LU Decomposition

Overview

In our study of linear algebra, the solution of systems of linear equations of the form $A\mathbf{x} = \mathbf{b}$ is a central theme. While methods such as Gaussian elimination provide a direct approach, they can be computationally inefficient, particularly when a system must be solved repeatedly with different right-hand side vectors. This chapter introduces a powerful and elegant alternative: matrix decomposition. We shall focus on LU decomposition, a method that factors a square matrix $A$ into the product of a lower triangular matrix $L$ and an upper triangular matrix $U$. This factorization effectively decouples the original problem into two simpler, sequential systems that can be solved efficiently.

The importance of this method in the context of the GATE examination cannot be overstated. Questions involving systems of linear equations are a staple of the Engineering Mathematics syllabus, and LU decomposition is frequently tested due to its algorithmic nature, which aligns well with the computational thinking required in computer science. A thorough understanding of this technique is essential not only for solving direct numerical problems but also for appreciating the computational advantages it offers. This chapter will provide a systematic treatment of the subject, beginning with the methods for obtaining the $L$ and $U$ factors and culminating in the application of these factors to find the solution vector $\mathbf{x}$ through forward and backward substitution.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Decomposition Methods | Factoring a matrix into L and U |
| 2 | Solving Systems with LU | Using L and U for substitution |

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Learning Objectives

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We now turn our attention to Decomposition Methods...
## Part 1: Decomposition Methods

Introduction

In the study of linear algebra, particularly in numerical analysis, we frequently encounter the need to solve systems of linear equations of the form $Ax = b$. While methods like Gaussian elimination are effective, they can be computationally inefficient if the system must be solved for multiple different vectors $b$ with the same coefficient matrix $A$. LU decomposition provides a powerful and efficient alternative by factorizing the matrix $A$ into the product of a lower triangular matrix $L$ and an upper triangular matrix $U$.

This factorization, once computed, allows for the rapid solution of $Ax = b$ through a two-step process of forward and backward substitution. This method is foundational in numerical computation and forms the basis for many advanced algorithms. For the GATE examination, understanding the process of decomposition and its application to solving linear systems is paramount.

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Key Concepts

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## 1. Doolittle's Method of LU Decomposition

The primary method for finding the LU factorization is Doolittle's method. Here, we seek a factorization $A = LU$ where $L$ is a unit lower triangular matrix and $U$ is an upper triangular matrix.

Let us consider a $3 \times 3$ matrix $A$. The decomposition will take the form:

By performing the matrix multiplication on the right-hand side and equating the corresponding elements with the matrix $A$, we can systematically solve for the unknown elements of $L$ and $U$. The process involves calculating the rows of $U$ and columns of $L$ alternately.

A
L
U






=


×

Worked Example:

Problem: Find the LU decomposition of the matrix $A = \begin{pmatrix} 2 & 1 & 4 \\ 8 & 7 & 18 \\ 6 & 8 & 23 \end{pmatrix}$ using Doolittle's method.

Solution:

We set $A = LU$:

Step 1: Determine the first row of $U$.
The first row of $A$ is obtained by multiplying the first row of $L$ with all columns of $U$.

Step 2: Determine the first column of $L$.
The first column of $A$ is obtained by multiplying all rows of $L$ with the first column of $U$.

Step 3: Determine the second row of $U$.
We use the elements from the second row of $A$.

Step 4: Determine the second column of $L$.
We use the element $a_{32}$ from the third row of $A$.

Step 5: Determine the third row of $U$.
We use the element $a_{33}$.

Result:
The decomposition is:

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## 2. Solving Systems of Linear Equations using LU Decomposition

The primary application of LU decomposition is in solving the system $Ax = b$. Once $A$ is factored into $LU$, the system becomes $LUx = b$. We can solve this in two stages.

Let $Ux = y$. The system becomes $Ly = b$.

