1
$\begingroup$

I'm currently working on a linear algebra problem and I'm having trouble understanding why my Gaussian elimination process is not yielding the same answer as the given solution.

The matrix I'm working with is:

$$\begin{pmatrix}1&0&-2\\ 3&1&-2\\ -5&-1&9\end{pmatrix}$$

And I performed Gaussian elimination on the matrix by doing the following steps:

  • R2-3R1
  • Switched row R1 with R3
  • R3 + R1/5
  • -R1/5
  • R3 + R2/5
  • R1 - R2/5
  • 5R3 / 11

And I ended up with:

$$\begin{pmatrix}1&0&-\frac{1}{5} \\ 0&1&-8\\ 0&0&1\end{pmatrix}$$

But this is not the same as the answer \begin{pmatrix}1&0&-2\\ 0&1&4\\ 0&0&3\end{pmatrix} that was given. I would really appreciate it if someone could help me figure out where I went wrong in my calculations.

Thank you in advance for your help!

$\endgroup$
1
  • $\begingroup$ You made an arithmetic mistake on the first step. You should have $-2--6=4$ but it looks like you did $-2-6=-8$. $\endgroup$
    – John Douma
    Commented May 11, 2023 at 23:48

1 Answer 1

1
$\begingroup$

The difference comes from the fact that you swapped rows when you didn't have to: you first use the operation $R2\to R2-3R_1$, but then swap R3 and R1. I expect you were supposed to replace R3 with R3+5R1, which gives you $\left(\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & -1 & -1\end{array}\right)$. If you then replace $R3$ with $R3+R2$ you get the matrix $\left(\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & 3\end{array}\right)$.

You should check to see exactly what are the steps of the algorithm you are required to use, and in particular when do you use the operation of swapping rows? Hopefully that will clear things up.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .