Dottie Parra

2021-05-31

Compute the LU factorization of each of the following matrices.
$\left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]$

Arham Warner

Step 1
Given: Matrix
$A=\left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]$
To find: LU decomposition of the given matrix
Procedure: We are trying to find two matrices L and U such that A = LU
Here, L is a lower triangular matrix whose all diagonal entries are 1 and entries above the main diagonal are 0
Similarly, U is an upper triangular matrix whose all entries below the main diagonal are 0
Hence we start by finding U, by making entries below main diagonal 0 using row operations and then we will find L using opposites of the multiples used in building U
Note that L is of the form:
$L=\left[\begin{array}{ccc}1& 0& 0\\ & 1& 0\\ & & 1\end{array}\right]$
Step 2
Computation of U and L
We need to make entry in the first column below -2 as 0
Apply row operation to get second row of U: ${R}_{2}\to 2{R}_{1}+{R}_{2}$ and now the multiplier is 2 and so we get -2 below first main diagonal element of L
$⇒A\sim \left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ -6& -3& 4\end{array}\right]L=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ & & 1\end{array}\right]$
Now we want to get 0 in place of -6 in the first column
Apply row operation, ${R}_{3}\to 3{R}_{1}+{R}_{3}$ and now the multiplier is -3, hence we get 3 below the entry -2 in L
$⇒A\sim \left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]L=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]$
Call the matrix obtained from A by applying row operations as U
Hence,
$⇒U=\left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]L=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]$
$⇒\left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]\left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]$
Step 3
$\left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]\left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]$