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mathmari
Gold Member
MHB
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Hey!
I want to determine the LU decomposition of
$A=\begin{pmatrix}0 & 2 & 1\\1 & 10 & 1 \\1 & 1 & 1\end{pmatrix}$ with total pivoting. I have done the following:
The biggest element of the whole matrix is $10$, so we exchange the first two rows and the first two columns and then we get $\begin{pmatrix}10 & 1 & 1 \\2 & 0 & 1\\1 & 1 & 1\end{pmatrix}$.
Applying now the Gauss algorithm we get $\begin{pmatrix}10 & 1 & 1 \\0 & -\frac{1}{5} & \frac{4}{5}\\0 & \frac{9}{10} & \frac{9}{10}\end{pmatrix}$.
The biggest element of the submatrix is $\frac{9}{10}$ and so we exchange the last two rows and get: $\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & -\frac{1}{5} & \frac{4}{5} \end{pmatrix}$. Now we apply the Gauss algorithm and get: $\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.
The matrix $U$ is the resulting matrix, $U=\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.
The matrix $L$ is $L=P\cdot P_0\cdot G_1^{-1}\cdot P_1\cdot G_2^{-1}$, or not? (Wondering)
The matrices $G_i^{-1}$ are defined as:
$$G_1^{-1}=\begin{pmatrix}1 & 0 & 0 \\\frac{2}{10} & 1 & 0\\\frac{1}{10} & 0 & 1\end{pmatrix} \ \text{ and } \ G_2^{-1}=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & -\frac{2}{9} & 1\end{pmatrix}$$ or not? (Wondering)
Are the matrices $P_i$ defined as follows?
$$P_0=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$$ since this describes the step at which we exchanged the first two rows and the first two columns. (Wondering)
$$P_1=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}$$ since this describes the step at which we exchanged the two last rows. (Wondering) If these are correct, it doesn't hold that $LU=PA$, does it? (Wondering)
I want to determine the LU decomposition of
$A=\begin{pmatrix}0 & 2 & 1\\1 & 10 & 1 \\1 & 1 & 1\end{pmatrix}$ with total pivoting. I have done the following:
The biggest element of the whole matrix is $10$, so we exchange the first two rows and the first two columns and then we get $\begin{pmatrix}10 & 1 & 1 \\2 & 0 & 1\\1 & 1 & 1\end{pmatrix}$.
Applying now the Gauss algorithm we get $\begin{pmatrix}10 & 1 & 1 \\0 & -\frac{1}{5} & \frac{4}{5}\\0 & \frac{9}{10} & \frac{9}{10}\end{pmatrix}$.
The biggest element of the submatrix is $\frac{9}{10}$ and so we exchange the last two rows and get: $\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & -\frac{1}{5} & \frac{4}{5} \end{pmatrix}$. Now we apply the Gauss algorithm and get: $\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.
The matrix $U$ is the resulting matrix, $U=\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.
The matrix $L$ is $L=P\cdot P_0\cdot G_1^{-1}\cdot P_1\cdot G_2^{-1}$, or not? (Wondering)
The matrices $G_i^{-1}$ are defined as:
$$G_1^{-1}=\begin{pmatrix}1 & 0 & 0 \\\frac{2}{10} & 1 & 0\\\frac{1}{10} & 0 & 1\end{pmatrix} \ \text{ and } \ G_2^{-1}=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & -\frac{2}{9} & 1\end{pmatrix}$$ or not? (Wondering)
Are the matrices $P_i$ defined as follows?
$$P_0=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$$ since this describes the step at which we exchanged the first two rows and the first two columns. (Wondering)
$$P_1=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}$$ since this describes the step at which we exchanged the two last rows. (Wondering) If these are correct, it doesn't hold that $LU=PA$, does it? (Wondering)
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