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Chapter 7: Problem 26

Find the best approximation to a solution of the given system of equations. $$\begin{aligned} 2 x+3 y+z &=21 \\ x+y+z &=7 \\ -x+y-z &=14 \\ 2 y+z &=0 \end{aligned}$$

Short Answer

Expert verified
The solution is \(x = 21\), \(y = 14\), \(z = -28\).

Step by step solution

01

Write the augmented matrix

The given system of equations can be represented as an augmented matrix: \[\begin{bmatrix} 2 & 3 & 1 & | & 21 \1 & 1 & 1 & | & 7 \-1 & 1 & -1 & | & 14 \0 & 2 & 1 & | & 0 \end{bmatrix}\]
02

Use Gaussian elimination to simplify

We will use row operations to simplify the matrix.- Subtract row 2 from row 1 to create a zero in the x column of row 2.- Add row 3 to row 1 to create a zero in row 3.- Simplify row 4 as well. This leads to the following matrix:\[\begin{bmatrix}2 & 3 & 1 & | & 21 \0 & 1 & 0 & | & 14 \0 & 4 & 0 & | & 7 \0 & 2 & 1 & | & 0\end{bmatrix}\]
03

Solve for one variable

Inspect the simplified matrix and solve for one of the variables, starting with those having a column of zeros beneath them. Here we solve for \( y \) using row 2:\[y = 14\]
04

Back-substitute to solve for remaining variables

Substitute \( y = 14 \) into another simplified equation. From row 4:\[2(14) + z = 0 \] Solve for \( z \):\[z = -28\] Substitute \( y = 14 \) and \( z = -28 \) into the original second row equation:\[x + 14 - 28 = 7 \]\[x = 21\]
05

Verify the solution in all equations

Plug \(x = 21, y = 14, z = -28\) back into the original equations to ensure all are satisfied.For the first equation:\[2(21) + 3(14) - 28 = 21\] which is true.For the third:\[-21 + 14 + 28 = 14\] which holds true. Thus, the solution set \(x = 21, y = 14, z = -28\) appropriately satisfies all equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Augmented Matrix
When dealing with systems of equations, an augmented matrix can be a powerful tool. Think of an augmented matrix as a compact and organized way to represent an entire system of linear equations.
It consists of the coefficients of the variables and the constants from each equation, arranged neatly in a grid.To create an augmented matrix, follow these basic steps:
  • Write the coefficients of each variable in the first columns.
  • In the last column, include the constants from each equation (the values on the right side of the equations).
In our example, the system of equations was converted into the following augmented matrix:\[\begin{bmatrix} 2 & 3 & 1 & | & 21 \1 & 1 & 1 & | & 7 \-1 & 1 & -1 & | & 14 \0 & 2 & 1 & | & 0 \\end{bmatrix}\]This format is convenient for applying methods like Gaussian elimination to solve the system.
System of Equations
A system of equations is like a puzzle where you're trying to find the values of variables that make all equations true at once.
These equations include the same variables, and your task is to find a set of variable values that satisfy each equation in the system. The solution to a system of linear equations can be approached using multiple methods:
  • Substitution, which involves solving one equation for a variable and then substituting that expression into another equation.
  • Elimination, where you eliminate one variable by adding or subtracting equations.
  • Matrix methods like augmented matrices and Gaussian elimination, an efficient way to handle multiple equations and variables.
In this exercise, we used an augmented matrix and Gaussian elimination to systematically reduce the system and find the variable values that solve all equations.
Back-Substitution
Back-substitution is a crucial step when solving systems of equations using the Gaussian elimination method.
After simplifying the augmented matrix, back-substitution allows us to solve for variables step-by-step, starting from the bottom of the matrix.Here's how back-substitution generally works:
  • Look at the equations from the bottom of the matrix upwards. You'll often start solving from the equation with only one variable first.
  • Use the value of the solved variable to find the next one by substituting it back into the previous equation.
  • Continue this process until all variable values are determined.
In our specific exercise, we started with the row from the reduced matrix to find \( y \), from there calculated \( z \) using previously found values, and finally determined \( x \).
This process helps to break down complex calculations into manageable parts, ensuring accuracy and clarity.

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Most popular questions from this chapter

Find the best linear approximation to f on the interval [-1,1]. $$f(x)=x^{3}$$

Find the best quadratic approximation to f on the interval [0,1]. $$f(x)=\sqrt{x}$$

If we multiply the Legendre polynomial of degree \(n\) by an appropriate scalar we can obtain a polynomial \(L_{n}(x)\) such that \(L_{m}(1)=1\) (a) Find \(L_{0}(x), L_{1}(x), L_{2}(x),\) and \(L_{3}(x)\) (b) It can be shown that \(L_{n}(x)\) satisfies the recurrence relation \\[L_{m}(x)=\frac{2 n-1}{n} x L_{n-1}(x)-\frac{n-1}{n} L_{n-2}(x)\\] for all \(n \geq 2 .\) Verify this recurrence for \(L_{2}(x)\) and \(L_{3}(x)\) Then use it to compute \(L_{4}(x)\) and \(L_{5}(x)\).

Find \(A^{+}\) and use it to compute the minimal length least squares solution to \(A \mathbf{x}=\mathbf{b}\) $$A=\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right], \mathbf{b}=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$$

Find the best linear approximation to fon the interval [-1,1]. $$f(x)=x^{2}+2 x$$
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