Precalculus Systems of Equations Gaussian Elimination How Do You Know Which Equation to Replace?

Learning Objectives

By the finish of this section, you will be able to:

• Write the augmented matrix of a system of equations.
• Write the arrangement of equations from an augmented matrix.
• Perform row operations on a matrix.
• Solve a system of linear equations using matrices.

Figure 1. German mathematician Carl Friedrich Gauss (1777–1855).

Carl Friedrich Gauss lived during the late 18th century and early on 19th century, merely he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such equally algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries.

We first encountered Gaussian emptying in Systems of Linear Equations: Two Variables. In this department, we will revisit this technique for solving systems, this time using matrices.

The Augmented Matrix of a System of Equations

A matrix can serve as a device for representing and solving a organisation of equations. To express a system in matrix grade, we excerpt the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to divide the coefficient entries from the constants, substantially replacing the equal signs. When a system is written in this form, nosotros call it an augmented matrix.

For example, consider the following [latex]2\times 2[/latex] organisation of equations.

[latex]\begin{assortment}{l}3x+4y=7\\ 4x - 2y=5\end{array}[/latex]

We can write this system as an augmented matrix:

[latex]\left[\begin{assortment}{rr}\hfill 3& \hfill 4\\ \hfill iv& \hfill -2\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 7\\ \hfill 5\finish{array}\right][/latex]

Nosotros can also write a matrix containing just the coefficients. This is called the coefficient matrix.

[latex]\left[\begin{array}{cc}iii& iv\\ 4& -2\end{array}\right][/latex]

A three-past-iii system of equations such equally

[latex]\begin{array}{l}3x-y-z=0\hfill \\ \text{ }x+y=five\hfill \\ \text{ }2x - 3z=2\hfill \end{array}[/latex]

has a coefficient matrix

[latex]\left[\begin{array}{rrr}\hfill 3& \hfill -ane& \hfill -1\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill -iii\cease{assortment}\right][/latex]

and is represented by the augmented matrix

[latex]\left[\brainstorm{array}{rrr}\hfill iii& \hfill -one& \hfill -one\\ \hfill 1& \hfill ane& \hfill 0\\ \hfill ii& \hfill 0& \hfill -3\terminate{array}\text{ }|\text{ }\begin{array}{r}\hfill 0\\ \hfill five\\ \hfill 2\end{assortment}\correct][/latex]

Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first cavalcade, y-terms in the 2nd column, and z-terms in the third cavalcade. It is very of import that each equation is written in standard form [latex]ax+past+cz=d[/latex] so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.

How To: Given a organization of equations, write an augmented matrix.

1. Write the coefficients of the x-terms every bit the numbers down the starting time cavalcade.
2. Write the coefficients of the y-terms equally the numbers down the second cavalcade.
3. If at that place are z-terms, write the coefficients as the numbers down the third cavalcade.
4. Draw a vertical line and write the constants to the correct of the line.

Example 1: Writing the Augmented Matrix for a Organization of Equations

Write the augmented matrix for the given system of equations.

[latex]\brainstorm{assortment}{l}\text{ }10+2y-z=3\hfill \\ \text{ }2x-y+2z=six\hfill \\ \text{ }x - 3y+3z=4\hfill \end{array}[/latex]

Solution

The augmented matrix displays the coefficients of the variables, and an additional column for the constants.

[latex]\left[\brainstorm{assortment}{rrr}\hfill ane& \hfill 2& \hfill -i\\ \hfill 2& \hfill -1& \hfill 2\\ \hfill i& \hfill -3& \hfill 3\stop{array}\text{ }|\text{ }\begin{array}{r}\hfill iii\\ \hfill vi\\ \hfill four\end{assortment}\right][/latex]

Try It 1

Write the augmented matrix of the given system of equations.

