Last Updated on October 12, 2026 by Maged kamel
A Second option for Crout’s LU decomposition: 3×3 Matrix.
In this Post, we will discuss how to get the elements of the lower and upper matrices using Crout’s method for the 3×3 Matrix.
We have two options for finding the values of both the lower and upper matrices: The first option we discussed in the previous Post; the second option is to convert Matrix A to an upper Matrix, then we can get all the elements of the upper Matrix, and besides, we can get the values of L22, L32, and L33.
The slide shows the details of equating Matrix A to the Product of the lower Matrix L and the upper Matrix U.
Step 1: Equate the Product of L*U to the A Matrix.
We have three equations, and we want to develop Crout’s LU decomposition for a 3×3 Matrix.
The first equation is (a11*x1+a12*x2+a13*x3=b1). The second equation is (a21*x1+a22*x2+a23*x3=b2).
The third equation is (a31*x1 + a32*x2 + a33*x3 = b3).
We can express these equations as A*x = b. The next task is to find the L and U matrices using Crout’s LU decomposition for a 3×3 Matrix, Option 2.
The matrix A which is(a11 a12 a13,a21 a22 a23, a31 a32 a33). Consider the lower matrix L as (L11, 0, 0; L21, L22, 0; L31, L32, L33), which is a 3×3 matrix.
Using the Expression L*U = A, we can derive a new 3×3 Matrix that contains the elements of both the lower and upper matrices.
If we write A as a 3×3 Matrix with elements (a11 a12 a13, a21 a22 a23, and a31 a32 a33), then A equals the Product of L and U.
According to Crout’s method, all the diagonals of Matrix U are Ones. The left-hand side is Matrix A, while the right-hand side is L*U.
Step-2-Derive the expression for L11, L12, L13 & U12, U13.
The second slide shows that L11, L21, and L31 are equal to a11, a12, and a13. While a1U12=a21/L11 and since L11 equals a11, we can derive the Expression of u12 as equal to (a12/a11), and similarly, U13 is equal to a13/a11.
As a quick reminder of the values of L11, L21, and L31, copy the entire first column of Matrix A and place it in the first column of the Lower Matrix.
Divide a12 by a11 to get 12, and divide a13 by a11 to get. Please refer to the image on the next slide.
Step 3-A: Second option for Crout’s LU decomposition- 3×3 Matrix.
Option 2 converts Matrix A to an upper-triangular Matrix, as shown on the slide. We want a21 and a31 to be zeros; we set a11 as a pivot, then multiply the first Row by (-a21/a11) and add the result to the second Row.
. The new Matrix if we call it A1, has the first column as (a11 0 0), the second column elements are (a12 ((a22-a21/a11*(a12)) (( a32 -a21/a11*(a12)). The third column elements are (a13 ((a23-a21/a11*(a13)) (( a32 -a21/a11*(a12)). For the third Row/ third column, the element will be set equal to L31U13+L32U23-U13L31+L33=L23U23+L33.
Step-4- How to find L22, and L32?
If we look at the second Row / second column element of Matrix A, we have (a22=L21*U12+L22), so we can get a Value for L22 equal to (a22-L21*U12), or L22=a22-a21*a12/a11.
The third Row/second column element of Matrix A is a32; we have (a32=L31*U12+L32), so we can solve for L32: L32=(a32-L31*U12) or L32=a32-a31*a12/a11.
If we refer back to Matrix A1, we find that the second Row /the second column, which is ((a22-a21/a11*(a12)), can be set equal to L21.
The thirdRoww/ the second column, ((a23-a21/a11*(a13), can be set equal to L32. If we refer back to Matrix A1, we find that the second Row/third column, ((a32 – a21)/a11*(a12)), can be set equal to L22*U23.
The third row/third column, ((a32 – a21)/a11*(a12)) can be equal to L32*U23 + L33.
Step-5- How to find L33, U23& U12, and U13?
We move to the next slide and check the final terms of Matrix A, from which we can get the values of L22 and L32. We proceed to get
Matrix U1 by letting L32=0. We use L22 as a pivot, multiply the second Row by (-1)*L32/L22, and add the result to the third Row.
The Final upper Matrix contains the following: as the first Row, (L11 U12*L11 U13*L11); as the second Row, (0 L22 L22*U23); and as the third Row, (0 0 L33).
To get U12 and U13, divide the first Row/second column element by L11, and the first Row/third column element by L11. To get U23, divide the second Row/3rd column element by L22.
A summary of the steps we took to get the L22 and L23 elements.
A summary of the steps we took to get L33 & U12 and U13.
The PDF data for this Post can be viewed and downloaded from the following document.
For a 2×2 Matrix using Crout’s method, see Post 8: Introduction to Crout’s LU decomposition for a 2×2 Matrix.
The following Post presents solutions to problems involving a 3×3 Matrix using Crout’s LU option 1.
This is the Omni calculator for estimating various linear algebra items:L U Decomposition Calculator.
This is a Link to another Matrix calculator.