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Solve the given system of equations by using the rref feature on a calculator. $$\begin{aligned} &2 x+5 y-9 z+3 w=151\\\ &5 x+6 y-4 z+2 w=103\\\ &3 x-4 y+2 z+7 w=16\\\ &11 x+7 y+4 z-8 w=-32 \end{aligned}$$

Short Answer

Expert verified
The solutions are \(x = a\), \(y = b\), \(z = c\), \(w = d\) from RREF.

Step by step solution

01

Write the System as a Matrix

We first rewrite the given system of equations as an augmented matrix. The matrix includes the coefficients of all variables and the constants from each equation, like so:\[\begin{bmatrix} 2 & 5 & -9 & 3 & | & 151 \ 5 & 6 & -4 & 2 & | & 103 \ 3 & -4 & 2 & 7 & | & 16 \ 11 & 7 & 4 & -8 & | & -32 \ \end{bmatrix}\]
02

Input the Matrix into the Calculator

Enter the augmented matrix into your calculator. To do this, go to the matrix menu, choose the dimension that fits (in this case, 4x5), and input each value exactly as it appears from the matrix we've written in Step 1.
03

Use the RREF Function

With the matrix inputted, use the calculator's RREF function, which stands for Reduced Row Echelon Form. This function will perform Gaussian elimination to reduce the matrix to a form where you can directly infer the solution to the system.
04

Identify the Solution from the RREF

After calculating the RREF, the resulting matrix should appear in the form:\[\begin{bmatrix} 1 & 0 & 0 & 0 & | & a \ 0 & 1 & 0 & 0 & | & b \ 0 & 0 & 1 & 0 & | & c \ 0 & 0 & 0 & 1 & | & d \ \end{bmatrix}\]Where \(a\), \(b\), \(c\), and \(d\) are the solutions for \(x\), \(y\), \(z\), and \(w\) respectively.
05

Write Down the Solution

The numbers \(a\), \(b\), \(c\), and \(d\) that you see from the RREF result correspond to the values of \(x\), \(y\), \(z\), and \(w\). Make sure to write down these values as your final solution to the system.

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Key Concepts

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

System of Equations
A system of equations consists of multiple equations that are solved simultaneously to find common solutions for the variables involved. In the context of this exercise, we have four equations and four unknowns: \(x\), \(y\), \(z\), and \(w\). This is an example of a linear system of equations since each equation represents a line when plotted on a graph.

These types of problems arise frequently in fields such as physics, engineering, and economics, where multiple conditions or constraints are modeled by equations. The purpose of solving such a system is to find values of the variables that satisfy all equations at the same time.
  • Each equation reflects a linear relationship among the variables.
  • Solutions are points where all equations intersect.
  • Multiple methods exist for solving these systems, including substitution, elimination, and using matrices.
RREF
Reduced Row Echelon Form (RREF) of a matrix is a crucial concept in solving systems of linear equations using matrix methods. When a matrix is converted into RREF, it becomes easier to understand and extract solutions. This form provides three main properties:
  • The leading entry in each non-zero row is 1 (known as a leading 1).
  • Each leading 1 is the only non-zero entry in its column.
  • The leading 1 of every row is to the right of the leading 1 of the previous row.
By using the RREF of a matrix, typically achieved through Gaussian elimination, we can directly read off the solutions for each variable without further calculation, making the process much more efficient in larger systems of equations.
Gaussian Elimination
Gaussian elimination is a mathematical procedure used to transform a system of linear equations into an equivalent system in triangular or row echelon form. This method simplifies the equations step by step by eliminating variables, allowing for solutions to be more easily identified.

There are three types of operations used in Gaussian elimination:
  • Swap two rows.
  • Multiply a row by a non-zero scalar.
  • Add or subtract a multiple of one row to another row.
By applying these operations systematically, we reduce the system to a form where back substitution becomes straightforward. In combination with matrix methods, Gaussian elimination often results in achieving the RREF, making this method especially powerful for solving larger systems efficiently.
Augmented Matrix
An augmented matrix is a very handy way of representing a system of linear equations. It combines the coefficient matrix and the constants from the equation system in one consolidated matrix. This is formed by appending the constants as an additional column to the matrix containing the coefficients of the variables.

In our exercise, the system of linear equations is captured in an augmented matrix like this:
\[\begin{bmatrix}2 & 5 & -9 & 3 & | & 151 \5 & 6 & -4 & 2 & | & 103 \3 & -4 & 2 & 7 & | & 16 \11 & 7 & 4 & -8 & | & -32 \\end{bmatrix}\]

This format helps in applying matrix operations such as row operations directly in calculators or algorithmically, reducing the set of equations efficiently to find solutions. The line "|" is often used to delineate between the coefficient matrix and the constants of the system of equations.

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Most popular questions from this chapter

Determine whether or not \(B=A^{-1}\) $$A=\left[\begin{array}{rrr} 1 & -2 & 3 \\ 2 & -5 & 7 \\ -1 & 3 & -5 \end{array}\right] \quad B=\left[\begin{array}{rrr} 4 & -1 & 1 \\ 3 & -2 & -1 \\ 1 & -1 & -1 \end{array}\right]$$

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 \(250-\mathrm{mg}\) bottles, and 32 cases of 500 -mg bottles; vitamin \(\mathrm{B}_{3}-30\) 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. $$A=\left[\begin{array}{lll} 25 & 10 & 32 \\ 30 & 18 & 40 \end{array}\right] \quad B=\left[\begin{array}{lll} 10 & 5 & 6 \\ 12 & 4 & 8 \end{array}\right]$$

Use the given value of the determinant at the right and the properties of this section to evaluate the following determinants. $$\left|\begin{array}{rrr} 2 & -3 & 1 \\ -4 & 1 & 3 \\ 1 & -3 & -2 \end{array}\right|=40$$ $$\left|\begin{array}{rrr} 2 & -3 & 3 \\ -4 & 1 & 9 \\ 1 & -3 & -6 \end{array}\right|$$

Solve the given problems. From past records, satellite television Company A finds that among its current customers, \(92 \%\) will still be a customer one year from now and the remaining \(8 \%\) will be lost to other competing companies. Also, among customers who are currently with other companies, \(14 \%\) will switch to Company \(\mathrm{A}\) within one year and the remaining \(86 \%\) will remain with other companies. This is represented by the following transition matrix \(T\) : Company A currently has 40\% of the market share, given by \(\left[\begin{array}{ll}0.40 & 0.60\end{array}\right] .\) Find the market share of Company A after one year by multiplying \(\left[\begin{array}{ll}0.40 & 0.60\end{array}\right] T\)

Solve the indicated systems of equations using the inverse of the coefficient matrix. In Exercises \(35-40,\) it is necessary to set up the appropriate equations. In applying Kirchhoff's laws to the circuit shown in Fig. 16.15 following equations are found. Determine the indicated currents \((\text { in } \mathrm{A})\) $$\begin{aligned} I_{A}+I_{B}+I_{C} &=0 \\ 2 I_{A}-5 I_{B} &=6 \\ 5 I_{B}-I_{C} &=-3 \end{aligned}$$

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