LU Decomposition Calculator

This calculator will help you to calculate the LU decomposition of a matrix of any order at a time with the steps shown.
Related Calculators:QR Factorization Calculator

Your Input :-

Your input can be in the form of an Integer, FRACTION or Real Number

Given Matrix: -

$\sf{2}$	$\sf{0}$	$\sf{-1}$	$\sf{5}$	$\sf{-2}$
$\sf{-4}$	$\sf{0}$	$\sf{0}$	$\sf{-8}$	$\sf{1}$
$\sf{0}$	$\sf{-5}$	$\sf{-5}$	$\sf{1}$	$\sf{8}$
$\sf{-6}$	$\sf{0}$	$\sf{7}$	$\sf{-3}$	$\sf{1}$

Formatted User input Display

\sf{Given\space Matrix:\space \begin{bmatrix}2 & 0 & -1 & 5 & -2\\ \\-4 & 0 & 0 & -8 & 1\\ \\0 & -5 & -5 & 1 & 8\\ \\-6 & 0 & 7 & -3 & 1\end{bmatrix}}

Question

\sf{Find \space the \space LU \space decomposition \space of \space the \space given \space matrix \space A \space as}

\sf{\begin{bmatrix}2 & 0 & -1 & 5 & -2\\ \\-4 & 0 & 0 & -8 & 1\\ \\0 & -5 & -5 & 1 & 8\\ \\-6 & 0 & 7 & -3 & 1\end{bmatrix}}

Step By Step Solution :-
LU decomposition for a matrix A is defined as the decomposition of a
matrix A into the product of two lower and upper triangular matrices
denoted as L and U.
A = LU
Step-1
Given the number of rows of matrix A = 4

\sf{Assume \space an \space Identity \space matrix \space of \space order \space 4 \space as \space L \space \& \space P \space = \space \begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 1\end{bmatrix}}

\sf{Now \space we \space will \space find \space the \space row \space echelon \space form \space (ref) \space of \space the \space given \space matrix \space A}

\sf{Row \space echelon \space form \space of \space the \space matrix \space A \space will \space be \space considered \space as \space upper \space triangular \space matrix.}

\sf{So \space U \space = \space {\begin{bmatrix}2 \space & \space 0 \space & \space -1 \space & \space 5 \space & \space -2\\ \space \\0 \space & \space -5 \space & \space -5 \space & \space 1 \space & \space 8\\ \space \\0 \space & \space 0 \space & \space -2 \space & \space 2 \space & \space -3\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 16 \space & \space -11\end{bmatrix}}}

to see the Steps to calculate the Row Echelon Form, click here

\sf{List \space of \space the \space row \space operations \space used \space to \space obtain \space above \space matrix \space U \space are}

\sf{Considering \space the \space column \space 1 \space elements \space :}

\sf{For \space the \space element \space A_{21}:- \space \space \bold{R_{2} \space → \space R_{2} \space - \space (\frac{-4}{2})R_{1}} \space }

\sf{For \space the \space element \space A_{31}:- \space \space \bold{R_{3} \space → \space R_{3} \space - \space (\frac{0}{2})R_{1}} \space }

\sf{For \space the \space element \space A_{41}:- \space \space \bold{R_{4} \space → \space R_{4} \space - \space (\frac{-6}{2})R_{1}} \space }

\sf{Swap \space \bold{R_{2} \space ↔ \space R_{3}}}

\sf{Considering \space the \space column \space 2 \space elements \space :}

\sf{For \space the \space element \space A_{32}:- \space \space \bold{R_{3} \space → \space R_{3} \space - \space (\frac{0}{-5})R_{2}} \space }

\sf{For \space the \space element \space A_{42}:- \space \space \bold{R_{4} \space → \space R_{4} \space - \space (\frac{0}{-5})R_{2}} \space }

\sf{Considering \space the \space column \space 3 \space elements \space :}

\sf{For \space the \space element \space A_{43}:- \space \space \bold{R_{4} \space → \space R_{4} \space - \space (\frac{4}{-2})R_{3}} \space }

Step-2

\sf{Now, \space we \space will \space replace \space the \space same \space position \space element \space in \space the \space matrix \space L \space with \space the \space }

\sf{negative \space of \space the \space multiplier \space coefficient \space used \space in \space the \space row \space operations \space }

\sf{L \space = \space {\begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\-2 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\-3 \space & \space 0 \space & \space -2 \space & \space 1\end{bmatrix}} \space \space and \space P=\begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 1\end{bmatrix}}

\sf{Swap \space R_{1} \space ↔ \space R_{2} \space (consider \space elements \space below \space the \space diagonal \space for \space left \space matrix)}

\sf{L \space = \space {\begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\-2 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\-3 \space & \space 0 \space & \space -2 \space & \space 1\end{bmatrix}} \space \space and \space P=\begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\0 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 1\end{bmatrix}}

\sf{And \space we \space have \space obtained \space both \space matrices \space L \space and \space U.}

Final Answer
The LU decomposition of the given matrix is

\sf{L \space = \space \begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\-2 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\-3 \space & \space 0 \space & \space -2 \space & \space 1\end{bmatrix} \space or \space \begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\-2 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\-3 \space & \space 0 \space & \space -2 \space & \space 1\end{bmatrix}}

\sf{U \space = \space \begin{bmatrix}2 \space & \space 0 \space & \space -1 \space & \space 5 \space & \space -2\\ \space \\0 \space & \space -5 \space & \space -5 \space & \space 1 \space & \space 8\\ \space \\0 \space & \space 0 \space & \space -2 \space & \space 2 \space & \space -3\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 16 \space & \space -11\end{bmatrix} \space or \space \begin{bmatrix}2 \space & \space 0 \space & \space -1 \space & \space 5 \space & \space -2\\ \space \\0 \space & \space -5 \space & \space -5 \space & \space 1 \space & \space 8\\ \space \\0 \space & \space 0 \space & \space -2 \space & \space 2 \space & \space -3\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 16 \space & \space -11\end{bmatrix}}

\sf{ \space and \space P=\begin{bmatrix}1 \space & \space 0 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 1 \space & \space 0\\ \space \\0 \space & \space 1 \space & \space 0 \space & \space 0\\ \space \\0 \space & \space 0 \space & \space 0 \space & \space 1\end{bmatrix}}

Note :- If you find any computational or Logical error in this calculator, then you can write your suggestion by clicking the below button or in the comment box.

