8-Introduction to Crout’s LU decomposition for a 2×2 Matrix.

Last Updated on October 21, 2026 by Maged kamel

Introduction to Crout’s LU decomposition for a 2×2 Matrix.

Using Crout’s LU decomposition for a 2×2 Matrix, you can obtain the elements of the 2×2 upper and lower matrices using the Gauss elimination method. Next, we show how to solve for x and y in two equations using the lower-upper triangular method through an example.

Video for the content of the Post

The Timestamps for the video are as follows:

• 00:00 Introduction to LU decomposition using Crout’s method for a 2×2 Matrix.
• 01:47 How can we determine the values of the lower and upper matrices using Crout’s method?
• 03:25 The four equations to solve for the L and U matrices’ values.
• 06:40 Option-1-Turn Matrix A into a lower Matrix to find L11, L21 and L22 values.
• 09:10 Option-1-Turn Matrix A into a lower Matrix using an elementary Matrix.
• 11:28 Option-2-Turn Matrix A into an upper Matrix.
• 13:50 Option-2-Turn Matrix A into an upper Matrix using an elementary Matrix.
• 14:34 Solve for x and y values for two simultaneous equations.
• 15:48 How can we use Matrix C to find the values of the two unknowns?
• 17:38 The final values for x and Y.
• 19:07 Use the inverses of L and U to get x and y values.

A brief description of the Post’s content.

Step 1: Derive the Expression for Crout’s LU decomposition for a 2×2 Matrix.

If we have a 2×2 square Matrix, we call it Matrix A. If it is invertible, it can be expressed as the Product of two matrices. One of these is called the lower Matrix. The other Matrix is the Upper Matrix.

The lower Matrix is expressed as L = [l_ij]. where i represents the Row, and j represents the column. We have 6 unknowns in total. The lower Matrix has three unknowns: L11, L21, and L22. The upper Matrix has 3 unknown elements: U11, U12, and U22.

In Crout’s method, U11 = U22 = 1. In a 2×2 Matrix, the lower Matrix has nonzero elements L11, L22, and L21. Unlike the previous method, Doolittle’s LU decomposition, Crout’s LU decomposition for the 2×2 Matrix assumes U11 and U22 are 1, whereas the L Matrix’s diagonal elements are not 1. Our unknowns become only four. Please refer to the slide image below.

Step 2: The values for the L and U matrices for Crout’s LU decomposition for a 2×2 Matrix.

Since A is the Product of the lower and upper matrices, we multiply L and U and equate the result to the elements of A.

 The Product of the first Row and the first column can beis11 = a11.

The Product of the First Row and the second column can be expressed as (L11*U12) +0 = a12.

If we multiply the second Row by the first column, we get (L21)*U11 + 0 = a21. If we multiply the second Row by the second column, we get (L21*U12) + L22*U22 = a22.

From equations I and II, we have L11 = a11 and L21 = a12. From equation III, we get U12 = a12/a11. Finally,L22=a22-(a21*(a12/a11)). Please refer to the slide image below.

Based on the previous equations, derive techniques for computing the matrices L and U from Matrix A. If we keep a11 and a21 as elements of L11 and L21, and set U11 = a11/a11 (which equals 1) and U12 = a12/a11 for the elements of the upper Matrix U, then U22 = 1. L22 will be set equal to a22-U12*L21.

Please refer to the slide image below.

We can use two options to obtain the elements of the lower and upper matrices; we will start with the first option.

Step-3-Option-1: Convert A to a lower Matrix.

To Option-1: Convert A to a lower Matrix. If we express L by U as a Matrix that has elements L11, L11*U12, L21, L21*U12+L22), converting such a Matrix to a lower Matrix is achieved by multiplying by a permutation Matrix that has ( 1-l11U12/L11, 0, 1), which will create the lower Matrix L. Please refer to the slide image below.

We use numerical values for the matrix A; we write A = (2 3 3 4), meaning a11, a12, a21, and a22.

We apply this information to the given Matrix A, which is (2, 3, 3, 4). We place the first column as the first column of the L Matrix. The L11= a11=2; the L21=a21=3. To set L12 to zero, we use a11 as the pivot. We multiply the second column by (-3/2) times C1 and add the result to it. The Value of L22=-(3/2)*3+4=-1/2.

We need to have the Value of L22. For the U Matrix, we have U11 = U22 = 1, while U12 = a12/a11 = 3/2. Perform Row echelon to convert Matrix A to an upper Matrix, using U22 as L22; the final Value of L22 is -1/2. Please refer to the slide image below.

Step 4: Multiply Matrix A by an elementary Matrix E1 to get the L Matrix.

To get the L Matrix, we multiply Matrix A on the right by an elementary Matrix E1 with elements (1, -3/2, 0, 1). To obtain the upper Matrix, we compute E1 inverse.

The inverse is (1, 3/2, 0, 1). We can estimate the Value of Matrix C by multiplying L-1 by B, where L-1 is the (2×2) Matrix we just estimated and B is the (2×1) Matrix with entries (13, 18). As we can see, C = (13/2 3) is a 2×1 Matrix.

The upper Matrix is the inverse of the E1 Matrix, as shown on the next slide.

Step A- Use option 2- Convert Matrix A to an Upper Matrix for Crout’s LU decomposition for a 2×2 Matrix.

To get the U Matrix, multiply Matrix A from the left by an elementary Matrix E with elements (1, 0, -L21/L11, 1). The elements of the U Matrix are (L11, L11*U12, 0, L33). Then multiply by -C21*C11/(C11)+C21 to obtain U12, as we will see in the next example.

Use Row operations to get the Upper Matrix U.

We will use the Matrix A = (2, 3, 3, 4) and perform Row operations. Keep R1 unchanged, use a11 as a pivot, multiply R1*-a21/a11, and add the result to the second Row to eliminate a21.

L11=2, L11*u12=3, L22=-1/2, U12=3/2. Please refer to the slide image below.

Use the elementary Matrix E1 to get the U Matrix.

We will do the Row operations using the elementary Matrix E1 that we multiply Matrix A by from the left side; the E1 elements are (1,0, -3/2,1); the previous values for L11, L22, and L11u12 are obtained.

Get the values for x and y from Matrix C for Crout’s LU decomposition for a 2×2 Matrix.

We have two equations for which we want to develop Crout’s LU decomposition for a 2×2 Matrix. The first equation is (2x + 3y = 13), and the second equation is (3x + 4y = 18).

As we can see, the determinant of the Matrix equals -1, not zero. We use the same L and U matrices to represent A.
The final Value of X, the Matrix,, can be estimated as a (2×1) Matrix, 2 3). We will check whether our given equation satisfies the

Find the values of the C Matrix for Crout’s LU decomposition of a 2×2 Matrix.

We use the Expression for C-U*X, then L*C = B. Multiply both sides by L-1 to get C = L-1*B. The inverse of Matrix L is (1/2, 0, 3, -1/2). The C Matrix is (13/2, 3). We can find the inverse of the U Matrix as (1,-3/2,0,1)

For the Expression U*x = c, multiply both sides by U-1 to get x = U-1*c. Finally, we get x = 2, y = 3. We check these values in the linear equation, and they satisfy it.

We can use LL*U*X = b; multiplying both sides by U-1*L-1* gives X = U-1*L-1*b. Please refer to the slide image below.

The full data for the Unknowns, using the multiplication U-1*L-1*B, are shown in the next slide.

Final check of the solution for the two simultaneous equations.

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This Post covers Crout’s LU decomposition for a 3×3 Matrix.

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