Forward Substitution: Solve $Ly = b$ for the vector $y$. Since $L$ is lower triangular, this is straightforward.

Backward Substitution: Solve $Ux = y$ for the vector $x$. Since $U$ is upper triangular, this is also straightforward.

This two-step process is computationally much faster than re-applying Gaussian elimination for each new vector $b$.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="For the matrix $A = \begin{pmatrix} 3 & 2 & 1 \\ 6 & 6 & 3 \\ 9 & 10 & 6 \end{pmatrix}$, its LU decomposition using Doolittle's method is $A = LU$. What is the value of the element $u_{22}$?" answer="2" hint="First, find the first row of U and the first column of L. Then use the formula for $a_{22}$." solution="
Step 1: Set up the decomposition.

Step 2: Find the first row of $U$.

Step 3: Find the element $l_{21}$ from the first column of $L$.

Step 4: Calculate $u_{22}$ using the element $a_{22}$.
The formula for $a_{22}$ is $l_{21} u_{12} + u_{22} = a_{22}$.

Result:
The value of $u_{22}$ is 2.
"
:::

:::question type="MCQ" question="Let $A = LU$ be the LU decomposition of $A = \begin{pmatrix} 4 & -2 \\ 12 & -3 \end{pmatrix}$. Which of the following is the correct matrix $L$?" options=["$\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$","$\begin{pmatrix} 1 & 0 \\ 1/3 & 1 \end{pmatrix}$","$\begin{pmatrix} 4 & 0 \\ 12 & 1 \end{pmatrix}$","$\begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$"] answer="$\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$" hint="Use Doolittle's method where L is a unit lower triangular matrix. Calculate $u_{11}$ first, then $l_{21}$." solution="
Step 1: Set up the decomposition for the $2 \times 2$ matrix.

Step 2: Determine the first row of $U$.

Step 3: Determine the first column of $L$.

Step 4: Assemble the matrix $L$.

Result:
The correct matrix $L$ is $\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$.
"
:::

:::question type="MSQ" question="If a non-singular matrix $A$ has an LU decomposition $A=LU$ using Doolittle's method, which of the following statements are ALWAYS true?" options=["$det(A) = det(U)$","$L$ is a singular matrix","$U$ must be non-singular","The diagonal elements of $U$ can be zero"] answer="A,C" hint="Consider the properties of determinants for triangular matrices and the conditions for a matrix to be non-singular." solution="

• Option A: In Doolittle's method, $L$ is a unit lower triangular matrix, meaning its diagonal elements are all 1. The determinant of any triangular matrix is the product of its diagonal elements. Therefore, $det(L) = 1$. Since $det(A) = det(L) \times det(U)$, it follows that $det(A) = 1 \times det(U) = det(U)$. This statement is true.


• Option B: The determinant of $L$ is 1, which is non-zero. A matrix is singular if and only if its determinant is zero. Therefore, $L$ is a non-singular matrix. This statement is false.


• Option C: We are given that $A$ is non-singular, so $det(A) \neq 0$. Since $det(A) = det(U)$, it must be that $det(U) \neq 0$. Therefore, $U$ must be non-singular. This statement is true.


• Option D: The determinant of $U$ is the product of its diagonal elements. Since $det(U) \neq 0$, none of its diagonal elements can be zero. This statement is false.

Thus, the correct statements are A and C.
"
:::

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Summary

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What's Next?

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Part 2: Solving Systems with LU

Introduction

The solution of systems of linear equations of the form $Ax = b$ is a fundamental problem in computational science and engineering. While methods such as Gaussian elimination provide a direct approach, they can be inefficient if we need to solve the system for multiple right-hand side vectors, $b$. For instance, consider solving $Ax = b_1, Ax = b_2, \dots, Ax = b_k$ for a fixed matrix $A$. Repeating Gaussian elimination for each $b_i$ would be computationally redundant.