[latex]\begin{assortment}{l}4x - 3y=xi\\ 3x+2y=4\terminate{assortment}[/latex]

Writing a System of Equations from an Augmented Matrix

Nosotros can apply augmented matrices to assistance u.s.a. solve systems of equations considering they simplify operations when the systems are not encumbered by the variables. Still, it is important to sympathise how to movement back and forth betwixt formats in order to make finding solutions smoother and more than intuitive. Here, nosotros volition use the information in an augmented matrix to write the system of equations in standard form.

Example ii: Writing a System of Equations from an Augmented Matrix Course

Find the system of equations from the augmented matrix.

[latex]\left[\begin{assortment}{rrr}\hfill ane& \hfill -three& \hfill -5\\ \hfill ii& \hfill -5& \hfill -4\\ \hfill -3& \hfill 5& \hfill 4\terminate{array}\text{ }|\text{ }\begin{array}{r}\hfill -2\\ \hfill 5\\ \hfill half-dozen\stop{array}\correct][/latex]

Solution

When the columns stand for the variables [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex],

[latex]\left[\begin{assortment}{rrr}\hfill 1& \hfill -3& \hfill -five\\ \hfill 2& \hfill -v& \hfill -four\\ \hfill -iii& \hfill five& \hfill 4\cease{array}\text{ }|\text{ }\begin{array}{r}\hfill -2\\ \hfill 5\\ \hfill 6\end{array}\right]\to \begin{array}{l}ten - 3y - 5z=-2\hfill \\ 2x - 5y - 4z=5\hfill \\ -3x+5y+4z=6\hfill \end{array}[/latex]

Try It 2

Write the system of equations from the augmented matrix.

[latex]\left[\brainstorm{array}{ccc}1& -1& 1\\ two& -i& 3\\ 0& 1& i\end{array}|\begin{array}{c}five\\ 1\\ -9\end{array}\right][/latex]

 Performing Row Operations on a Matrix

At present that nosotros can write systems of equations in augmented matrix grade, we will examine the various row operations that can be performed on a matrix, such every bit addition, multiplication by a abiding, and interchanging rows.

Performing row operations on a matrix is the method we apply for solving a organisation of equations. In club to solve the organization of equations, we want to convert the matrix to row-echelon form, in which in that location are ones down the chief diagonal from the upper left corner to the lower right corner, and zeros in every position below the principal diagonal as shown.

[latex]\begin{array}{c}\text{Row-echelon form}\\ \left[\begin{array}{ccc}i& a& b\\ 0& 1& d\\ 0& 0& 1\finish{array}\right]\end{assortment}[/latex]

Nosotros apply row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Hither are the guidelines to obtaining row-echelon form.

1. In any nonzero row, the get-go nonzero number is a 1. It is called a leading 1.
2. Whatever all-naught rows are placed at the bottom on the matrix.
3. Any leading 1 is beneath and to the right of a previous leading ane.
4. Any cavalcade containing a leading 1 has zeros in all other positions in the column.

To solve a system of equations we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-exchange to find the solution.

1. Interchange rows. (Note: [latex]{R}_{i}\leftrightarrow {R}_{j}[/latex] )
2. Multiply a row by a abiding. (Annotation: [latex]c{R}_{i}[/latex] )
3. Add the product of a row multiplied past a constant to another row. (Note: [latex]{R}_{i}+c{R}_{j}[/latex])

Each of the row operations corresponds to the operations we have already learned to solve systems of equations in iii variables. With these operations, in that location are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the commencement entry so that row one can be used to convert the remaining rows.

A General Note: Gaussian Elimination

The Gaussian emptying method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix [latex]A[/latex] with the number i equally the entry down the main diagonal and accept all zeros below.

[latex]A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\\ \hfill {a}_{31}& \hfill {a}_{32}& \hfill {a}_{33}\finish{assortment}\right]\stackrel{\text{After Gaussian elimination}}{\to }A=\left[\begin{array}{rrr}\hfill one& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill 0& \hfill ane& \hfill {b}_{23}\\ \hfill 0& \hfill 0& \hfill ane\finish{array}\correct][/latex]

The first stride of the Gaussian strategy includes obtaining a ane every bit the offset entry, so that row one may exist used to alter the rows below.

