An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

# A linear system

mark

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.

ctyra

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.

sgerall

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.

amccann1

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
$$x1 = 1$$

mark

@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?

gbrock

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.

NateRicks

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.