Last Updated on October 21, 2026 by Maged kamel
First solved Problem: LU decomposition for the 3×3 Matrix.
I have created a video to accompany this Post.
The video timestamps are shown below.
- 00:00 Introduction and a summary of the content of the video.
- 01:12 Do matrices always have LU decomposition?
- 02:05 Verify whether the given Matrix is invertible or not.
- 04:12 Find the element values for the L Matrix.
- 05:40 Create an upper Matrix.
- 07:25 Find the Value of L32 and the final U Matrix.
- 08:10 The final L and U matrices.
- 08:40 Verify that the L and U values are correct.
- 11:35 The second option is to use elementary matrices for L and U values.
- 13:11 Create E1 and E2 matrices.
- 15:10 Find E3 matrix, the product of E2 by E1.
- 16:00 Prove that the Inverse of E3 is the lower Matrix.
- 16:43 A review of Minors and cofactor matrices for the Inverse of a Matrix.
- 18:35 Find cofactor values.
- 22:35 Find the lower Matrix.
- 23:08 Use an augmented Matrix to create the Lower Matrix as the Inverse of E3.
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. HELM is a valuable resource; please refer to Item #30—Introduction to Numerical Methods, article 30.3, which I used 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 that it is sometimes not possible to write the Matrix as a Lower triangular Matrix multiplied by an Upper triangular Matrix. Why not?
An invertible Matrix has an LU decomposition if all leading submatrices have nonzero determinants.
We will move to the following slide to apply this statement.
We check the sub-matrices’ determinants as follows: the first sub-matrix is the corner element a11; it has a Value of 3, which is nonzero. The second sub-matrix is A2, a nonzero 2×2 Matrix with elements (3, 1, -6, 0). The determinant is nonzero; it equals +6.
The Value of subnnonzeros should be nonzero to ensure the Matrix has 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, we need 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 from -6 to 3 give L21 = -3, and the second arrow from 0 to 3 gives L31 = 0/3.
The next step is to apply Gauss elimination to get the upper Matrix U1. Multiply L21*(-1)R1 and add it to the second Row.
Multiply L31 by (-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, 0, 0, 8, -17).
The next step is to find L32, the last unknown in the lower Matrix L.
Step 2: Derive the Expression for the final U and L matrices.
From the previous slide, we have U1. Use a22 as the pivot and divide a32 by a22 to find L32. Check the arrow from 8 to 2; dividing 8 by 2 gives 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 U1 and add it 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 check 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) · (3 0 0) = 3; this is a11.
We multiply Row 1 by column 2: (1 0 0) · (1 2 0) = 1; this 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) to get 0, which is a22.
We multiply Row2 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 as Matrix A.
The video includes a second method for obtaining the L and U matrices using Elementary matrices. In this Post, I haven’t included the slide images to keep it from getting too long.
I will Link to the PDF for this Post, which you can view or download by clicking the button below.
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.