How To: Given an augmented matrix, perform row operations to reach row-echelon form.

1. The commencement equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.
2. Use row operations to obtain zeros down the first column below the commencement entry of 1.
3. Use row operations to obtain a ane in row two, column 2.
4. Use row operations to obtain zeros downwardly column 2, beneath the entry of 1.
5. Utilize row operations to obtain a 1 in row 3, column 3.
6. Keep this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below.
7. If whatever rows contain all zeros, place them at the bottom.

Example 2: Solving a [latex]2\times 2[/latex] Arrangement past Gaussian Emptying

Solve the given system by Gaussian elimination.

[latex]\brainstorm{assortment}{l}2x+3y=six\hfill \\ \text{ }x-y=\frac{ane}{ii}\hfill \end{array}[/latex]

Solution

Beginning, we write this as an augmented matrix.

[latex]\left[\brainstorm{array}{rr}\hfill ii& \hfill 3\\ \hfill 1& \hfill -1\cease{array}\text{ }|\text{ }\begin{array}{r}\hfill 6\\ \hfill \frac{1}{ii}\end{array}\right][/latex]

We want a i in row 1, column 1. This can be accomplished by interchanging row 1 and row two.

[latex]{R}_{i}\leftrightarrow {R}_{2}\to \left[\begin{assortment}{rrr}\hfill 1& \hfill -i& \hfill \\ \hfill 2& \hfill three& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill half dozen\end{array}\correct][/latex]

We now have a ane as the start entry in row 1, column 1. Now allow'southward obtain a 0 in row two, cavalcade 1. This tin can exist accomplished by multiplying row i by [latex]-ii[/latex], and then adding the result to row 2.

[latex]-two{R}_{1}+{R}_{2}={R}_{two}\to \left[\brainstorm{array}{rrr}\hfill 1& \hfill -ane& \hfill \\ \hfill 0& \hfill 5& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{ii}\\ \hfill & \hfill five\cease{assortment}\right][/latex]

Nosotros simply have one more than step, to multiply row 2 by [latex]\frac{ane}{5}[/latex].

[latex]\frac{1}{v}{R}_{2}={R}_{ii}\to \left[\begin{array}{rrr}\hfill one& \hfill -1& \hfill \\ \hfill 0& \hfill 1& \hfill \finish{array}|\begin{array}{cc}& \frac{1}{ii}\\ & i\terminate{array}\right][/latex]

Apply back-substitution. The second row of the matrix represents [latex]y=ane[/latex]. Back-substitute [latex]y=one[/latex] into the first equation.

[latex]\begin{array}{l}x-\left(1\right)=\frac{ane}{2}\hfill \\ \text{ }x=\frac{iii}{2}\hfill \end{assortment}[/latex]

The solution is the point [latex]\left(\frac{3}{2},1\right)[/latex].

Endeavour It 3

Solve the given system by Gaussian elimination.

[latex]\begin{array}{fifty}4x+3y=11\hfill \\ \text{ }\text{}\text{}x - 3y=-ane\hfill \terminate{array}[/latex]

Example 3: Using Gaussian Elimination to Solve a System of Equations

Use Gaussian elimination to solve the given [latex]two\times ii[/latex] organisation of equations.

[latex]\brainstorm{assortment}{l}\text{ }2x+y=ane\hfill \\ 4x+2y=6\hfill \end{array}[/latex]

Solution

Write the system every bit an augmented matrix.

[latex]\left[\begin{array}{ll}two\hfill & 1\hfill \\ 4\hfill & 2\hfill \end{array}\text{ }|\text{ }\brainstorm{array}{l}1\hfill \\ 6\hfill \end{array}\correct][/latex]

Obtain a 1 in row one, column 1. This can be accomplished by multiplying the first row by [latex]\frac{1}{two}[/latex].

