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Chapter 9: Problem 36

Use the Gauss-Jordan method to solve each system of equations. For systems in two variables with infinitely many solutions, give the solution with y arbitrary. For systems in three variables with infinitely many solutions, give the solution with z arbitrary. $$\begin{array}{l}3 x+y+3 z-1=0 \\\x+2 y-z-2=0 \\\2 x-y+4 z-4=0\end{array}$$

Short Answer

Expert verified
The solution is \(x = -\frac{77}{20},\ y = \frac{43}{20},\ z = \frac{11}{4}\).

Step by step solution

01

Write the augmented matrix

Start by writing the augmented matrix for the given system of equations:\[\begin{bmatrix}3 & 1 & 3 & | & 1 \1 & 2 & -1 & | & 2 \2 & -1 & 4 & | & 4\end{bmatrix}\]
02

Perform row operations to get a leading 1 in the first row, first column

Since the first element of the first row is already 3, divide the whole row by 3:\[R1 = \frac{1}{3}R1: \begin{bmatrix} 1 & \frac{1}{3} & 1 & | & \frac{1}{3} \1 & 2 & -1 & | & 2 \2 & -1 & 4 & | & 4 \end{bmatrix}\]
03

Eliminate the first column under the first row

Perform row operations to make all other elements in the first column 0:\[R2 = R2 - R1: \begin{bmatrix} 1 & \frac{1}{3} & 1 & | & \frac{1}{3} \0 & \frac{5}{3} & -2 & | & \frac{5}{3} \2 & -1 & 4 & | & 4 \end{bmatrix}\]\[R3 = R3 - 2R1: \begin{bmatrix} 1 & \frac{1}{3} & 1 & | & \frac{1}{3} \0 & \frac{5}{3} & -2 & | & \frac{5}{3} \0 & -\frac{7}{3} & 2 & | & \frac{10}{3} \end{bmatrix}\]
04

Perform row operations to get a leading 1 in the second row, second column

To get a leading 1 in the second row, divide the second row by \(\frac{5}{3}\):\[R2 = \frac{3}{5}R2: \begin{bmatrix} 1 & \frac{1}{3} & 1 & | & \frac{1}{3} \0 & 1 & -\frac{6}{5} & | & 1 \0 & -\frac{7}{3} & 2 & | & \frac{10}{3} \end{bmatrix}\]
05

Eliminate the second column elements in other rows

Perform row operations to make all other elements in the second column 0:\[R1 = R1 - \frac{1}{3}R2: \begin{bmatrix} 1 & 0 & \frac{7}{5} & | & 0 \0 & 1 & -\frac{6}{5} & | & 1 \0 & -\frac{7}{3} & 2 & | & \frac{10}{3} \end{bmatrix}\]\[R3 = R3 + \frac{7}{3}R2: \begin{bmatrix} 1 & 0 & \frac{7}{5} & | & 0 \0 & 1 & -\frac{6}{5} & | & 1 \0 & 0 & \frac{4}{5} & | & \frac{11}{5} \end{bmatrix}\]
06

Get a leading 1 in the third row, third column

Divide the third row by \(\frac{4}{5}\):\[R3 = \frac{5}{4}R3: \begin{bmatrix} 1 & 0 & \frac{7}{5} & | & 0 \0 & 1 & -\frac{6}{5} & | & 1 \0 & 0 & 1 & | & \frac{11}{4} \end{bmatrix}\]
07

Eliminate the third column elements in other rows

Perform row operations to make all other elements in the third column 0:\[R1 = R1 - \frac{7}{5}R3: \begin{bmatrix} 1 & 0 & 0 & | & -\frac{77}{20} \0 & 1 & -\frac{6}{5} & | & 1 \0 & 0 & 1 & | & \frac{11}{4} \end{bmatrix}\]\[R2 = R2 + \frac{6}{5}R3: \begin{bmatrix} 1 & 0 & 0 & | & -\frac{77}{20} \0 & 1 & 0 & | & \frac{43}{20} \0 & 0 & 1 & | & \frac{11}{4} \end{bmatrix}\]
08

