Question

For each of the following linear systems find a permuted $L U$ factorization of the coefficient matrix and then use it to solve the system by Forward and Back Substitution. $$ \begin{aligned} 4 x_1-4 x_2+2 x_3 & =1, \\ -3 x_1+3 x_2+x_3 & =3, \\ -3 x_1+x_2-2 x_3 & =-5 \end{aligned} $$ (a) $-3 x_1+3 x_2+x_3=3$, $$ y-z+w=0, $$ $$ y+z=1, $$ (b) $$ \begin{aligned} & x-y+z-3 w=2, \\ & x+2 y-z+w=4 . \end{aligned} $$ $$ \begin{aligned} x-y+2 z+w & =0, \\ -x+y-3 z & =1, \end{aligned} $$ (c) $$ \begin{aligned} x-y+4 z-3 w & =2, \\ x+2 y-z+w & =4 . \end{aligned} $$ 0\end{array}\right)$. Do you always obtain the same result? Explain why or why not. 1.4.24. (a) Find three different permuted $L U$ factorizations of the matrix $A=\left(\begin{array}{rrr}0 & 1 & 2 \\ 1 & 0 & -1 \\ 1 & 1 & 3\end{array}\right)$. (b) How many different permuted $L U$ factorizations does $A$ have?

   For each of the following linear systems find a permuted $L U$ factorization of the coefficient matrix and then use it to solve the system by Forward and Back Substitution.
$$
\begin{aligned}
4 x_1-4 x_2+2 x_3 & =1, \\
-3 x_1+3 x_2+x_3 & =3, \\
-3 x_1+x_2-2 x_3 & =-5
\end{aligned}
$$
(a) $-3 x_1+3 x_2+x_3=3$,
$$
y-z+w=0,
$$
$$
y+z=1,
$$
(b)
$$
\begin{aligned}
& x-y+z-3 w=2, \\
& x+2 y-z+w=4 .
\end{aligned}
$$
$$
\begin{aligned}
x-y+2 z+w & =0, \\
-x+y-3 z & =1,
\end{aligned}
$$
(c)
$$
\begin{aligned}
x-y+4 z-3 w & =2, \\
x+2 y-z+w & =4 .
\end{aligned}
$$
0\end{array}\right)$.
Do you always obtain the same result? Explain why or why not.
1.4.24. (a) Find three different permuted $L U$ factorizations of the matrix $A=\left(\begin{array}{rrr}0 & 1 & 2 \\ 1 & 0 & -1 \\ 1 & 1 & 3\end{array}\right)$. (b) How many different permuted $L U$ factorizations does $A$ have?
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Applied Linear Algebra (Undergraduate Texts in Mathematics)
Applied Linear Algebra (Undergraduate Texts in Mathematics)
Peter J. Olver,… 2nd Edition
Chapter 1, Problem 22 ↓

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For each of the following linear systems find a permuted $L U$ factorization of the coefficient matrix and then use it to solve the system by Forward and Back Substitution. $$ \begin{aligned} 4 x_1-4 x_2+2 x_3 & =1, \\ -3 x_1+3 x_2+x_3 & =3, \\ -3 x_1+x_2-2 x_3 & =-5 \end{aligned} $$ (a) $-3 x_1+3 x_2+x_3=3$, $$ y-z+w=0, $$ $$ y+z=1, $$ (b) $$ \begin{aligned} & x-y+z-3 w=2, \\ & x+2 y-z+w=4 . \end{aligned} $$ $$ \begin{aligned} x-y+2 z+w & =0, \\ -x+y-3 z & =1, \end{aligned} $$ (c) $$ \begin{aligned} x-y+4 z-3 w & =2, \\ x+2 y-z+w & =4 . \end{aligned} $$ 0\end{array}\right)$. Do you always obtain the same result? Explain why or why not. 1.4.24. (a) Find three different permuted $L U$ factorizations of the matrix $A=\left(\begin{array}{rrr}0 & 1 & 2 \\ 1 & 0 & -1 \\ 1 & 1 & 3\end{array}\right)$. (b) How many different permuted $L U$ factorizations does $A$ have?
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Key Concepts

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LU Factorization
LU Factorization is the process of decomposing a given square matrix into the product of a lower triangular matrix (L) and an upper triangular matrix (U). This decomposition simplifies the solution of linear systems, as it allows the system to be solved in two stages: first solving Ly = b by forward substitution and then solving Ux = y by back substitution. LU Factorization is a fundamental concept in numerical linear algebra and aids in understanding matrix operations, inverses, and determinants.
Permuted LU Factorization
Permuted LU Factorization involves including a permutation matrix (P) in the factorization so that the original coefficient matrix is expressed as PA = LU. Permutations reorder the rows of the matrix to place a suitable pivot element on the diagonal, which is critical when the original matrix has zero or near?zero pivot elements. This technique not only ensures the existence of the LU decomposition but also enhances numerical stability during computations.
Pivoting
Pivoting is the process of rearranging the rows (or columns) of a matrix to position a suitable nonzero (or large in magnitude) element as the pivot during the elimination process. In the context of LU factorization, pivoting helps avoid division by zero and minimizes numerical errors, leading to a more robust solution method, especially when dealing with ill-conditioned matrices.
Forward and Back Substitution
Forward and back substitution are the sequential processes used to solve the triangular systems that arise from LU or permuted LU factorizations. Forward substitution is applied to the lower triangular system (Ly = b) where the solution is computed sequentially for each equation from top to bottom. Back substitution is then used on the upper triangular system (Ux = y), solving the equations from bottom to top. This two-step approach simplifies solving linear systems that have been decomposed into triangular matrices.
Uniqueness of LU Factorizations
The uniqueness of LU factorizations is influenced by the choices made during the factorization process, such as the order of pivoting and the scaling of rows. Different row permutations can lead to different LU factorizations, even though they ultimately yield the same solution for a linear system if performed correctly. This phenomenon is important when analyzing algorithmic stability and the sensitivity of numerical solutions to the details of the factorization process.

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