LU decomposition presents a more elegant and efficient strategy for such scenarios. It is a method of matrix factorization that decouples the elimination process, which depends only on $A$, from the handling of the vector $b$. The central idea is to factor the coefficient matrix $A$ into the product of a lower triangular matrix $L$ and an upper triangular matrix $U$. Once this decomposition is achieved, solving the original system is reduced to two simpler problems: one involving forward substitution and another involving backward substitution. This separation of concerns is the primary advantage of the method.

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Key Concepts

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## 1. The Two-Step Solution Process

Once we have decomposed the matrix $A$ into $L$ and $U$, we can substitute this factorization into the original system of equations.

The system $Ax = b$ becomes:

By the associativity of matrix multiplication, we can write this as:

This form suggests a two-step approach. We introduce an intermediate vector $y \in \mathbb{R}^{n \times 1}$, defined as:

Substituting this into the previous equation, we obtain our first system:

This strategy transforms the single, complex problem of solving $Ax=b$ into two simpler, sequential problems involving triangular matrices.

Solve $Ly = b$ for $y$
(Forward Substitution)

y

Solve $Ux = x$ for $x$
(Backward Substitution)

Step 1: Forward Substitution
We first solve the system $Ly = b$. Since $L$ is a lower triangular matrix, this is straightforward. For a $3 \times 3$ system:

We can solve for $y_1$ directly from the first equation, then substitute its value into the second equation to find $y_2$, and finally use both $y_1$ and $y_2$ to find $y_3$.

Step 2: Backward Substitution
After finding the vector $y$, we solve the system $Ux = y$. As $U$ is an upper triangular matrix, this is also computationally simple.

We solve for $x_3$ from the last equation, substitute its value into the second-to-last equation to find $x_2$, and proceed upwards until all components of $x$ are found.

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## 2. Finding the L and U Matrices (Doolittle's Method)

The primary task is to find the matrices $L$ and $U$ such that $A = LU$. Doolittle's method is a popular algorithm for this, which specifies that $L$ must be a unit lower triangular matrix (i.e., $l_{ii} = 1$ for all $i$).

Let us consider a $3 \times 3$ matrix $A$. We wish to find $L$ and $U$ such that:

By performing the matrix multiplication on the right-hand side and equating corresponding entries with $A$, we can derive a sequence of equations to solve for the unknown elements of $L$ and $U$. The process generally proceeds by computing one row of $U$, followed by one column of $L$, and repeating.

Worked Example:

Problem: Solve the following system of equations using LU decomposition.
$2x_1 + x_2 + 3x_3 = 1$
$4x_1 + 4x_2 + 7x_3 = 1$
$2x_1 + 5x_2 + 9x_3 = 3$

Solution:

The system is $Ax = b$, where:

Part A: Find the LU Decomposition

We set $A = LU$ with $L$ being a unit lower triangular matrix.

Step 1: Determine the first row of $U$.
From the first row of the product: $u_{11} = a_{11} = 2$, $u_{12} = a_{12} = 1$, $u_{13} = a_{13} = 3$.

Step 2: Determine the first column of $L$.
From the first column of the product:
$l_{21} u_{11} = a_{21} \implies l_{21} \times 2 = 4 \implies l_{21} = 2$.
$l_{31} u_{11} = a_{31} \implies l_{31} \times 2 = 2 \implies l_{31} = 1$.

Step 3: Determine the second row of $U$.
$l_{21} u_{12} + u_{22} = a_{22} \implies (2)(1) + u_{22} = 4 \implies u_{22} = 2$.
$l_{21} u_{13} + u_{23} = a_{23} \implies (2)(3) + u_{23} = 7 \implies u_{23} = 1$.

Step 4: Determine the second column of $L$.
$l_{31} u_{12} + l_{32} u_{22} = a_{32} \implies (1)(1) + l_{32}(2) = 5 \implies 2l_{32} = 4 \implies l_{32} = 2$.

Step 5: Determine the third row of $U$.
$l_{31} u_{13} + l_{32} u_{23} + u_{33} = a_{33} \implies (1)(3) + (2)(1) + u_{33} = 9 \implies 5 + u_{33} = 9 \implies u_{33} = 4$.

Result (Decomposition):

Part B: Solve the System

Step 1: Solve $Ly = b$ using forward substitution.

From row 1: $y_1 = 1$.
From row 2: $2y_1 + y_2 = 1 \implies 2(1) + y_2 = 1 \implies y_2 = -1$.
From row 3: $y_1 + 2y_2 + y_3 = 3 \implies 1 + 2(-1) + y_3 = 3 \implies -1 + y_3 = 3 \implies y_3 = 4$.
So, $y = [1, -1, 4]^T$.

Step 2: Solve $Ux = y$ using backward substitution.

From row 3: $4x_3 = 4 \implies x_3 = 1$.
From row 2: $2x_2 + x_3 = -1 \implies 2x_2 + 1 = -1 \implies 2x_2 = -2 \implies x_2 = -1$.
From row 1: $2x_1 + x_2 + 3x_3 = 1 \implies 2x_1 + (-1) + 3(1) = 1 \implies 2x_1 + 2 = 1 \implies 2x_1 = -1 \implies x_1 = -0.5$.

Answer: The solution is $x_1 = -0.5, x_2 = -1, x_3 = 1$.

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## 3. Properties and Conditions for LU Decomposition

The existence of an LU decomposition and its properties are directly linked to the properties of the original matrix $A$.

We observe that the determinant of a triangular matrix is the product of its diagonal elements.
For Doolittle's method, $L$ is unit lower triangular, so $l_{ii} = 1$ for all $i$. It follows that $\det(L) = 1$.
Thus, for Doolittle's decomposition, we have a simpler relation:

This relationship leads to two critical conclusions for GATE.

Invertibility:
A matrix $A$ is invertible (non-singular) if and only if its determinant is non-zero. From the property above, $\det(A) \neq 0$ if and only if $\det(U) \neq 0$. Since $\det(U)$ is the product of its diagonal elements, this means all diagonal elements $u_{ii}$ must be non-zero. If $A$ is invertible, then $L$ and $U$ must also be invertible. For $L$ in Doolittle's method, this is always true as $\det(L) = 1$. For $U$, its invertibility depends on its diagonal elements being non-zero.

Singularity:
A matrix $A$ is singular if and only if $\det(A) = 0$. This implies that $\det(U) = 0$. Consequently, at least one of the diagonal elements of $U$, $u_{ii}$, must be zero. This is a very common test point.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="A $3 \times 3$ matrix $A$ has an LU decomposition where $L$ is a unit lower triangular matrix and $U = \begin{pmatrix} 2 & -1 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 4 \end{pmatrix}$. Which of the following statements is TRUE?" options=["A is invertible.","The system $Ax=b$ has a unique solution for any vector $b$.","The determinant of A is 8.","The rank of A is less than 3."] answer="The rank of A is less than 3." hint="Relate the diagonal elements of U to the properties of matrix A." solution="
Step 1: Analyze the matrix $U$.
The diagonal elements of $U$ are $u_{11}=2$, $u_{22}=0$, and $u_{33}=4$.

Step 2: Relate the diagonal of $U$ to the determinant of $A$.
The determinant of $A$ is the product of the diagonal elements of $U$ (since $\det(L)=1$).

Step 3: Evaluate the options based on $\det(A)=0$.

• If $\det(A)=0$, the matrix $A$ is singular (not invertible). So, option A is false.


• If $A$ is singular, the system $Ax=b$ does not have a unique solution. It either has no solution or infinitely many solutions, depending on $b$. So, option B is false.


• The determinant of A is 0, not 8. So, option C is false.


• Since $A$ is a $3 \times 3$ matrix and it is singular ($\det(A)=0$), its rank must be less than 3. So, option D is true.

Result: The correct statement is that the rank of A is less than 3.
"
:::

:::question type="NAT" question="Consider the matrix $A = \begin{pmatrix} 3 & 1 & 2 \\ 6 & 3 & 5 \\ 9 & 5 & 6 \end{pmatrix}$. In the LU decomposition of $A$ where $L$ is a unit lower triangular matrix, what is the value of the element $u_{33}$?" answer=" -2" hint="Systematically compute the elements of L and U until you reach $u_{33}$." solution="
Step 1: Set up the decomposition $A=LU$.

Step 2: Find the first row of $U$.

Step 3: Find the first column of $L$.

Step 4: Find the second row of $U$.

Step 5: Find the second column of $L$.

Step 6: Find the third row of $U$, specifically $u_{33}$.

Result: The value of $u_{33}$ is -2.
"
:::

:::question type="MSQ" question="Let the system $Ax=b$ be solved using LU decomposition, where $A=LU$. Which of the following statements is/are ALWAYS true?" options=["The computational complexity of finding $L$ and $U$ is $O(n^3)$.","If $A$ is invertible, then all diagonal elements of $L$ must be non-zero.","The intermediate vector $y$ is found by solving $Uy=b$.","If $\det(A) \neq 0$, then both $L$ and $U$ are invertible."] answer="The computational complexity of finding $L$ and $U$ is $O(n^3).$,If $A$ is invertible, then all diagonal elements of $L$ must be non-zero.,If $\det(A) \neq 0$, then both $L$ and $U$ are invertible." hint="Evaluate each statement based on the definitions and properties of LU decomposition." solution="

• Option A: The process of decomposing $A$ into $L$ and $U$ is equivalent to Gaussian elimination and has a time complexity of $O(n^3)$. This statement is true.


• Option B: The matrix $L$ is lower triangular. Its determinant is the product of its diagonal elements. If any diagonal element of $L$ were zero, $\det(L)$ would be 0. Since $\det(A)=\det(L)\det(U)$, this would imply $\det(A)=0$, contradicting the invertibility of $A$. Therefore, all diagonal elements of $L$ must be non-zero. This statement is true. (In Doolittle's method, they are all 1, which is non-zero).


• Option C: The intermediate vector $y$ is defined by $Ux=y$. It is found by solving the system $Ly=b$. The statement says it's found by solving $Uy=b$, which is incorrect. This statement is false.


• Option D: If $\det(A) \neq 0$, then since $\det(A) = \det(L)\det(U)$, it must be that both $\det(L) \neq 0$ and $\det(U) \neq 0$. A matrix is invertible if and only if its determinant is non-zero. Thus, both $L$ and $U$ are invertible. This statement is true.

"
:::

:::question type="NAT" question="A system of equations $Ax=b$ is defined by $L = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & -1 & 1 \end{pmatrix}$, $U = \begin{pmatrix} 2 & 1 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{pmatrix}$ and $b = \begin{pmatrix} 2 \\ -1 \\ 8 \end{pmatrix}$. Calculate the value of $x_1 + x_2 + x_3$." answer="4" hint="First solve $Ly=b$ for $y$, then solve $Ux=y$ for $x$. Finally, sum the components of $x$." solution="
Step 1: Solve $Ly=b$ for $y$.

From row 1: $y_1 = 2$.
From row 2: $-2y_1 + y_2 = -1 \implies -2(2) + y_2 = -1 \implies y_2 = 3$.
From row 3: $3y_1 - y_2 + y_3 = 8 \implies 3(2) - 3 + y_3 = 8 \implies 3 + y_3 = 8 \implies y_3 = 5$.
So, $y = [2, 3, 5]^T$.

Step 2: Solve $Ux=y$ for $x$.

From row 3: $2x_3 = 5 \implies x_3 = 2.5$.
From row 2: $x_2 - x_3 = 3 \implies x_2 - 2.5 = 3 \implies x_2 = 5.5$.
From row 1: $2x_1 + x_2 + 4x_3 = 2 \implies 2x_1 + 5.5 + 4(2.5) = 2 \implies 2x_1 + 5.5 + 10 = 2 \implies 2x_1 = -13.5 \implies x_1 = -6.75$.

Step 3: Calculate the required sum.

Wait, let me recheck the calculation.
Step 1:
y1 = 2
-2(2) + y2 = -1 => -4 + y2 = -1 => y2 = 3
3(2) - (3) + y3 = 8 => 6 - 3 + y3 = 8 => 3 + y3 = 8 => y3 = 5
y = [2, 3, 5]^T. Correct.

Step 2:
2x3 = 5 => x3 = 2.5
x2 - x3 = 3 => x2 - 2.5 = 3 => x2 = 5.5
2x1 + x2 + 4x3 = 2 => 2x1 + 5.5 + 4(2.5) = 2 => 2x1 + 5.5 + 10 = 2 => 2x1 + 15.5 = 2 => 2x1 = -13.5 => x1 = -6.75
x = [-6.75, 5.5, 2.5]^T. Correct.

Sum:
-6.75 + 5.5 + 2.5 = -6.75 + 8 = 1.25.

Ah, let me design a question with an integer answer. Let's change $b$.
Let $b = [2, -2, 5]^T$.
Step 1:
y1 = 2
-2(2) + y2 = -2 => -4 + y2 = -2 => y2 = 2
3(2) - (2) + y3 = 5 => 6 - 2 + y3 = 5 => 4 + y3 = 5 => y3 = 1
y = [2, 2, 1]^T.

Step 2:
2x3 = 1 => x3 = 0.5
x2 - x3 = 2 => x2 - 0.5 = 2 => x2 = 2.5
2x1 + x2 + 4x3 = 2 => 2x1 + 2.5 + 4(0.5) = 2 => 2x1 + 2.5 + 2 = 2 => 2x1 = -2.5 => x1 = -1.25

Sum = -1.25 + 2.5 + 0.5 = 1.75. Still not integer.

Let me design the problem backwards.
Let $x = [1, 2, 1]^T$.
$y = Ux = \begin{pmatrix} 2 & 1 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2+2+4 \\ 0+2-1 \\ 0+0+2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \\ 2 \end{pmatrix}$.
$b = Ly = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & -1 & 1 \end{pmatrix} \begin{pmatrix} 8 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ -16+1 \\ 24-1+2 \end{pmatrix} = \begin{pmatrix} 8 \\ -15 \\ 25 \end{pmatrix}$.

Ok, new question.
:::question type="NAT" question="A system of equations $Ax=b$ is defined by $L = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & -1 & 1 \end{pmatrix}$, $U = \begin{pmatrix} 2 & 1 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{pmatrix}$ and $b = \begin{pmatrix} 8 \\ -15 \\ 25 \end{pmatrix}$. Calculate the value of $x_1 + x_2 + x_3$." answer="4" hint="First solve $Ly=b$ for $y$, then solve $Ux=y$ for $x$. Finally, sum the components of $x$." solution="
Step 1: Solve $Ly=b$ for the intermediate vector $y$.

From row 1: $y_1 = 8$.
From row 2: $-2y_1 + y_2 = -15 \implies -2(8) + y_2 = -15 \implies -16 + y_2 = -15 \implies y_2 = 1$.
From row 3: $3y_1 - y_2 + y_3 = 25 \implies 3(8) - 1 + y_3 = 25 \implies 23 + y_3 = 25 \implies y_3 = 2$.
So, $y = [8, 1, 2]^T$.

Step 2: Solve $Ux=y$ for the solution vector $x$.

From row 3: $2x_3 = 2 \implies x_3 = 1$.
From row 2: $x_2 - x_3 = 1 \implies x_2 - 1 = 1 \implies x_2 = 2$.
From row 1: $2x_1 + x_2 + 4x_3 = 8 \implies 2x_1 + 2 + 4(1) = 8 \implies 2x_1 + 6 = 8 \implies 2x_1 = 2 \implies x_1 = 1$.
So, $x = [1, 2, 1]^T$.

Step 3: Calculate the required sum.

Result: The value of the sum is 4.
"
:::

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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="A system of linear equations $Ax=b$ is to be solved, where the Doolittle LU decomposition of the matrix $A$ is given by $L = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & 1 & 1 \end{bmatrix}$ and $U = \begin{bmatrix} 2 & -1 & 3 \\ 0 & 3 & 1 \\ 0 & 0 & -2 \end{bmatrix}$. If the vector $b = \begin{bmatrix} 8 \\ 19 \\ 2 \end{bmatrix}$, what is the value of $x_3$ in the solution vector $x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$?" options=["-3.5","3.5","7.0","-2.0"] answer="A" hint="First, solve the system $Ly=b$ for the intermediate vector $y$ using forward substitution. Then, solve the system $Ux=y$ for the final solution vector $x$ using backward substitution." solution="The problem is to solve $Ax=b$, which is equivalent to solving $LUx=b$. We break this into two steps.

Step 1: Solve $Ly=b$ for $y$.
We have the system:

Using forward substitution:

• From the first row: $y_1 = 8$.


• From the second row: $2y_1 + y_2 = 19 \implies 2(8) + y_2 = 19 \implies 16 + y_2 = 19 \implies y_2 = 3$.


• From the third row: $-y_1 + y_2 + y_3 = 2 \implies -8 + 3 + y_3 = 2 \implies -5 + y_3 = 2 \implies y_3 = 7$.

Thus, the intermediate vector is $y = \begin{bmatrix} 8 \\ 3 \\ 7 \end{bmatrix}$.

Step 2: Solve $Ux=y$ for $x$.
We now solve the system:

Using backward substitution, we start from the last row to find $x_3$:

• From the third row: $-2x_3 = 7 \implies x_3 = -\frac{7}{2} = -3.5$.

The question only asks for $x_3$, so we can stop here. The correct answer is -3.5.
"
:::

:::question type="NAT" question="The determinant of the matrix $A = \begin{bmatrix} 2 & 1 & 1 \\ 4 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}$ is calculated using its LU decomposition. The value of the determinant is ____." answer="8" hint="Recall that $\det(A) = \det(U)$, assuming a Doolittle decomposition where the diagonal elements of $L$ are 1. You must first find the matrix $U$." solution="To find the determinant using LU decomposition, we first need to find the upper triangular matrix $U$. We use Gaussian elimination to transform $A$ into $U$.

Given matrix $A$:

Step 1: Perform row operations to create zeros in the first column below the diagonal.

• $R_2 \to R_2 - (4/2)R_1 \implies R_2 \to R_2 - 2R_1$


• $R_3 \to R_3 - (-2/2)R_1 \implies R_3 \to R_3 + R_1$

This yields the intermediate matrix:

Step 2: Perform row operations to create a zero in the second column below the diagonal.

• $R_3 \to R_3 - (3/-1)R_2 \implies R_3 \to R_3 + 3R_2$

This yields the final upper triangular matrix $U$:

The determinant of $A$ is the product of the diagonal elements of $U$.

