9- Crout’s LU decomposition for a 3×3 matrix-Option 1

Last Updated on October 19, 2026 by Maged kamel

Crout’s LU decomposition for a 3×3 matrix-Option 1.

The elements of the upper and lower 3×3 matrices can be obtained using Gaussian elimination. The U Matrix for Crout’s LU decomposition of a 3×3 Matrix is a (3×3) Matrix composed of (u11, u21, u13) for the first Row and (0, 1, u23) for the second Row. The last Row is (0, 0, 1).

Option 1 reduces Matrix A to a lower Matrix.

Video illustrating this Post.

The Timestamps for the video are as follows:

• 00:00 Introduction to LU decomposition using Crout’s method for a 3x3matrix.
• 00:58 Introduction to the three simultaneous equations.
• 03:28 Find a relation between the elements of Matrix A and the Matrix elements for L and U.
• 04:44 Option 1: Turn Matrix A into a lower Matrix.
• 05:37 Option-1: Find u12 and u13 values,
• 08:48 Option 1: Find u23 Value and the final L Matrix.

The following steps explain how to find the components of both the lower L and upper U matrices.

Step 1: Equate the Product of L*U to the A Matrix.

We have three equations, and we want to develop Crout’s LU decomposition for a 3×3 Matrix.
The first equation is (a11*x1+a12*x2+a13*x3=b1). The second equation is (a21*x1+a22*x2+a23*x3=b2).

The third equation is (a31*x1 + a32*x2 + a33*x3 = b3). We can express these equations as A*x = b. The next task is to find the L and U matrices using Crout’s LU decomposition for a 3×3 Matrix (Option 1).

The Matrix A which is(a11 a12 a13,a21 a22 a23, a31 a32 a33). Consider the lower Matrix L as (L11 0 0, L21 L22 0, L31 L32 L33), a 3×3 Matrix.

Using the Expression L*U =A, we can derive a new 3×3 Matrix that contains the elements of both the lower and upper matrices.

By writing down the Matrix A as  3×3 (a11  a12 a13, a21  a22  a23 and  (a31 a32 a33) =the multiplication of l by U, or the multiplication of ( L11 0  0, L21 L22  0, L31 L32  0) multiplied by  ( 1 U12 U13, 0  1  U23, 0  0 1), as we know that according to Crout’s method, we have the diagonal of Matrix U is ones. The left-hand side is Matrix A, while the right-hand side is L*U.

Step 2: Derive the Expression for L11, L12, L13, U12, and U13 for Crout’s LU decomposition of a 3×3 Matrix.

From the second slide, we can see that by multiplying the lower Matrix’s First Row by the upper Matrix’s First Column, we get (L11*1)  equal to the first element a11 of Matrix A.

Multiply the third Row of the lower Matrix by the First Column of the upper Matrix; we get (L31), which equals a31.

At first sight, we see that the first column of the lower triangular Matrix is identical to the first column of the original Matrix since L11 = a11. L21=a21, L31=a31.

We will proceed to multiply the first Row of the lower Matrix by the second column of the upper Matrix. We can write that (l11*u12 + 0*1 + 0*0) = a12.

We have L11 equals to a11, so U12 will be set=a12/a11, since L11*U12=a12.

Suppose we multiply the first Row of the lower Matrix by the third column of the upper Matrix. We can write that L11*U13 = a13. So U13 will be set=a12/a13, since L11*U13=a13. We have L11 equals to a11, so U12 will be set=a12/a11, since L11*U12=a12.

As a quick reminder of the values of L11, L21, and L31, please copy the entire first column of Matrix A and place it in the first column of the Lower Matrix. Divide a12 by a11; this is1 to get. Divide a13 by a11; this is L31. Refer to the image on the next slide.

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Step 3: Convert the A Matrix to a lower Matrix using option 1: Crout’s LU decomposition for a 3×3 Matrix.

We want a12 and a13 to be zeros; we set a11 as a pivot, then multiply the first column by (-a12/a11) and add the result to the second column. Similarly, we can multiply the first column by (-a13/a11) and add the result to the third column.

If we have a look at the second Row / second column element, we have (a22-a21/a11*a21). Please refer to the upper-left corner of this slide. If we compare Matrix A with the Product of L and U, we can see that a22 = L21U12 + L22.

Since we already know that a21/a11=U12 and L21=a12, we can rewrite the second-row/second-column element, which is (a22-a12/a11*a21), as L21*U12+L22-U12*L21=L22. We can also rewrite the third Row/second column element, (a32-a12/a11*a31), as L31*U12+L32-U12*L31=L32.

For the second Row/third column, the element equals L21*U13+L22*U23-U13*L21=L22*U23. For the third Row/ third column, the element will be equal to L31*U13+L32*U23-U13*L31+L33=L23*U23+L33.

Step 4: Derive the Value of U23 for Crout’s LU decomposition for a 3×3 Matrix.

After adjusting its rows, we move to the next slide and check the lower Matrix L1.

We can find U23 by dividing the second-row, third-column element by the second-row, second-column element. To obtain the final lower Matrix L, we set L22 U23 to zero. To obtain the final lower Matrix L, we want to set L22 U23 to zero.

Consider L22 as a pivot. Divide L22*U23/L22, multiply by (-1) *second column, and add the result to the third column.

Step 5: The final Crout’s LU decomposition for a 3×3 Matrix.

We can get the Value of U23 from the summary in Matrix L1. The final L Matrix gives all elements of the lower Matrix L, including L33.

The following two slides summarize the process of Crout’s method (option 1) for a 3×3 Matrix.

The following slides show the upper Matrix elements. Thanks, and I hope this Post is helpful. The next Post shows another option for Crout’s method for a 3×3 Matrix.

You can view or download the PDF for this Post from the following Link.

Previous Post (Link): Step-by-step Introduction to Crout’s LU decomposition for a 2×2 Matrix.

The next Post is a section on Crout’s decomposition for a 3×3-3×3 Matrix.

This is the Omni calculator for estimating various linear algebra items: LU Decomposition Calculator.

Here is a Link to another Matrix calculator.