Last Updated on September 17, 2025 by Maged kamel
How to derive the expression for L-U values for-3×3 matrix?
I have created a video to illustrate the content of this post.
This is an explanation for the terms for both the upper and lower L-u values for the 3 x 3 matrix-Doolittle’s method.
The matrix A of 3×3 is represented by(aij) where i is the row number and j is the column number.
The matrix a 3×3 can be represented by the multiplication of the Lower triangular matrix L( with 3 rows and 3 columns), which is to be multiplied by the upper triangular matrix (3×3) U.
For the terms of the L matrix, we use the notation aij as well. For that L matrix, the diagonal is =1, and we have three diagonals a11=a22=a33=1.
While for a12, the element in the first row at the second column =a13=0. Below the diagonal, all remaining elements are nonzeros.
On the contrary, the upper triangular matrix has U11, U12, and U13, and all the diagonals are non-zero, while all elements below the diagonal represented by U11, U22, and U33 are all zeros.
Step-1: Equating the product of L-U values for-3×3 matrix by the value of matrix A.
Since matrix A is the product of LU matrices, we can equate the value of the elements of A with the product of LU for the corresponding items.
The product of L*u is represented by a 3×3 matrix, as we can see in the next slide image. The first row contains (U11, U12, and U13), which are equal to the first row of matrix A, hence we can proceed to the next step.
Step-2 Get the expression for L-U elements for the upper and lower matrix for a 3 x 3 matrix.
1-Equate the first element of (L*U) decomposition to the first element a11 of the matrix A as follows: U11=a11. Similarly, we can equate the second and the third element to the corresponding elements of matrix A.
So U12=a12 and U13=a13.
2-We will move to the second row of the L*U matrix to get the expression for L-U values.
3- We equate L21*U11 to a21, also, L21*U12+U22 to a22, and L21*U13+U23 to a23, we will use the derived values for U11 as a11, U12 as a12 from which we have estimated from step 1.
4- We can get the values of L21 and L32.
We equate L21*U12+U22 to a22, we substitute the values of u12=a12, L21 as equal to a21/a11.
5- We can get U22, and U23 as shown in the next slide image.
The value of L32 is indicated in the following slide image in more detail.
To get the value of U33, we equate L31*U13+L32*U23+U33 to a33, and we use the values that we have estimated to get the expression of U33. Please refer to the following slide image.
The following slide shows the component of the lower matrix L, we need to create what is called U1 matrix to find the value of L32.
We need to create what is termed as U1, matrix; we multiply the second row of matrix A by -a21/a11*R1 and add to the second row. We multiply the third row of matrix A by -a31/a11*R1 and add to the third row. In that matrix, the U21=U31=0.
The element u22=a22-a21/a11*a12, this is the povot element we need to get the value of L32.
The next slide image shows how to get the value of L32. It is equal to ((a32-(a31/a11)*a12/a22-a21*(a12/a11). We take U21 as a pivot.
The following slide shows how we can get the value of U33.
Thus, we have come to the end of the method to derive the expression for L-U values for an A 3×3 matrix.
In the next post, we will use a solved problem to apply this method. A discussion on whether all matrices will have LU decomposition or not.
This is the next post, which is the First solved problem-LU decomposition for the 3×3 matrix.
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