Solve the system

$$

\begin{align}

10^{-10}x_1 + x_2 &= 1 \\

x_1 + x_2 &= 2,

\end{align}

$$

using strict Gaussian elimination with and without pivoting. Comment on possible sources of error.

# A linear system

x1 = 1 and x2 = 1 with slight error

Through Gaussian elimination x2 equals (2-(10^10))/(1-(10^10)) which is super close to 1, and x1 = 2 - x1. So x1 is roughly 1 as well.

The system can be rewritten in the form $Ax = b$ such that matrix $A$ is the coefficients of $x_1$ and $x_2$, $x$ is $x_1$ and $x_2$ and b is comprised of the solutions, $1$ and $2$. Thus, it would look like this:

$$ A =

\left(

\begin{matrix}

10^{−10} & 1 \\

1 & 1

\end{matrix}

\right)

+

\left(

\begin{matrix}

x_1 \\

x_2\\

\end{matrix}

\right)

=

\left(

\begin{matrix}

1 \\

2\\

\end{matrix}

\right).

$$

This can be rewritten in augmented form with the terms flipped so that it is in upper triangular form (note $10^-10$ is close to zero--this is a possible source of error).

$$\left(

\begin{matrix}

1 & 1 & | & 2 \\

10^{-10} & 1 & | & 1\\

\end{matrix}

\right).

$$

Therefore:

$$ x_1 + x_2 = 2

and

x_2 = 1$$

So, we can easily see, $$ x_1 = 2

and

x_2 = 1$$

Though, I think the claim that $10^{-10}$ is close to $zero$ may be too strong, so this answer may be incorrect.

Strict Gaussian Elimination without swapping rows

$$\begin{bmatrix}

10^{-10} & 1 & 1 \\

1 & 1 & 2 \\

\end{bmatrix}$$

Row reduce to upper triangular

$$\begin{bmatrix}

10^{-10} & 1 & 1 \\

0 & 1-10^{-10} & 2-10^{-10} \\

\end{bmatrix}$$

Back substitution

$$x2 = 2-10^{-10}/1-10^{-10}$$

and

$$x1 = -1/10^{-10}$$

with $$10^{-10}$$ being so close to zero this creates a nasty results

With swapping rows

$$\begin{bmatrix}

1 & 1 & 2 \\

10^{-10} & 1 & 1 \\

\end{bmatrix}$$

Row reduce to upper triangular

$$\begin{bmatrix}

1 & 1 & 2 \\

0 & 1-10^{-10} & 1-2(10^{-10}) \\

\end{bmatrix}$$

Back substitution

$$x2 = 1-2(10^{-10})/1-10^{-10}$$

and

$$x1 = 1$$

@amccann1 is closest, but having a little trouble with the the typesetting, I think.

Here's the key issue: You've got to solve the system *exactly*. Using straight Gaussian elimination without changing the order of the rows, I get:

$$\begin{align}

x_1 &= 10^{10}\left(1-\frac{2-10^{10}}{1-10^{10}}\right) \\

x_2 &= \frac{2-10^{10}}{1-10^{10}}

\end{align}.$$

Using Gaussian elimination *after* a change in the order of the rows I get

$$\begin{align}

x_1 &= 2-\frac{2-10^{10}}{1-10^{10}} \\

x_2 &= \frac{2-10^{10}}{1-10^{10}}

\end{align}.$$

You should be able to verify both these computations. Also, note that the expressions are algebraically equivalent. One of them yields a better numerical approximation when evaluated on a computer, however. The questions are which and why?

I don't mean to sound like a genius, but the second one is better. However, the reason for this is too big to fit in a comment box.

Im confused. I ended up getting x1=10^10/(10^10-1) and x2=(10^10-2)/(10^10-1) without changing the order of the rows and wolframalpha even agrees with my solution.