Last Updated on December 14, 2025 by Maged kamel
First solved problem-LU decomposition for the 3×3 matrix.
I have created a video for the content of this post.
For the First solved problem, LU decomposition, check that the given matrix is invertible.
First, I will refer to one of the math sites, HELM, which stands for helping engineers learn mathematics. There is avaluable source HELM, please refer to Item# 30. Introduction to Numerical Methods, article 30.3, from which I have used the information to create this post. We can use the term LU factorization. This solved problem is the LU factorization of a 3×3 matrix.
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 matrix multiplied by an Upper triangular matrix. 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 sub-matrices’ determinants to be non-zero, as follows: the first sub-matrix 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 sub-matrix is A2, which is a 2×2 matrix with elements (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 an LU decomposition.
Finally, the entire matrix, A3, has a determinant of -6, not 0. Since all the submatrices satisfy the requirements, the given matrix can be decomposed into an LU factorization.
Step 1: Develop an expression for the U1 and L1 matrices.
We will proceed to estimate the L and U matrices for the given 3×3 matrix. For the lower matrix L, by definition, we have L11 = L22 = L33 = 1. Therefore, 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 that 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 view the matrix u1 as a 3×3 matrix with the first row (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 the final U and L matrices.
From the previous slide, we have U1. Consider a22 as a pivot and divide a32 by a22 to get the value of L32. Check the arrow that goes from 8 to 2; the division of 8/2 is equal to 4, which is the value of L32.
We have now 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 by multiplying -l32*R2 and adding it to R3. Then, we multiply -4 by the second row of the 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 on the product of the L and U matrices.
We multiply row 1 by column 1 (1 0 0) by (3 0 0), which gives 3; this is a11.
We multiply row 1 by column 2 (1 0 0) by (1 2 0), which gives the value of 1, which is a12.
We multiply row 1 by column 3 (1 0 0) by (6 -4 -1), which gives a13 = 6. We multiply row 2 by column 1 (-2 1 0) by (3 0 0), which gives -6. This is a 21.
We multiply row 2 by column 2 (-2 1 0) by (1 2 0), which gives 0, which is a22.
We multiply row 2 by column 3 (-2 1 0) by (6 -4 -1) to get -16, which is a23.
We multiply row 3 by column 1 (0 4 1) by (3 0 0), which gives 0. This is a31. We multiply row 3 by column 2 (0 4 1) by (1 2 0), which gives 8, which is a32.
Finally, we multiply row 3 by column 3 (0 4 1) by (6 -4 -1), which gives the value -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.
The video includes a second method to get L and U matrices by using Elementary matrices. In this post, I have not included the images for the slides to keep it from being too long.
I will put a link to the PDF document of this post.
This is the following 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 linear algebra items – LU Decomposition Calculator.
LU Decomposition Calculator – eMathHelp
https://www.emathhelp.net/calculators/linear-algebra/lu-decomposition-calculator/
This is a link to another Matrix calculator, –Calculator for matrices.