[latex]\frac{1}{2}{R}_{1}={R}_{1}\to \left[\begin{array}{cc}ane& \frac{1}{2}\\ 4& 2\terminate{array}\text{ }|\text{ }\begin{array}{c}\frac{i}{ii}\\ vi\end{assortment}\right][/latex]

Next, nosotros want a 0 in row ii, column 1. Multiply row 1 by [latex]-four[/latex] and add row 1 to row ii.

[latex]-4{R}_{one}+{R}_{ii}={R}_{2}\to \left[\begin{array}{cc}one& \frac{1}{two}\\ 0& 0\stop{array}\text{ }|\text{ }\begin{array}{c}\frac{1}{2}\\ 4\terminate{array}\right][/latex]

The second row represents the equation [latex]0=4[/latex]. Therefore, the arrangement is inconsistent and has no solution.

Instance 4: Solving a Dependent System

Solve the organization of equations.

[latex]\begin{array}{l}3x+4y=12\\ 6x+8y=24\finish{array}[/latex]

Solution

Perform row operations on the augmented matrix to try and achieve row-echelon form.

[latex]A=\left[\begin{assortment}{llll}3\hfill & \hfill & 4\hfill & \hfill \\ 6\hfill & \hfill & eight\hfill & \hfill \finish{array}|\begin{array}{ll}\hfill & 12\hfill \\ \hfill & 24\hfill \terminate{array}\correct][/latex]

[latex]\begin{array}{l}\hfill \\ \begin{assortment}{l}-\frac{1}{2}{R}_{2}+{R}_{i}={R}_{1}\to \left[\brainstorm{assortment}{llll}0\hfill & \hfill & 0\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \terminate{array}|\begin{array}{ll}\hfill & 0\hfill \\ \hfill & 24\hfill \end{array}\right]\hfill \\ {R}_{1}\leftrightarrow {R}_{two}\to \left[\begin{array}{llll}half dozen\hfill & \hfill & 8\hfill & \hfill \\ 0\hfill & \hfill & 0\hfill & \hfill \cease{array}|\begin{array}{ll}\hfill & 24\hfill \\ \hfill & 0\hfill \cease{array}\correct]\hfill \end{array}\hfill \end{array}[/latex]

The matrix ends upward with all zeros in the concluding row: [latex]0y=0[/latex]. Thus, at that place are an infinite number of solutions and the organization is classified as dependent. To find the generic solution, return to one of the original equations and solve for [latex]y[/latex].

[latex]\begin{array}{l}3x+4y=12\hfill \\ \text{ }4y=12 - 3x\hfill \\ \text{ }y=three-\frac{three}{4}x\hfill \end{array}[/latex]

So the solution to this system is [latex]\left(x,three-\frac{three}{4}10\right)[/latex].

Example 5: Performing Row Operations on a three×3 Augmented Matrix to Obtain Row-Echelon Form

Perform row operations on the given matrix to obtain row-echelon form.

[latex]\left[\begin{array}{rrr}\hfill 1& \hfill -3& \hfill 4\\ \hfill two& \hfill -5& \hfill vi\\ \hfill -three& \hfill three& \hfill iv\end{array}\text{ }|\text{ }\begin{assortment}{r}\hfill iii\\ \hfill 6\\ \hfill 6\stop{array}\right][/latex]

Solution

The first row already has a 1 in row 1, cavalcade one. The next stride is to multiply row 1 by [latex]-2[/latex] and add information technology to row 2. Then replace row 2 with the consequence.

[latex]-ii{R}_{ane}+{R}_{two}={R}_{ii}\to \left[\brainstorm{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill four& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill -3& \hfill & \hfill 3& \hfill & \hfill iv& \hfill \cease{array}|\brainstorm{assortment}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 6\end{array}\correct][/latex]

Next, obtain a zilch in row iii, column 1.

