Find an LU factorization of $A=\left[\begin{array}{cccc}h& -4& -2& 10\\ h& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& \frac{h}{2}\end{array}\right]$ .

h=102

h=102

Rui Baldwin
2021-02-13
Answered

Find an LU factorization of $A=\left[\begin{array}{cccc}h& -4& -2& 10\\ h& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& \frac{h}{2}\end{array}\right]$ .

h=102

h=102

You can still ask an expert for help

yagombyeR

Answered 2021-02-14
Author has **92** answers

Step 1

$A=\left[\begin{array}{cccc}102& -4& -2& 10\\ 102& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& 51\end{array}\right]$

In this problem, we have to find matrices L(lower triangular) and U(upper triangular) for which A=LU

Initial Matrix:$\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}102& -4& -2& 10\\ 102& -9& 4& 2\\ 0& 0& -4& 2\\ 0& 1& 4& 4\\ 0& 0& 0& 51\end{array}\right]$

Using some row transformations we convert this initial matrix into LU factorization,

where L will the matrix having all diagonal entries 1, and entries above diagonal 0.

Step 2

by some row transformations, we get the matrix L and matrix U

$L=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 1& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& -\frac{1}{5}& -\frac{13}{10}& 1& 0\\ 0& 0& 0& \frac{51}{5}& 1\end{array}\right]$

$U=\left[\begin{array}{cccc}102& -4& -2& 10\\ 0& -5& 6& -8\\ 0& 0& -4& 2\\ 0& 0& 0& 5\\ 0& 0& 0& 0\end{array}\right]$

If we multiply L and U, we will get the same matrix A.

In this problem, we have to find matrices L(lower triangular) and U(upper triangular) for which A=LU

Initial Matrix:

Using some row transformations we convert this initial matrix into LU factorization,

where L will the matrix having all diagonal entries 1, and entries above diagonal 0.

Step 2

by some row transformations, we get the matrix L and matrix U

If we multiply L and U, we will get the same matrix A.

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