0 & 0 & -1 & | & -2\end{pmatrix}\)

\(\displaystyle \begin{array}{ccccc}\quad\\ \quad \\ \quad \\ (-1/2)(R_2)\\ (-1)R_3\end{array}\;
\begin{pmatrix}1 & 1 & 2 & | & 6 \\ 0 & 1 & \frac{7}{2} & | & 9 \\ 0 & 0 & 1 & | & 2\end{pmatrix}\)


\(\displaystyle \begin{array}{ccccccc}R_1\,-\,R_2 \\\quad \\\quad \\ \quad\\ \quad\\ \quad \\ \qquad \end{array}\;
\begin{pmatrix}1 & 0 & -\frac{3}{2} & | & -3 \\ 0 & 1 & \frac{7}{2} & | & 9 \\ 0 & 0 & 1 & | & 2\end{pmatrix}\)


\(\displaystyle \begin{array}{ccccccc}R_1\,+\,(3/2)R_3 \\ R_2\,-\,(7/2)R_3\\ \quad \\ \quad \\ \quad \\ \quad\end{array}\;
\begin{pmatrix}1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 2\end{pmatrix}\)

 
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