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Solve the given systems of equations by Gaussian elimination. If there is an unlimited number of solutions, find two of them. $$\begin{aligned} &s+2 t-3 u=2\\\ &6 t+3 s-9 u=6\\\ &7 s+14 t-21 u=13 \end{aligned}$$

Short Answer

Expert verified
The system has no solution due to a contradiction in its reduced form.

Step by step solution

01

Write the system in augmented matrix form

First, we will express the given system of equations as an augmented matrix. The system of equations is:\[ \begin{aligned} &s + 2t - 3u = 2\ &3s + 6t - 9u = 6 \ &7s + 14t - 21u = 13 \end{aligned} \]This corresponds to the augmented matrix:\[ \begin{bmatrix} 1 & 2 & -3 & | & 2 \ 3 & 6 & -9 & | & 6 \ 7 & 14 & -21 & | & 13 \end{bmatrix} \]Here, each row corresponds to an equation, with the vertical line separating the coefficients from the constants.
02

Simplify the matrix using Gaussian elimination

Next, we aim to get a row-echelon form where there are leading ones and zeros below them in the matrix. First, manipulate Row 2 by subtracting 3 times Row 1 from Row 2:\[ R_2 = R_2 - 3R_1 \]This gives:\[ \begin{bmatrix} 1 & 2 & -3 & | & 2 \ 0 & 0 & 0 & | & 0 \ 7 & 14 & -21 & | & 13 \end{bmatrix} \]Now, simplify Row 3 by subtracting 7 times Row 1 from Row 3:\[ R_3 = R_3 - 7R_1 \]This gives:\[ \begin{bmatrix} 1 & 2 & -3 & | & 2 \ 0 & 0 & 0 & | & 0 \ 0 & 0 & 0 & | & -1 \end{bmatrix} \]
03

Interpret the resulting matrix

In the resulting matrix, the second row consists entirely of zeros, indicating a possibility of infinite solutions, but the zero row corresponds correctly to this system without any contradictions in the constant terms.However, the third row ends with a non-zero constant term on the right-hand side of the augmented matrix, leading to the equation:\[ 0s + 0t + 0u = -1 \]This results in a contradiction, showing that no solution exists because an equation with only zeros on the left cannot equal a non-zero number.

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Key Concepts

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

Solving Systems of Equations
When we talk about solving systems of equations, we are trying to find the values for the variables that make each equation in the system true simultaneously. It might sound tricky, but it's a powerful method for understanding relationships between different quantities. For the given problem, we're looking at a system of three equations with three variables. Solving such systems can reveal a single solution, infinite solutions, or sometimes no solution at all.

To effectively tackle these types of problems, we use methods like substitution, elimination, or matrices. Matrices, in particular, can simplify the process by organizing and manipulating the equations systematically. Gaussian Elimination, which utilizes matrices, is especially useful for this as it practically walks us through reducing the numbers of variables step by step, making solutions more straightforward.
Augmented Matrix
An augmented matrix is a fundamental tool when working with systems of equations. It's a rectangular array that incorporates all coefficients and constants from the equations into a single grid-like form. This setup aids in simplifying the process through matrix manipulations, like those in Gaussian elimination.

In our problem, the given equations are represented as an augmented matrix:
  • Each row stands for a single equation.
  • The vertical line in the matrix separates the coefficients of the variables from the constant terms of each equation.
By converting equations into this form, we can more easily apply row operations to simplify and solve the system.
Row-Echelon Form
Row-echelon form is a specific arrangement of a matrix that simplifies solving systems of equations. It resembles an upper triangular arrangement in matrices. In this form:
  • Leading elements in each non-zero row, known as "leading ones," are 1, and they appear to the right of any leading one in the row right above it.
  • Any rows consisting all zeros, if present, are at the bottom of the matrix.
Achieving this structure using Gaussian Elimination involves systematic row operations like substitutions and row-swapping. Once in row-echelon form, further tweaking might lead to reduced row-echelon form if needed, offering a clearer view of solutions or contradictions in the system.
Contradiction in Equations
Contradictions in systems of equations occur when an impossible situation arises, such as a false equation like 0=1. In this exercise, when reducing the matrix, we encountered one such contradiction in the form of the equation: \[ 0s + 0t + 0u = -1 \] This equation is problematic because no possible values for \(s\), \(t\), or \(u\) could satisfy an equation where zero equals a nonzero number.

This contradiction helps us conclude that the system of equations has no solution. Understanding contradictions is crucial as it tells us that the system simply has no set of solutions that satisfy all equations simultaneously. This scenario can often happen in systems with parallel lines or planes when graphed, indicating they never meet or intersect.

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Most popular questions from this chapter

Solve the given systems of equations by using the inverse of the coefficient matrix. Use a calculator to perform the necessary matrix operations and display the results and the check. $$\begin{aligned} &2 x-y-z=7\\\ &4 x-3 y+2 z=4\\\ &3 x+5 y+z=-10 \end{aligned}$$

Perform the indicated matrix operations. The inventory of a drug supply company shows that the following numbers of cases of bottles of vitamins \(\mathrm{C}\) and \(\mathrm{B}_{3}\) (niacin) are in stock: Vitamin \(\mathrm{C}-25\) cases of \(100-\mathrm{mg}\) bottles, 10 cases of 250-mg bottles, and 32 cases of 500-mg bottles; vitamin B \(_{3}-\) 30 cases of \(100-\mathrm{mg}\) bottles, 18 cases of \(250-\mathrm{mg}\) bottles, and 40 cases of \(500-\mathrm{mg}\) bottles. This is represented by matrix \(A\) below. After two shipments are sent out, each of which can be represented by matrix \(B\) below, find the matrix that represents the remaining inventory. $$A=\left[\begin{array}{lll} 25 & 10 & 32 \\ 30 & 18 & 40 \end{array}\right] \quad B=\left[\begin{array}{lll} 10 & 5 & 6 \\ 12 & 4 & 8 \end{array}\right]$$

Perform the indicated matrix multiplications. Evaluate \(x, y, z\) and \(t\) such that \(\left[\begin{array}{rr}1 & -1 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}x & y \\ z & t\end{array}\right]=\left[\begin{array}{rr}2 & -3 \\ 7 & 0\end{array}\right]\).

Solve the given systems of equations by using the inverse of the coefficient matrix. Use a calculator to perform the necessary matrix operations and display the results and the check. $$\begin{aligned} &4 x+2 y-2 z=2\\\ &3 x-2 y-8 z=-3\\\ &x+3 y+z=10 \end{aligned}$$

Perform the indicated matrix multiplications. For matrices \(A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]\) and \(B=\left[\begin{array}{ll}c & d \\ d & c\end{array}\right],\) show that \(A B=B A\)

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