37 lines
1.4 KiB
TeX
37 lines
1.4 KiB
TeX
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\subsection{Problem 4}
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Solve the system of linear equations:
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\begin{equation*}
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\systeme{0.835x_1 + 0.667x_2 = 0.168,0.333x_1 + 0.266x_2 = 0.067}
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\end{equation*}
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Then slightly perturb $b_2$ from $0.067$ to $0.066$ and compute the solution to the perturbed system. Explain the change in the solution by computing the condition number of the system matrix.
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\subsubsection*{Solution}
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\begin{equation*}
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\begin{amatrix}{1}{1}
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\matr{A} & \matr{b}
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\end{amatrix} =
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\begin{amatrix}{2}{1}
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0.835 & 0.667 & 0.168\\
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0.333 & 0.266 & 0.067
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\end{amatrix} \xrightarrow[R_2-0.333R_1]{\frac{R_1}{0.835}}
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\begin{amatrix}{2}{1}
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1 & 0.7988 & 0.2012\\
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0 & 0 & 0
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\end{amatrix}
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\end{equation*}
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which yields infinite many solutions. Perturbing $b_2$ we obtain:
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\begin{equation*}
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\begin{amatrix}{1}{1}
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\matr{A} & \matr{b'}
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\end{amatrix} \xrightarrow[R_2-0.333R_1]{\frac{R_1}{0.835}}
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\begin{amatrix}{2}{1}
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1 & 0.7988 & 0.2012\\
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0 & 0 & -0.001
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\end{amatrix}
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\end{equation*}
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which yields none solutions. Performing the same calculations with Algorithm~\ref{algorithm:5}:
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\lstinputlisting[style=Matlab-editor]{problems/Problem4.m}
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With enough precision \MATLAB presented two valid solutions. Both of them differed by a factor of hundreds, which may be explained by a huge coefficient number (way over a million).
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