[latex]3{R}_{1}+{R}_{3}={R}_{iii}\to \left[\begin{assortment}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill -half-dozen& \hfill & \hfill 16& \hfill \stop{array}|\begin{assortment}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill xv\end{assortment}\right][/latex]

Side by side, obtain a zero in row 3, cavalcade 2.

[latex]6{R}_{two}+{R}_{3}={R}_{three}\to \left[\begin{assortment}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill iv& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill iv& \hfill \end{array}|\brainstorm{assortment}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill fifteen\end{array}\right][/latex]

The final step is to obtain a i in row 3, cavalcade 3.

[latex]\frac{1}{2}{R}_{3}={R}_{three}\to \left[\brainstorm{assortment}{rrr}\hfill i& \hfill -three& \hfill 4\\ \hfill 0& \hfill 1& \hfill -2\\ \hfill 0& \hfill 0& \hfill 1\stop{assortment}\text{ }|\text{ }\begin{assortment}{r}\hfill three\\ \hfill -vi\\ \hfill \frac{21}{2}\end{assortment}\right][/latex]

Try Information technology 4

Write the system of equations in row-echelon course.

[latex]\begin{array}{50}\text{ }x - 2y+3z=ix\hfill \\ \text{ }-ten+3y=-4\hfill \\ 2x - 5y+5z=17\hfill \end{array}[/latex]

 Solving a System of Linear Equations Using Matrices

We have seen how to write a organization of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now, we will take row-echelon form a footstep farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all only one variable using row operations and then back-substitute to solve for the other variables.

Example half dozen: Solving a Organisation of Linear Equations Using Matrices

Solve the system of linear equations using matrices.

[latex]\begin{array}{c}\brainstorm{assortment}{l}\hfill \\ \hfill \\ x-y+z=8\hfill \end{array}\\ 2x+3y-z=-two\\ 3x - 2y - 9z=ix\end{array}[/latex]

Solution

Showtime, we write the augmented matrix.

[latex]\left[\begin{array}{rrr}\hfill ane& \hfill -one& \hfill ane\\ \hfill two& \hfill 3& \hfill -1\\ \hfill 3& \hfill -2& \hfill -ix\stop{array}\text{ }|\text{ }\begin{array}{r}\hfill 8\\ \hfill -ii\\ \hfill ix\end{assortment}\right][/latex]

Next, nosotros perform row operations to obtain row-echelon form.

[latex]\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{two}\to \left[\begin{array}{rrrrrr}\hfill ane& \hfill & \hfill -i& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill five& \hfill & \hfill -3& \hfill \\ \hfill 3& \hfill & \hfill -2& \hfill & \hfill -nine& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill viii\\ \hfill & \hfill -xviii\\ \hfill & \hfill 9\end{array}\correct]& \hfill & \hfill & \hfill & \hfill -3{R}_{one}+{R}_{3}={R}_{iii}\to \left[\begin{assortment}{rrrrrr}\hfill 1& \hfill & \hfill -one& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 0& \hfill & \hfill i& \hfill & \hfill -12& \hfill \end{assortment}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill -xv\cease{array}\right]\terminate{array}[/latex]

The easiest style to obtain a 1 in row ii of column 1 is to interchange [latex]{R}_{2}[/latex] and [latex]{R}_{three}[/latex].