Interpret the results

The system is now in reduced row echelon form. This means:\[x = -\frac{77}{20},\ y = \frac{43}{20},\ z = \frac{11}{4}\]

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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 is a set of two or more equations with the same set of variables. In our exercise, we have a system comprising three equations with three variables: x, y, and z. Solving a system of equations involves finding the values of the variables that satisfy all equations simultaneously. These systems can have a unique solution, infinitely many solutions, or no solution at all. In our case, we aim to find the unique solution using the Gauss-Jordan method.
augmented matrix
An augmented matrix is a compact way to handle systems of equations. It includes both the coefficients of the variables and the constants from the equations. For the given system of equations:
3x + y + 3z - 1 = 0
x + 2y - z - 2 = 0
2x - y + 4z - 4 = 0
We convert this into an augmented matrix as follows:
 \[\begin{bmatrix} 3 & 1 & 3 & | & 1 \ 1 & 2 & -1 & | & 2 \ 2 & -1 & 4 & | & 4 \ \end{bmatrix}\]
The bar separates the constants on the right side of the equations.
row operations
Row operations are techniques used to simplify matrices. These operations help transform augmented matrices into a form where solutions are easily identified. The three types of row operations are:
  • Swapping two rows
  • Multiplying a row by a nonzero scalar
  • Adding or subtracting multiples of one row from another row
In this exercise, these operations are used iteratively to reach a more straightforward form known as reduced row echelon form (RREF).
reduced row echelon form
Reduced row echelon form (RREF) is a specific form of an augmented matrix that makes it easy to interpret the solutions of a system of equations. A matrix is in RREF if it satisfies the following conditions:
  • Every leading entry (first nonzero number from the left, in a row) is 1.
  • The leading entry in each row is the only nonzero entry in its column.
  • The leading entry in each row is to the right of the leading entry in the row above it.
  • Any rows of all zeros are at the bottom of the matrix.
Following these conditions, the RREF of our matrix:\[\begin{bmatrix} 1 & 0 & 0 & | & -\frac{77}{20} \ 0 & 1 & 0 & | & \frac{43}{20} \ 0 & 0 & 1 & | & \frac{11}{4} \ \end{bmatrix}\]This tells us directly that x = -\frac{77}{20}, y = \frac{43}{20}, and z = \frac{11}{4}.

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Most popular questions from this chapter

Supply and Demand In many applications of economics, as the price of an item goes up, demand for the item goes down and supply of the item goes up. The price where supply and demand are equal is the equilibrium price, and the resulting sup. ply or demand is the equilibrium supply or equilibrium demand. Suppose the supply of a product is related to its price by the equation $$p=\frac{2}{3} q$$ where \(p\) is in dollars and \(q\) is supply in appropriate units. (Here, \(q\) stands for quantity.) Furthermore, suppose demand and price for the same product are related by $$p=-\frac{1}{3} q+18$$ where \(p\) is price and \(q\) is demand. The system formed by these two equations has solution \((18,12),\) as seen in the graph. (GRAPH CANNOT COPY) Suppose the demand and price for a certain model of electric can opener are related by \(p=16-\frac{5}{4} q\), where \(p\) is price, in dollars, and \(q\) is demand, in appropriate units. Find the price when the demand is at each level. (a) 0 units (b) 4 units (c) 8 units

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{array}{r} 2 x-y+4 z=-2 \\ 3 x+2 y-z=-3 \\ x+4 y+2 z=17 \end{array}$$

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{aligned} &4 x-3 y+z+1=0\\\ &\begin{array}{l} 5 x+7 y+2 z+2=0 \\ 3 x-5 y-z-1=0 \end{array} \end{aligned}$$

Use the determinant theorems to find the value of each determinant. $$\left|\begin{array}{rrrr} 3 & -6 & 5 & -1 \\ 0 & 2 & -1 & 3 \\ -6 & 4 & 2 & 0 \\ -7 & 3 & 1 & 1 \end{array}\right|$$

$$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right], \quad B=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right], \quad \text { and } \quad C=\left[\begin{array}{ll} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array}\right] $$ where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \((A B) C=A(B C)\) (associative property)
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