5c- Solved problem for x-y-z values by LU decomposition.

Last Updated on December 17, 2025 by Maged kamel

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A solved problem for x-y-z values by LU decomposition for a 3×3 matrix.

A solved problem for x-y-z values by LU decomposition for a 3×3 matrix.

We have three simultaneous linear equations; we need to find x, y, and z using LU decomposition. In this example, we will use a C matrix as the product of the upper matrix U by the matrix of unknowns X.

We start by converting matrix A into U1 matrix by letting a21,a31 values are zero to constitute an upper matrix.

Reverse the signs of the estimated two factors to get the value of L21 as=(+4) and for L31 as=(+2). The l32 is unknown.

For the a33 since it is already 0, we need not create any multiplication. This zero will be placed as the l33 value in our created L1 lower matrix.

Solved problem for x-y-z values by LU decomposition.

The factors shown are (-a21/a11) and (-a11/a31). The first factor value=-(8/2)=-4, while the second factor=(-4/2)=-2.

The U1 matrix is formed for the first row (2 1 4) which are the values of u11,u12, and u13.

For the second row of the U1-matrix( 0 -7-14) which are the values for U21, U22 U23, the last row is(0 9 -9).

Step 2: Derive the expression for U2 and L2 matrices.

Convert a32 which is (9) to zero by multiplying by (-9/-7)R2 adds the value to R3, and we have used -7 as a second pivot, and the result which is (+9/7)-7+9=0 will be the value of u32, the corresponding L32 value is the reversed sign factor which is(-)*(9/7)=-9/7. The detailed calculations are shown in the next slide image.

The final shape of U2 and l2 will constitute the L and U matrices both of these matrices are 3×3 matrices, hence A can be set to be the product (L*U). A routine check of the product of L*u will produce the same elements of matrix A to be performed.

Step-3-Derive the expression for the inverse matrix L-1 matrix for the solved problem for x-y-z.

We will use the augmented matrix to get the inverse matrix L-1 value which will be (3×3 matrix). The process will continue till the left side of the matrix is an identity matrix and the right side of the augmented matrix will be the required inverse matrix of the lower matrix, denoted as Ll-1.

Derive the expression for L-1 matrix for the solved problem for x-y-z.

Step 5: Get the value of the C-matrix.

The C matrix is set to the multiplication of the lower matrix inverse by the matrix B, and expressed as =L-1*B, both matrices’ values are known.

The C matrix is a (3×1). The detailed calculations for multiplying rows by columns are shown in the next slide images.

We have U*x=C. We can write the matrices U and X, and equate their product to the matrix C.

Step 6- Get the value of the X-matrix for the solved problem for x-y-z.

C-matrix is =(12 -28 -27) it is 3×1 matrix. We have U*X = C, and the inverse matrix U^-1 needs to be estimated to obtain x = U^-1*C. So we can estimate the final values of x,y, and z, by multiplying by the inverse matrix of U, which is (U-1)

We will derive the expression for the inverse matrix -U-1, by using the augmented matrix method. The full details of the necessary calculations are shown in the following two slide images.

The steps for obtaining the inverse of the U matrix are shown in the next slide.

A step by step to get the inverse of U matrix.

The inverse of U is also an upper matrix, called U-1 and written as (1/2 1/14 1/27) for the first row, (0 -1/7 12/27) for the second row, and (0 0 -1/27) for the third row. The inverse of the Upper matrix is to be multiplied by the matrix C, to get the values of x,y, and z.

The product of multiplying inverse of U by the C matrix to get the unknown values

The last values for x, y, and z are for the solved problem in x-y-z.

The multiplication of the inverse of matrix u by matrix c is shown, and the final values for x, y, and z are (3, 2, 1). The routine check of the estimated values for x, y, and z will be performed to satisfy the given three equations. The calculations are shown in the last slide image.