Last Updated on October 10, 2026 by Temp User
Second solved Problem for a 3×3 Matrix by Crout’s LU (Option 2).
We will start with the Second solved Problem. The following slide shows the steps that we use for the Second solved Problem: how to find the L and U matrices using Option 2, and how to use elementary matrices to find L and U.
The Second solved Problem is a 3×3 Matrix using Crout’s LU decomposition (Option 2).
The Second Solved Problem for the 3×3 Matrix is Crout’s LU decomposition. It is required to find both the Lower and upper Matrix decomposition (LU) using Crout’s method for the given Matrix A. The given Matrix is 3×3.
Step 1: Convert Matrix A to an upper Matrix (Option 2).
First, Matrix A is (10, 3, 4; 2, -10, 3; 3, 2, -10). We estimate the determinant of Matrix A as +1163. Since the determinant is not zero, the Matrix is invertible, and we can proceed to find the L and U matrices using Crout’s method. The details for estimating the determinant are shown on the next slide.
The three elements in the first column of the A Matrix, which are 1, 4, and 3, will be set as L11, L21, and L31. L11=10, L21=2, L31 =3 by comparing the given Matrix A with the Product of the L and U matrices.
We can find U12 and U13 by comparing the first Row of Matrix A with the element-wise Product of L and U. U12 = +3/10 and U13 = +4/10.
We use a11 as the pivot and calculate the multipliers a21/a11 and a31/a11. We change their signs to eliminate a21 and a33. There are two operations: first, multiply R1 by -2/10 and add it to R2. The elements of the new Matrix will be (10, 3, 4; 0, -10.6, 2.2; 0, 1.1, -11.2). Please refer to the following slide for more details.
Step-2:Find values of L22, L32 and U23-Option-2.
The following slide shows the elements we need to find: L22, L32, L33, and U23.
The Value of L22 is -10.6 (second Row/second column of Matrix A1), while L32 is the third Row/second column of Matrix A1, with a Value of (1.10). For the Value of U23, divide -2.2/10.60 = -1.1/5.3.
Next, use a22 as the pivot and eliminate the a32 element by multiplying (-1.1/-10.60) by R2 and adding it to the third Row. The elements of the new Matrix will be (10, 3, 4; 0, -10.6, 2.2; 0, 0, -1163/106). Please refer to the following slide for more details.
Step 4: Derive the Value for L33.
L33 = -1163/106, which is the element in the 3rd Row/third column.
Step 5: Find the elements of a 3×3 U Matrix.
Divide the first Row by L11, the second Row by L22, and the third Row by L33.
On the next slide, the elements of the U Matrix can be written as (1, 3/10, 4/10; 0,+1, 11/53; 0, 0, +1).
We can write both the lower and upper matrices and check whether multiplying the lower by the upper yields the final A Matrix. Using row-by-column multiplication, we can see that L*U equals A.
The following slide shows that the Product of L and U gives the elements of Matrix A.
Elementary matrices for Row operations, second solved Problem.
We can perform the same steps using elementary matrices; the elementary Matrix E1 to the right of Matrix A will eliminate the a12 and a13 elements of Matrix A. The multipliers are -2/10 and -3/10. The elementary Matrix E2 will eliminate the a23 element of the Matrix formed by multiplying E1 by A. The multiplier is -1.1/10.60. We can get the Value of the E2 Matrix; please refer to the following slide.
Multiply E2*(E1A); the diagonal of the new Matrix should equal ones. Then, for E3, we can find the elements of E3 are(1/10,0,0; 0-10.60,0; 0, 0,-106/1163).
Multiply E3 by E2*E1*/A from left to right to get the U Matrix elements.
How can we get the inverse of the L Matrix?
The Product of E3, E2, and E1 will give the inverse of L. See the proof on the next slide.
The final image shows the inverse of the lower Matrix, formed by multiplying E3*E2*E1.
Check that L, when multiplied by its inverse, gives the identity Matrix.
You can view and download the PDF for the Second Problem from this Link.
This is a Link to the Previous Post: 11-First solved Problem for a 3×3 Matrix by Crout’s LU-option 2.
This is the Omni calculator for estimating various linear algebra items: LU Decomposition Calculator.
Link to the the Omni calculator:: LU Decomposition Calculator.
Another calculator to use is the Calculator for matrices.