The value of the determinant is 8.
"
:::

:::question type="MCQ" question="Which of the following statements regarding LU Decomposition is FALSE?" options=["For a fixed matrix $A$, solving $Ax=b$ for multiple vectors $b$ is more efficient using LU decomposition than by repeatedly applying Gaussian elimination.","A non-singular matrix $A$ has a unique LU decomposition.","LU decomposition may not exist for a matrix if one of its leading principal minors is zero.","The determinant of a matrix can be computed as the product of the diagonal entries of its $L$ and $U$ matrices."] answer="B" hint="Consider the conditions required to guarantee a unique decomposition, such as those specified in the Doolittle and Crout methods." solution="Let us analyze each statement:

• A: For a fixed matrix $A$, solving $Ax=b$ for multiple vectors $b$ is more efficient using LU decomposition than by repeatedly applying Gaussian elimination. This is TRUE. LU decomposition separates the $O(n^3)$ factorization step from the $O(n^2)$ substitution step. For multiple $b$ vectors, the expensive factorization is done only once. Repeated Gaussian elimination would perform the $O(n^3)$ work for each vector $b$.

• B: A non-singular matrix $A$ has a unique LU decomposition. This is FALSE. The decomposition $A=LU$ is not unique without additional constraints. For example, if $A=LU$, then for any non-zero scalar $\alpha$, $A = (L\alpha^{-1})(\alpha U)$ would be another valid decomposition. Uniqueness is achieved by imposing constraints, such as requiring the diagonal of $L$ to be all ones (Doolittle's method) or the diagonal of $U$ to be all ones (Crout's method). These two methods produce different (but unique) decompositions.

• C: LU decomposition may not exist for a matrix if one of its leading principal minors is zero. This is TRUE. The existence of an LU decomposition (without pivoting) is guaranteed if and only if all leading principal minors of the matrix are non-zero. A zero leading principal minor corresponds to a zero pivot element during the elimination process, which halts the standard algorithm.

• D: The determinant of a matrix can be computed as the product of the diagonal entries of its $L$ and $U$ matrices. This is TRUE. We know that $\det(A) = \det(LU) = \det(L)\det(U)$. Since $L$ and $U$ are triangular, their determinants are the products of their respective diagonal entries. Therefore, $\det(A) = (\prod l_{ii}) (\prod u_{ii})$.

The only false statement is B. " :::

:::question type="NAT" question="In the Doolittle LU decomposition ($L_{ii}=1$) of the matrix $A = \begin{bmatrix} 3 & -1 & 2 \\ 6 & -1 & 6 \\ -3 & 5 & 1 \end{bmatrix}$, the value of the element $l_{32}$ is ____." answer="4" hint="Calculate the elements of $L$ and $U$ sequentially. The element $l_{32}$ can be found after the first column of $L$ and the first two rows of $U$ are determined." solution="We are tasked with finding the element $l_{32}$ in the Doolittle decomposition of $A$, where $A=LU$ and $L$ is a unit lower triangular matrix.

We can determine the elements of $L$ and $U$ step-by-step.

Step 1: Determine the first row of $U$.
The first row of $U$ is the same as the first row of $A$.
$u_{11} = 3$, $u_{12} = -1$, $u_{13} = 2$.

Step 2: Determine the first column of $L$.
$l_{21} u_{11} = a_{21} \implies l_{21} \times 3 = 6 \implies l_{21} = 2$.
$l_{31} u_{11} = a_{31} \implies l_{31} \times 3 = -3 \implies l_{31} = -1$.

Step 3: Determine the second row of $U$.
$l_{21} u_{12} + u_{22} = a_{22} \implies (2)(-1) + u_{22} = -1 \implies -2 + u_{22} = -1 \implies u_{22} = 1$.
$l_{21} u_{13} + u_{23} = a_{23} \implies (2)(2) + u_{23} = 6 \implies 4 + u_{23} = 6 \implies u_{23} = 2$.

Step 4: Determine the element $l_{32}$.
The element $a_{32}$ is given by the matrix product:
$l_{31} u_{12} + l_{32} u_{22} = a_{32}$
Substituting the values we have found:
$(-1)(-1) + l_{32}(1) = 5$
$1 + l_{32} = 5$
$l_{32} = 4$

The value of the element $l_{32}$ is 4.
"
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