Last Updated on October 21, 2026 by Maged kamel
Two solved problems for a 3×3 Matrix using Crout’s LU option -1.
We have two solved problems for the 3×3 Matrix using Crout’s LU decomposition. For these, we will apply the same technique used to derive the elements of the lower and upper matrices. Option 1 is to find the lower Matrix first.
The following video covers the first solved Problem.
The first solved Problem for the 3×3 Matrix: Crout’s LU decomposition (Option 1) 1.
The first Problem is to perform Crout’s LU decomposition on a 3×3 Matrix. 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 obtain the L1 and U1 Matrix components as follows.
Step 1: Convert Matrix A to a lower Matrix (Option 1).
First, Matrix A is (1, 1, 1; 4, 3, -1; 3, 5, 3). We estimate the determinant of Matrix A to be +10. Since the determinant is not zero, the Matrix is invertible, and we can proceed to find the L and U matrices using Crout’s method.
The first-column elements of A, 1, 4, and 3, are set as L11, L21, and L31. L11=1, L21=4,L31=3. For U11=a11=1. U12, we will set a12/a11 = 1/1 = 1. While U13=a13/a11=1.
The detailed procedures are shown in the slide image.
Step-2-Derive the values for 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 2nd 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). Comparing the first Row of L1 to the following Matrix gives (L11 0 0), so L11 = 1. Comparing the second Row of L1 to the next Matrix gives (L21 L22 L22*U23), so L21 = 4, L22 = -1, and L22*U23 = -5.
If we compare the third row of L1 to the next matrix, we have (L31 L32 L32*U23 + L33), so L31 = 3 and L32 = 2 and (L32*U23+L33)=0.
Step 3: Derive the Value for L33.
To get the final L Matrix, we set -5 equal to zero. We use L22 as a pivot. Multiply column 2 by (-1)*(-5/-1) 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) + 1) + (-5). The third Row is (3, 2, -5*2 + (0—the Value of L33 = -10)). Please refer to the slide image below.
Step 4: Check the Product of (L*U)=A-matrix for a 3×3 Matrix.
As the final verification step in our series, we multiply L and U to confirm that it reconstructs our original Matrix.
On the next slide, all elements in the first column of the Matrix match those in the first column of the lower Matrix. The diagonals othe matrixix arU11=1, L22=-1, U23=+5, and L33=-10 1. L22=-1. U23=+5 while L33=-10. U12=1 and u13=1.
We can write both the lower and upper matrices and check whether multiplying the lower by the upper yields the final A Matrix. Using row-by-column multiplication, we can see that L*U equals A.
Elementary matrices for Row operations, first solved Problem.
We can perform the same steps using elementary matrices; the elementary Matrix E1 to the right of Matrix A will eliminate the a12 and a13 elements. The elementary Matrix E2 will eliminate the a23 element and be placed to the right of the A*E1 Matrix; we then obtain a Lower Matrix. Please refer to the slide image below.
We can use the matrices A*E1*E2=L to find the inverse of the upper Matrix U. If we multiply both sides by L-1, we get E1*E2=U-1.
We can verify that U multiplied by U-1 equals I using the U data we estimated earlier.
You can view and download the PDF for the first solved Problem from this Link.
The second solved Problem: Crout’s LU (Option 1).
Convert Matrix A to a lower Matrix (Option 1).
The following video covers the first solved Problem.
First, Matrix A is (10 3 4, 2 -10 3, 3 2 -10). We estimate the determinant of Matrix A to be +1163. Since the determinant is not zero, the Matrix is invertible, and we can proceed to find the L and U matrices using Crout’s method.
The second solved Problem, the detailed procedure for estimating the determinant, is shown in the next slide.
Step 1: Convert Matrix A to a lower triangular Matrix.
The three elements in the first column of Matrix A, 10, 2, and 3, will be set as L11, L21, and L31. L11=10, L21=2,L31=3. For U12, we will divide a21/a11 = 3/10. While U13=a13/a11=4/10.
Step 2: Derive the values for L22, L23, and U23.
We will use a11 as a pivot, we multiply (-1)*(a12/a11)*column 1 and add the result to the 2nd 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, -1, 0, 3). The third Row is (3, 2, -10). Comparing the first Row of L1 to the next Matrix gives (L11 0 0), so L11 = 1. If we compare the second Row of L1 to the following Matrix, we have (L21 L22 L22*U23=(2 ,-10.6, 8.20).L21=2, L22=-10.6, L22*U23=2.20.
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 L33 in Crout’s LU decomposition.
To obtain the final L Matrix, we set 2.2 to zero. We multiply (-1)*(2.2/-10.60) by column 2 and add the result to column 3. The final elements of the Lower Matrix L will be as follows:
The first Row is (10, 0, 0). The second Row is (2, -10.6, 0, 0). The third Row is (3, 1.1, -1163/106). As we can see, L33 equals -1163/106.
Step 4: Check the Product of(L*U) = A for a 3×3 Matrix.
Using row-by-column multiplication, we can see that L*U equals A.
The complete detailed steps are shown in the previous slide images. At the end, we will check the Product L*U against the Matrix A.
Elementary matrices for Row operations, second solved Problem.
We can perform the same steps using elementary matrices; the elementary Matrix E1 to the right of Matrix A will eliminate the a12 and a13 elements. The elementary Matrix E2 will eliminate the a23 element and be placed to the right of the A*E1 Matrix; we then obtain a Lower Matrix. Please refer to the slide image below.
Use elementary matrices to find the inverse of Matrix U, since E1*E2 = U^-1.
Verify that U*U-1=I.
You can view and download the PDF for the second solved Problem from this Link.
This is a Link to the next Post, 6b- Practice problems for permutation matrices.
References
This is the Omni calculator for estimating various linear algebra items – LU Decomposition Calculator.
This is a Link to another –Calculator for matrices.