and to the one that you get in MATLAB by typing $\frac{A}{\frac{?}{?}}b($which $d_{0es}{d}_0$

partial pivoting$).$

Transcribed Image Text: and to the one that you partial pivoting$).$ CHAPTER $7$ t in MATLAB by typing A$)$b $($which does do

Let

$A=([{10}^{-16},1],[1,1])$ and $b=([2],[3])$

Here we will see the effect of using a tiny element as a pivot. $\%$ Gaussian elimination without pivoting

$\%$ for solving linear systems Ax$=$b

$\%$

$\%$A $=[123$; $456$; $780]$; b $=[102]\text{'}$;

$\%$A $=[2110$; $4331$; $8795$; $6798]$; b $=[2345]\text{'}$; $\%$ another example

n $=$ size$($A$,1)$;

$\%$

$\%---------------------------------------------\%$

$\%$ This is Step $1$ of Gaussian Elimination

$\%----------------------------------------------\%$

$\%$

for i$=2$:n $\%$ Loop over rows below row $1$

mult $=$ A$($i$,1)/$A$(1,1)$; $\%$ Subtract this multiple of row $1$ from

$\%$ row i to make A$($i$,1)=0.$

A$($i$,$:$)=$ A$($i$,$:$)-$mult$*$A$(1,$:$)$; $\%($this line is equivalent to the "for loop:

$\%$ for k$=1$:n$,$ A$($i$,$k$)=$ A$($i$,$k$)-$mult$*$A$(1,$k$)$; end; $")$

b$($i$)=$ b$($i$)-$ mult$*$b$(1)$;

end

U $=$ A $\%$ display U

$\%$

$\%$

$\%----------------------------------------------\%$

$\%$ All steps of Gaussian elimination

$\%----------------------------------------------\%$

$\%$

A $=[123$; $456$; $780]$; b $=[102]\text{'}$;

$\%$A $=[2110$; $4331$; $8795$; $6798]$; b $=[2345]\text{'}$; $\%$ another example

n $=$ size$($A$,1)$;

for j$=1$:n$-1\%$ Loop over columns.

for i$=$j$+1$:n $\%$ Loop over rows below j$.$

mult $=$ A$($i$,$j$)/$A$($j$,$j$)$; $\%$ Subtract this multiple of row j from

$\%$ row i to make A$($i$,$j$)=0.$

A$($i$,$:$)=$ A$($i$,$:$)-$ mult$*$A$($j$,$:$)$; $\%$ This does more work than necessary! WHY?

b$($i$)=$ b$($i$)-$ mult$*$b$($j$)$;

end

end $\%$ the resulting A is an upper triangular matrix

U $=$ A $\%$ display U

$\%$

$\%$

$\%----------------------------------------------\%$

$\%$ Modified Gaussian elimination

$\%($to avoid recomputing zeros$)$

$\%----------------------------------------------\%$

$\%$

A $=[10^-161$; $11]$; b $=[2$; $3]$;

$\%$A $=[123$; $456$; $780]$; b $=[102]\text{'}$;

$\%$A $=[2110$; $4331$; $8795$; $6798]$; b $=[2345]\text{'}$; $\%$ another example

n $=$ size$($A$,1)$;

for j$=1$:n$-1\%$ Loop over columns.

for i$=$j$+1$:n $\%$ Loop over rows below j$.$

mult $=$ A$($i$,$j$)/$A$($j$,$j$)$; $\%$ Subtract this multiple of row j from

$\%$ row i to make A$($i$,$j$)=0.$

A$($i$,$j:n$)=$ A$($i$,$j:n$)-$ mult$*$A$($j$,$j:n$)$; $\%$ modified! no more recomputing of

$\%$ of zeros in columns $1$ to j$-1$ and rows j to n$.$

b$($i$)=$ b$($i$)-$ mult$*$b$($j$)$;

end

end $\%$ the resulting A is an upper triangular matrix

U $=$ A $\%$ display U

$($a$)$ By hand, solve the linear system $Ax=b$ exactly. Write your answer in

a form where it is clear what the approximate values of $x_1$ and $x_2$ are.

$($b$)$ In MATLAB, enter the matrix A and type cond$($A$)$ to determine the

$2-$norm condition number of $A.$ Would you say that this matrix is well

conditioned or ill conditioned in the $2-$norm?

$($c$)$ Write a MATLAB code $($or use one from the text$)$ that does not do

partial pivoting to solve the linear system $Ax=b.$ Compare the

answer returned by this code to the one that you determined by hand