Last Updated on August 20, 2026 by Maged kamel
First solved problem for a 3×3 matrix by Crout’s LU-option 2.
We have two solved problems for the 3×3 matrix using Crout’s LU (option 2). For which we will apply the same technique used to derive the elements of the lower and upper matrices. Option 2 is finding the upper matrix first.
We will start with the first solved problem.
The following slide shows the content of the first solved problem: how can we get L and U matrices using option 2, and also how can we use elementary matrices to find L and U matrices.
The first solved problem for the 3×3 matrix by Crout’s LU-Option-2.
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.
Step 1: Convert matrix A to an upper matrix (Option 2).
First, we have the matrix A as (1, 1, 1; 4, 3, -1; 3, 5, 3). We will estimate the determinant of this matrix A, which we set to +10. Since the determinant value is not equal to zero, the matrix is invertible, and we can proceed to find L and U matrices based on Crout’s method. The details for estimating the determinant are shown on the next slide.
The three elements in the first column of the A matrix, which are 1, 4, and 3, will be set as L11, L21, and L31. L11=1, L21=4, L31=3.
We can find the values of U12 and U13 by comparing the first row of matrix A with the element-wise product of L and U. U12 = +1 and U13 = +1.
The following slide shows the elements we need to find: L22, L32, L33, and U23.
Minus 4 multiplied by 1, plus 3, will give us minus 1. Again, minus 4 multiplied by 1, plus minus 1, will give us minus 5. Regarding the third row, we have minus 3. Added to two plus three will give us zero. Minus three added to five will give us plus two. Minus three by one is minus three; adding three will give us zero.
The second row will be minus 4 multiplied by 1. The detailed procedures are shown in the slide image.
Step-2:Find values of L22, L32, option-2.
We will use a11 as a pivot to eliminate a21 and a31 elements. These multipliers are -a21/a11 = -4, to be multiplied by R1 and added to R2. The second multiplier is -a31/a11 = -3, to be multiplied by R1 and added to R3. These are the two operations: the matrix becomes (1, 1, 1, 0, -1, -5, 0, 2, 0). Compared with the matrix shown on the next slide, we find that L22 = -1 and L32 = +2.
Step 3: Find the value of U23 – option 2.
We will use a22 as a pivot to eliminate the a32 element; the multiplier is +2, which is multiplied by R2 and added to R3. The matrix elements will be (1 1 1,0 1 5, 0 0 1). Compared with the matrix shown on the next slide, we find that U23 = +5.
Step 4: Derive the value for L33.
We want to proceed to obtain the final L and u matrices; we use a22 as a pivot and divide -a32 by a22 = +2. This multiplier is multiplied by R2 and added to R3. We compare the two matrices; we find that L33 =-10:
The first row is (1, 1, 1). The second row is (0, -1, -5). The third row is (0, 0, -10).
Step 5: Find the elements of a 3×3 U matrix.
On the next slide, to obtain the elements of the U matrix, we will divide the first row (R1) by 1, the second Row (R2) by -1, and the third row by -10 to set the diagonals of the U matrix to 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.
The following slide shows that the product of L by U gives the elements of Matrix A.
The PDF data for the first part can be viewed or downloaded from the next button.
Elementary matrices to do row operations-first solved problem.
The same steps we have done can be performed by using elementary matrices; the elementary matrix E1 to the right of matrix A will eliminate the a12 and a13 elements. While the elementary matrix E2 will eliminate the a23 element and be placed to the right of the A*E1 matrix, we finally obtain a Lower matrix. Please refer to the following slide image.
We can get the value of the E2 matrix; please refer to the following slide.
We can verify the U matrix by multiplying E3*E2*E1*A from left to right.
How can we get the inverse of the L matrix?
The product of E3 by E2 by E1 will give the inverse of L.
Check that L, when multiplied by its inverse, gives the identity matrix.
The PDF for the first solved problem can be viewed and downloaded from this link.
This is a link to the following post: 10-Two solved problems for 3×3 matrix by Crout’s LU-option 1.
This is the Omni calculator for estimating various linear algebra items – LU Decomposition Calculator.
Link to Omni calculator -LU Decomposition Calculator.
Another calculator to use is the Calculator for matrices.