5A- Second solved problem-LU decomposition-3×3 matrix.

The second solved problem- LU Decomposition-3×3 matrix.

Step-1- Develop U1 and L1 matrices for the second solved problem-LU decomposition

This is the second solved problem for Lu decomposition is required to get (x₁,x₂, x₃) values of three linear simultaneous equations. matrix A is 3×3 and matrix b is the (3×1) matrix.

The first check is conducted for all sub-matrices values A1 A2, and A3, that their values are nonzeros as we can see that A1=1 and A2=+2 and A3=+6.

The matrix A is invertible and has L and U decomposition.

To get the Lower Matrix, we write Matrix A as shown in the slide. We use a11 as a pivot; we know that for the lower Matrix, L11=L22=L33=1. We need to find out the value of L21 and L31.
For L21 it is equal to 3/1, while l31 is equal to 2/1; please check the arrows.

The elements of L1 are completed. Still, we need to estimate the value of L32, but this depends on the pivot u22 of the matrix U1.

We proceed to find the matrix u1. The first three elements in the first Row of the U1 Matrix are the same as the first Row of matrix A ( 1 2 4).

We change the sign of L21 by multiplying by -1* the first Row and adding the value to the second Row. We change the sign of L31 by multiplying by -1* the first Row and adding the value to the third Row. We can see the elements of the matrix U1 as ( 1 2 4, 0 2 2, 0 2 5).

Step-2-Develop U and L matrices for the second solved problem-LU decomposition

The next step is finding L32, the last unknown for the lower matrix L. The elements of L1 matrix are (1  0  0, 3 1 0,2 L32 1).

From the previous slide, we have U1, consider u22 as a pivot divide U32/U22 to get the value of l32, and check the arrow which goes from 2 to 2, the division of 2/2 is equal to 1, which is the value of L32.

Now we have completed the elements of the lower Matrix. These elements are (1 0 0, 3 1 0, 2 1 1).

Step-3-Develop a new matrix C.

We have L and U matrices written as shown. We know that LU=A, for the expression A*x=b, we can rewrite it as (L)*(U)*x=b, and we can write U*x=c.

The final expression will be LC=b, for l and b are known values; hence we can get the value of matrix C, C=L-1*b, but we need to estimate the inverse of matrix L.

Step-4- Get the inverse matrix L-1.

For the second solved problem-LU decomposition, the determinant value of L=+1.We estimate the cofactor elements and consider the inverse equals adjugate/ determinant.

The adjugate of the lower Matrix is the cofactor transpose. The cofactor C11 is equal to +1; the cofactor C12 is equal to -3. The cofactor C13 is equal to +1; cofactor C21 is equal to 0.

The cofactor C22 is equal to +1; cofactor C23 is equal to -1. The cofactor C31 is equal to 0; cofactor C32 is equal to 0. The cofactor C33 is equal to 1.

Step 4-Get the value of the C matrix.

The inverse of matrix L is a 3×3 matrix with the first Row as ( 1  0 0). The Second Row is (-3 1 0), and the last is (1 -1 1).
We can estimate the C matrix by multiplying L-1 by b.

The product of the inverse of L by b will be (3×1) Matrix with the elements (3 4 -6). But U*X=c. We need to estimate the inverse of matrix U to get the column vector X.

Step 5-Get the (x₁, x₂, x₃) values for the 3 x 3 matrix.

The inverse of Matrix U is a 3×3 matrix with the first Row as ( 1  -1 -2/3). The Second Row is (0 ½- 1/3), and the last Row is (0 0 1/3). The product of the inverse of U by c will give the value of the column vector X. X1=3, x2=4, x3=-2.

It is essential to check that the value of the three unknowns is correct; we double-check by multiplying A*x and check that the product is equal to vector b, which can be found valid.

We can get the value of the X column vector by multiplying U-1L-1b to get the values of x1,x2, and x3 as shown in the next slide image.

This is the pdf file used for the illustration of this post.

The next post: Solved problem for x-y-z values by Lu decomposition.

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