## First solved problem-LU decomposition for the 3×3 matrix.

### The video I used in the illustration.

In the video, there is an introduction to the case for the matrix 3×3 has a matrix?

In the second part, there is a solved problem for a 3×3 matrix, for which it is required to get the Lower and upper matrix for it. The video has a subtitle and a closed caption in English.

### For the First solved problem-LU decomposition, check that the given matrix is invertible

First I will refer to one of the math sites which is, HELM, which is the abbreviation for helping engineers learn mathematics, there is a valuable source HELM, please refer to 30. Introduction to Numerical Methods, article 30.3, from which I have used the information to create this post.

You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.

Start with the first question, do all matrices have an LU decomposition? The answer is sometimes it is not possible to write the matrix in the form of a Lower triangular multiplied by the Upper triangular. Why not?

An invertible matrix has an LU decomposition provided that all the leading submatrices have non-zero determinants.

We will move to the second slide, to apply this statement.

We check the submatrices determinants to be nonzeros, as follows: the first submatrix is A1 which is the corner element of the matrix simply a11, it has a value of 3 which is not a zero value. The second submatrix is A2, which is a matrix of 2×2, elements are( 3 1 -6 0) the determinant value is not zero it equals +6.

The value of sub-matrices should be non-zero to check that the matrix will have Lu decomposition.

Finally, the entire matrix which is called A3, its determinant is -6 not zero. Since all the submatrices have fulfilled the requirements, then the given matrix can have LU decomposition.

### Step-1-Develop an expression for U1 and L1 matrices.

We will proceed to estimate L and U matrices for the given 3×3 matrix. For the lower matrix L by definition, we have L11=L22=L33=1, then it is required to find the values of L21 and L31.

L21=a21/a11 while L31=a31/a11, consider a11 as a pivot. We can see the two arrows going from -6 to 3 will give -3 for L21 and the second arrow from 0 to 3, then l31= 0/3.

The next step is to apply gauss elimination to get the upper matrix U1. Multiply the value of L21*(-1)R1 and add it to the second Row.

Multiply the value of L31*(-1)R1 and add it to the third Row. We can see the matrix u1 as a 3×3 matrix that has the first row of (3 1 16 ) which is identical to the first row of the original matrix. The second row is ( 0 2 -4). The

third row is (0 8 -17).

The next step is to find L32 which is the last unknown for the lower matrix L.

### Step-2-Derive the expression for final U and L matrices.

From the previous slide we have U1, consider a22 as a pivot and divide a32/a22 to get the value of l32, check the arrow which goes from 8 to 2, the division of 8/2 is equal to 4 which is the value of L32.

Now we have completed the elements of the lower matrix. These elements are ( 1 0 0 , -2 1 0, 0 4 1).

We proceed to find the final upper matrix, multiply -l32*R2 and add it to R3, then we multiply -4 to the second row of U1 matrix and add the value to the third row.

The final Upper matrix is ( 3 1 6, 0 2 4, 0 0 -1).

### Step-3-Check LU=A for the 3×3 matrix.

Now, we would like to check the expression of L*U=A. We will perform row-column operations of the multiplication of L by U matrices.

We multiply row 1 by column 1 (1 0 0 ) by (3 0 0) will give the value of 3 this is a11.

We multiply row 1 by column 2 (1 0 0 ) by (1 2 0) will give the value of 1 which is a12. We multiply row 1 by column 3 (1 0 0 ) by (6 -4 -1) will give the value of 6 which is a13. We multiply row 2 by column 1 (-2 1 0 ) by (3 0 0) will give the value of -6 this is a21. We multiply row 2 by column 2 (-2 1 0 ) by (1 2 0) will give the value of 0 which is a22.

We multiply row 2 by column 3 (-2 1 0 ) by (6 -4 -1) will give the value of -16 which is a23.

We multiply row 3 by column 1 ( 0 4 1) by (3 0 0) will give the value of 0 this is a31. We multiply row 3 by column 2 ( 0 4 1) by (1 2 0) will give the value of 8 which is a32.

Finally, we multiply row 3 by column 3 ( 0 4 1) by (6 -4 -1) will give the value of -17 which is a33. The final Upper matrix is ( 3 1 6, 0 2 4, 0 0 -1).

The matrix A can be set to be the product (L*U).

A routine check of the product of L*U will produce the same elements of matrix A to be performed.

This is the pdf file used for the illustration of this post and the next post.

This is the next post: The Second solved problem– LU decomposition for the 3×3 matrix.

HELM-Helping Engineers Learn Mathematics.

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

This is a link to another –Calculator for matrices.