I'm having hell of a time going around solving matrices to reduced row echelon form. My main issue is which row to start simplifying values and based on what? I have this example
so again, the questions are:
1.Which row to start simplifying values?
2.Based on what criteria?
Our professor solved it in the class with no fractions but I could not do it. Even though I know the 3 operations performed on matrices
$\endgroup$4 Answers
$\begingroup$Where you start is not really a problem.
My tip:
Always first make sure you make the first column: 1,0,0
Then proceed making the second one: 0,1,0
And lastly, 0,0,1
Step one:
$$\begin{pmatrix} 1&2&3&9 \\ 2&-1&1&8 \\ 3&0&-1&3\end{pmatrix}$$
row 3 - 3 times row 1
$$\begin{pmatrix} 1&2&3&9 \\ 2&-1&1&8 \\ 0&-6&-10&-24\end{pmatrix}$$
row 2 - 2 times row 1
$$\begin{pmatrix} 1&2&3&9 \\ 0&-5&-5&-10 \\ 0&-6&-10&-24\end{pmatrix}$$
Which simplifies to
$$\begin{pmatrix} 1&2&3&9 \\ 0&1&1&2 \\ 0&3&5&12\end{pmatrix}$$
Now you can proceed with step 2, and 3.
row 1 - 2 times row 2 and row 3 - 3 times row 2
$$\begin{pmatrix} 1&0&1&5 \\ 0&1&1&2 \\ 0&0&2&6\end{pmatrix}$$
Simplifies to
$$\begin{pmatrix} 1&0&1&5 \\ 0&1&1&2 \\ 0&0&1&3\end{pmatrix}$$
row 2 - row 3
$$\begin{pmatrix} 1&0&1&5 \\ 0&1&0&-1 \\ 0&0&1&3\end{pmatrix}$$
row 1 - row 3
$$\begin{pmatrix} 1&0&0&2 \\ 0&1&0&-1 \\ 0&0&1&3\end{pmatrix}$$
$\endgroup$ 5 $\begingroup$The order of steps doesn't matter. Here are some tips:
1/ If the first row doesn't have 1 as the leading entry, make it be!
2/ Go by columns when you want to make entries becoming 0's. Usually, start from the first column and make all entries in the first column (except the leading 1 of the first row) be 0's.
3/ Avoid fractions. They create a big confusion when you apply elementary row/column operations. Sometimes you can multiply the fractional entries to become whole numbers, apply ERO's or ECO's, then reduce it back to be fraction. It's a bit redundant, but it helps you see what you have done.
$\endgroup$ 4 $\begingroup$You can go in whatever order you want - it won't affect the final answer. If you wanted, you could always go about by eliminating the bottom left corner, then the bottom middle, then the middle left, etc. And I really do mean always.
At any point, if you get fractions and you don't want fractions, then you could multiply everything by the denominator of the fraction you're getting - that will remove all fractions and just leave integers.
$\endgroup$ 1 $\begingroup$practice with a 2x3 , look to get 1 or 0 in the first or sec. column . you my have to mul. first. EXAMP; 2(R1) + R2 , ect. to get started.
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