Problem 34 - Differential Equations and Linear Algebra

Question

(a) An \(n \times n\) system of linear equations whose matrix of coefficients is a lower triangular matrix is called a lower triangular system. Assuming that \(a_{i i} \neq 0\) for each \(i,\) devise a method for solving such a system that is analogous to the back substitution method. (b) Use your method from (a) to solve $$ \begin{aligned} x_{1} &=2 \\ 2 x_{1}-3 x_{2} &=1 \\ 3 x_{1}+x_{2}-x_{3} &=8 \end{aligned} $$

Step-by-Step Solution

Answer

To solve a lower triangular system of linear equations, we can use forward substitution. In this method, we solve the equations in their natural order, substituting the previously found variable values into the following equations. For the given system: $ \begin{aligned} x_{1} &=2 \\ 2 x_{1}-3 x_{2} &=1 \\ 3 x_{1}+x_{2}-x_{3} &=8 \end{aligned} $ We find \(x_1 = 2\), \(x_2 = 1\), and \(x_3 = -1\). So, the solution is \((x_1, x_2, x_3) = (2, 1, -1)\).

Step 1: Understanding the Lower Triangular System

A lower triangular matrix is a special type of square matrix where all the elements above the diagonal are zero. A lower triangular system of linear equations is one whose matrix of coefficients is a lower triangular matrix. The main feature of this system is that its variables are arranged in decreasing order of importance, which means that each equation has a variable that affects only it and no other variable depends on it. Since the coefficients matrix is lower triangular and nonzero on the diagonal of the matrix, we can preprocess a method to solve such systems quite easily.

Step 2: Devise a method for solving a lower triangular system

The method we will use to solve the lower triangular system is called forward substitution. It is analogous to back-substitution, but in this case, we substitute the variables in their natural order (from first to last). Here's how it works: 1. Start with the first equation. Since it contains only one variable (the first variable), we can solve it directly. 2. Move on to the next equation. Substitute the value of the first variable that we found in step 1 to solve for the second variable. 3. Continue this process, substituting the values of the previously solved variables into the following equations to find the next variable, until all variables are found.

Step 3: Applying the method to the given system of linear equations

We will now apply the forward substitution method to solve the given system of linear equations: $ \begin{aligned} x_{1} &=2 \\ 2 x_{1}-3 x_{2} &=1 \\ 3 x_{1}+x_{2}-x_{3} &=8 \end{aligned} $ 1. Solve for \(x_1\): From the first equation, we have: \(x_1 = 2\) 2. Solve for \(x_2\): Substitute the value of \(x_1\) into the second equation and solve for \(x_2\): \(2(2) - 3x_2 = 1\) \(4 - 3x_2 = 1\) \(3x_2 = 3\) \(x_2 = 1\) 3. Solve for \(x_3\): Substitute the values of \(x_1\) and \(x_2\) into the third equation and solve for \(x_3\): \(3(2) + 1(1) - x_3 = 8\) \(6 + 1 - x_3 = 8\) \(x_3 = -1\) So, the solution to the system of linear equations is: \((x_1, x_2, x_3) = (2,1,-1)\)

Key Concepts

Lower Triangular Matrix

Imagine staring at a square matrix, and you see that all the numbers above the diagonal running from the top left to the bottom right are zeroes. This isn't just any old matrix—it's a lower triangular matrix. What makes it special? It lays out a system of linear equations in a way that's a bit like a ladder. Each step (or row) leads to the next, only leaning on the steps below.

In more technical terms, if you were to write out a general lower triangular matrix, it would look something like this:
\[ \begin{pmatrix} a_{11} & 0 & \cdots & 0 \ a_{21} & a_{22} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ a_{n1} & \cdots & a_{n,n-1} & a_{nn} \end{pmatrix} \]
What's crucial here is that each diagonal element, \( a_{ii} \), is non-zero. Why is that a big deal? It's essential for the method we use to solve these systems, known as forward substitution. Each non-zero diagonal allows us to solve the equation at each step without worrying about division by zero, a definite 'no-no' in math.

Forward Substitution

Forward substitution is like solving a mystery by following breadcrumb clues in order. It's a method used specifically to handle lower triangular systems and works by moving, yep, you guessed it—forward!

Let's get a bit more detailed:

• You start with the first equation, which helpfully has only one unknown, so you solve it outright without a sweat.
• Next up, the second equation. It's a little trickier because it has two unknowns. But since you already found the first one in the previous step, you only have one new unknown to solve for.
• This process continues equation by equation, with each new step only introducing one new unknown. Just plug in the answers you already have, and solve for the new piece of the puzzle.

It's methodical and straightforward—exactly what's needed for equations that line up in an orderly queue.

System of Linear Equations

Linear equations are the bread and butter of algebra—the simple straight line relationships between x's and y's that form the basis for so much of what comes later in math. But when you've got more than one of them and they're all linked together, welcome to the world of a system of linear equations. These systems can seem daunting. Imagine multiple lines on a graph; where they meet is what you're looking to find.

Now, how do we solve these systems when they're all tangled up? If the equations form a lower triangular matrix, we've got a clear path forward—forward substitution. And remember, there might be unique, infinite, or no solutions at all, depending on how these lines relate to each other. So what's the beauty of lower triangular systems? They neatly sidestep the chaos—each equation peels away, layer by layer, to reveal the solution without any guesswork.