$\color{red} \bold{Related \space Calculators:}$

Rank of a matrix
Basis of a matrix
Solving a system of linear equation
Reduced row echelon form of a matrix
Linear Independence of vectors
Row space of a matrix.
Column space of a matrix.

$\bold{Table \space of \space Content}$

1. Introduction to the LU Decomposition calculator
2. What is the Formulae used & conditions required ?
3. How do I find the LU Decomposition of a matrix?
4. Why choose our LU Decomposition of a matrix Calculator?
5. A Video for explaining this concept
6. How to use this calculator ?
7. Solved Examples
8. Frequently Asked Questions (FAQs)
9. What are the real-life applications?
10. Conclusion

1. Introduction to LU Decomposition

Have you ever felt overwhelmed by the intricacies of matrix operations? LU decomposition is a friendly guide, simplifying the world of matrices. In this blog, we'll explore LU decomposition, shedding light on its definition, applications, and how it eases the complexities of linear algebra.
$\bold{Definition}$
LU decomposition, short for Lower-Upper decomposition, is a method that breaks down a matrix into the product of two separate matrices – a lower triangular matrix (L) and an upper triangular matrix (U). This decomposition simplifies solving linear systems and offers a clearer understanding of the matrix's structure.

2. What is the Formulae used & conditions required?

$\bold{Formula \space used}$
The formula for LU decomposition is A=LU, where A is the original matrix, L is the lower triangular matrix, and U is the upper triangular matrix. The process involves Gaussian elimination with pivoting.
$\bold{Conditions \space required}$
Conditions include having a square matrix and ensuring that the pivots do not become zero during the elimination process.

3. How do I calculate the LU Decomposition of a matrix?

Write the matrix U as the row echelon form of the matrixrow echelon form of the matrix.
Now, apply some operations to the identity matrix of the same order to convert it into a lower triangular form, i.e., L.

4. Why choose our LU Decomposition of a matrix Calculator?

$\bold{Easy \space \space to \space Use}$
Our calculator page provides a user-friendly interface that makes it accessible to both students and professionals. You can quickly input your square matrix and obtain the matrix of minors within a fraction of a second.

$\bold{Time \space Saving \space By \space automation}$
Our calculator saves you valuable time and effort. You no longer need to manually calculate each cofactor, making complex matrix operations more efficient.

$\bold{Accuracy \space and \space Precision}$
Our calculator ensures accurate results by performing calculations based on established mathematical formulas and algorithms. It eliminates the possibility of human error associated with manual calculations.

$\bold{Versatility}$
Our calculator can handle all input values like integers, fractions, or any real number.

$\bold{Complementary \space Resources}$
Alongside this calculator, our website offers additional calculators related to Pre-algebra, Algebra, Precalculus, Calculus, Coordinate geometry, Linear algebra, Chemistry, Physics, and various algebraic operations. These calculators can further enhance your understanding and proficiency.

5. A video based on how to find the LU Decomposition of a matrix.

6. How to use this calculator

This calculator will help you find the LU decomposition of any order in the given matrix.
In the given input boxes, you have to put the value of the matrix.
A step-by-step solution will be displayed on the screen after clicking the Calculate button.
You can access, download, and share the solution.

7. Solved Example

$\bold{Question}$
Convert the matrix A = $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ into LU decomposition form.
$\bold{Solution}$
$\bold{Step \space 1:}$ Let's find the row echelon form of matrix A = $\begin{bmatrix} 1 & 2 \\ 0 & -3 \end{bmatrix}$ and we will call it a U matrix.
$\bold{Step \space 2:}$ After applying certain transformation we can get L = $\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$

8. Frequently Asked Questions (FAQs):-

Why use LU decomposition?

LU decomposition simplifies solving linear systems, especially with multiple systems with the same coefficient matrix.

Can any matrix be decomposed using LU decomposition?

During the elimination process, LU decomposition is possible for square matrices that meet certain conditions, such as non-singularity and no division by zero.

Is LU decomposition unique for a given matrix?

LU decomposition is not unique. Different choices in the pivoting strategy may lead to different L and U matrices, but the product remains the same.

What is the role of pivoting in LU decomposition?

Pivoting is crucial to avoid division by zero during Gaussian elimination. It ensures the stability of the LU decomposition process.

How does LU decomposition aid in solving linear systems?

LU decomposition simplifies solving linear systems by breaking down the original matrix into two triangular matrices, making it easier to substitute and solve.

9. What are the real-life applications?

LU decomposition finds application in various fields, including engineering and numerical analysis. It is commonly used in solving systems of linear equations arising from physical models and simulations.

10. Conclusion

As we wrap up our journey through LU decomposition, remember that it's not just a mathematical tool but a practical approach to simplify problem-solving in linear algebra. Embrace the elegance of LU decomposition and witness how it transforms the complexities of matrices into a more manageable and intuitive form, making linear systems more accessible and applicable in diverse real-world scenarios.

This blog is written by Neetesh Kumar

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