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

Skilled2021-06-01Added 102 answers

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

$\Rightarrow A\sim \left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ -6& -3& 4\end{array}\right]L=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ & & 1\end{array}\right]$

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

$\Rightarrow A\sim \left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]L=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]$

Call the matrix obtained from A by applying row operations as U

Hence,

$\Rightarrow U=\left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]L=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]\left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]$

Step 3

Answer:

$\left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 3& -2& 1\end{array}\right]\left[\begin{array}{ccc}-2& 1& 2\\ 0& 3& 2\\ 0& 0& 2\end{array}\right]$

Given: Matrix

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:

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:

Now we want to get 0 in place of -6 in the first column

Apply row operation,

Call the matrix obtained from A by applying row operations as U

Hence,

Step 3

Answer: