How to derive the expression for L-U values for-3×3 matrix?
This is an explanation for the terms for both the upper and lower L-u value for the 3 x 3 matrix.
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, we have three diagonals a11=a22=a33=1.
While for a12, the element n 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, 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 by the value of matrix A for 3 x 3 matrix.
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 matrix 3×3 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 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 values of L21, and L31 concerning values and also the value of U22 as shown in the next slide image.
5- We can get values of U23, and L32 for values as shown in the next slide image.
The product of LU is equated to matrix A. The product of the LU matrix of 3×3 is written in symbols. From the equating of both Lu and A, we can find out the values of U11=a11, U22= a22, and U33=a33.
To create an upper matrix we need to let L21*U11=0 this product is equal to a21, to get the L21*U11 to be zero, we will subtract this value from a21. The details of the subtraction are shown in the slide image.
To create an upper matrix we need to let L31*U11=0 this product is equal to a31, to get the L21*U11 to be zero, we will subtract this value from a31. The details of the subtraction are shown in the slide image.
The details of the subtraction are shown in the slide image.
I have written the A matrix in the form aij, the process of subtraction for both a21 and a31 is shown in the next slide image. We can get the value of L21=a21/a11. The value of L31=a31/a11.
To get the expression for L-U elements with regards to the value of U22, we will subtract (L21*U12+U22) from a22, but we will substitute the value of L21 as a21/a11, while U12=a12,.
To get the values of U23, we will subtract (L21*U13+U23) from a23, but we will substitute the value of L21 as a21/a11, while U13=a13.
How to find the elements of the lower matrix L1 is shown in the next slide image where we divide a11/a12 and set it as equal to L21 and divide (a11) by a13 and set the division =L31.The equations used for the estimations are shown in the next slide image.
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.
Thus we have come to the end of the method to derive the expression for L-U values for 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 pdf file used for the illustration of this post.
This is the next post which is the First solved problem-LU decomposition for the 3×3 matrix.
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