## Solved problems for 3×3 matrix by Crout’s LU-option -1.

We have two solved problems for the 3×3 matrix by Crout’s LU decomposition. for which we will apply the same technique used to derive the elements of the lower matrix and the upper matrix.

## The first solved problem for the 3×3 matrix by Crout’s LU-Option-1.

The first problem for the 3×3 matrix is Crout’s LU decomposition. It is required to find both the Lower and upper matrix decomposition LU by using Crout’s method through a given matrix A. The given matrix is 3×3.

We will apply the previous post technique to get the L1 and U1 matrix components by using the following steps.

### Step-1 Convert matrix A to a lower matrix.

First, we have matrix A as (1 1 1, 4 3 -1, 3 5 3 ). We will estimate the determinant value of this matrix A, it is estimated as equal to +10. Since the determinant value is not equal to zero, then the matrix is invertible and we can proceed to find L and U matrices based on Crout’s method.

The three elements in the first column vector of the A matrix which are 1 4 3 will be set as L11 &L21 and L31. L11=1, L21=4,L31=3. While for U12, we will divide a21/a11=1/1=1. While U13=a13/a11=1.

As for the upper matrix U1, the u11=1, U12=a21/a11=1/1=1 and u13=a13/a11=1/1=1. The detailed procedures are shown in the slide image.

### Step-2-Derive the values for the L22, L23, and U23-option-1.

We will use a11 as a pivot, we multiply (-1)*(a12/a11)*column 1 and add the result to the 2^{nd} column, similarly, we multiply (-1)*(a13/a11)*column 1 and add the result to the 3rd column. If we call the new matrix L1, the elements of the matrix are as follows:

The first row is (1 0 0). The second row is (4 -1 -5). The third row is (3 2 0). If we compare the first row of L1 to the next matrix, we have (L11 0 0), then L11=1. If we compare the second row of L1 to the next matrix, we have (L21 L22 L22*U23), then L21=4.

If we compare the third row of L1 to the next matrix, we have (L31 L32 L32*U23+L33), then L31=3 and L32 =2.

### Step-3-Derive the value for the L33.

We want to proceed to get the final l matrix, we want to let -5 equal zero. We multiply (-1)*(-5/-1)*column 2 and add the result to the 3rd column. The final elements of the Lower matrix l will be as follows:

The first row is (1 0 0). The second row is ( 4 -1 -5*-1+(-5)). The third row is ( 3 2 -5*2+(0)).

### Step-4-Check the product of(L*U)=A-matrix for a 3×3 matrix.

In the final slide, we have all the elements in the first column of the matrix identical to the first column of the lower matrix. The diagonals of the U matrix U11=U22=U33=1. L22=-1. U23=+5 while L33=-10. U12=1 and u13=1.

We can write both the lower matrix and the upper matrix and check whether the multiplication of the lower matrix by the upper matrix will give us the final A matrix or not. Using the row-by-column multiplication, we can see that L*U multiplication will give the matrix A.

**The second solved problem-Crout’s LU-Option-1.**

First, we have matrix A as ( 10 3 4, 2 – 10 3, 3 2 -10 ). We will estimate the determinant value of this matrix A, which is estimated as equal to +1163 Since the determinant value is not equal to zero, then the matrix is invertible and we can proceed to find L and U matrices based on Crout’s method. The second solved problem, the detailed procedure to estimate the determinant is shown in the next slide image.

### Step-1 Convert matrix A to a lower matrix.

The three elements in the first column vector of the A matrix which are 10 2 3 will be set as L11 &L21 and L31. L11=10, L21=2,L31=3. While for U12, we will divide a21/a11=3/10=3/10. While U13=a13/a11=4/10.

### Step-2-Derive the values for the L22, L23, and U23.

We will use a11 as a pivot, we multiply (-1)*(a12/a11)*column 1 and add the result to the 2^{nd} column, similarly, we multiply (-1)*(a13/a11)*column 1 and add the result to the 3rd column. If we call the new matrix L1, the elements of the matrix are as follows:

The first row is (10 3 4). The second row is (2 -10 3). The third row is (3 2 -10). If we compare the first row of L1 to the next matrix, we have (L11 0 0), then L11=1. If we compare the second row of L1 to the next matrix, we have (L21 L22 L22*U23), then L21=2.

If we compare the third row of L1 to the next matrix, we have (L31 L32 L32*U23+L33), then L31=3 and L32 =.1. To get U23 divide ((2.2)/(-10.6)=-11/53.

### Step-3-Derive the value for the L33-Crout’s LU decomposition.

We want to proceed to get the final l matrix, we want to let 2.2 to be equal zero. We multiply (-1)*(2.2/-10.60)*column 2 and add the result to the 3rd column. The final elements of the Lower matrix l will be as follows:

The first row is (10 0 0). The second row is ( 2 -10.60 0). The third row is ( 3 1.1 -1163/106). As we can see L33 will be equal to -1163/106.

### Step-4-Check the product of(L*U)=A-matrix for a 3×3 matrix.

Using the row-by-column multiplication, we can see that L*U multiplication will give the matrix A.

The full detailed steps are shown in the previous slide images, at the end we will check the product of L*U against the Matrix A value.

This is a link to the next post is the introduction to the permutation matrix.

*References*

This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.

Another calculator to use is a Calculator for matrices.