[latex]\text{Interchange}{R}_{two}\text{and}{R}_{3}\to \left[\begin{assortment}{rrrrrrr}\hfill i& \hfill & \hfill -i& \hfill & \hfill 1& \hfill & \hfill viii\\ \hfill 0& \hfill & \hfill i& \hfill & \hfill -12& \hfill & \hfill -xv\\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill & \hfill -18\end{array}\right][/latex]

Then

[latex]\begin{array}{50}\\ \brainstorm{array}{rrrrr}\hfill -v{R}_{two}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill ane& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 57& \hfill \end{array}|\brainstorm{array}{rr}\hfill & \hfill eight\\ \hfill & \hfill -15\\ \hfill & \hfill 57\finish{array}\correct]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\brainstorm{array}{rrrrrr}\hfill i& \hfill & \hfill -one& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill one& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1& \hfill \end{array}|\begin{assortment}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 1\end{array}\right]\end{array}\end{array}[/latex]

The last matrix represents the equivalent system.

[latex]\begin{array}{l}\text{ }10-y+z=viii\hfill \\ \text{ }y - 12z=-xv\hfill \\ \text{ }z=1\hfill \end{array}[/latex]

Using dorsum-commutation, we obtain the solution as [latex]\left(4,-3,1\correct)[/latex].

Case 7: Solving a Dependent System of Linear Equations Using Matrices

Solve the post-obit system of linear equations using matrices.

[latex]\begin{array}{r}\hfill -x - 2y+z=-1\\ \hfill 2x+3y=2\\ \hfill y - 2z=0\cease{assortment}[/latex]

Solution

Write the augmented matrix.

[latex]\left[\begin{array}{rrr}\hfill -1& \hfill -2& \hfill ane\\ \hfill two& \hfill 3& \hfill 0\\ \hfill 0& \hfill one& \hfill -2\stop{array}\text{ }|\text{ }\begin{array}{r}\hfill -1\\ \hfill 2\\ \hfill 0\end{array}\right][/latex]

First, multiply row 1 past [latex]-1[/latex] to get a ane in row ane, column one. Then, perform row operations to obtain row-echelon form.

[latex]-{R}_{i}\to \left[\begin{array}{rrrrrrr}\hfill i& \hfill & \hfill 2& \hfill & \hfill -1& \hfill & \hfill ane\\ \hfill 2& \hfill & \hfill 3& \hfill & \hfill 0& \hfill & \hfill ii\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill & \hfill 0\end{array}\correct][/latex]

[latex]{R}_{2}\leftrightarrow {R}_{3}\to \left[\brainstorm{assortment}{rrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1\\ \hfill 0& \hfill & \hfill ane& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill iii& \hfill & \hfill 0\end{assortment}\text{ }|\begin{array}{rr}\hfill & \hfill 1\\ \hfill & \hfill 0\\ \hfill & \hfill 2\cease{array}\right][/latex]

[latex]-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill two& \hfill & \hfill -1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -two& \hfill \\ \hfill 0& \hfill & \hfill -1& \hfill & \hfill ii& \hfill \cease{array}|\brainstorm{assortment}{rr}\hfill & \hfill 1\\ \hfill & \hfill 0\\ \hfill & \hfill 0\cease{assortment}\right][/latex]

[latex]{R}_{2}+{R}_{iii}={R}_{3}\to \left[\brainstorm{array}{rrrrrr}\hfill one& \hfill & \hfill ii& \hfill & \hfill -1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 0& \hfill \end{array}|\begin{assortment}{rr}\hfill & \hfill two\\ \hfill & \hfill 1\\ \hfill & \hfill 0\end{array}\right][/latex]

The final matrix represents the following organization.

[latex]\begin{array}{50}\text{ }ten+2y-z=1\hfill \\ \text{ }y - 2z=0\hfill \\ \text{ }0=0\hfill \cease{array}[/latex]

We run into by the identity [latex]0=0[/latex] that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the 2nd equation for [latex]y[/latex] and substituting information technology into the get-go equation we can solve for [latex]z[/latex] in terms of [latex]x[/latex].

[latex]\begin{array}{l}\text{ }ten+2y-z=1\hfill \\ \text{ }y=2z\hfill \\ \hfill \\ x+2\left(2z\right)-z=1\hfill \\ \text{ }x+3z=1\hfill \\ \text{ }z=\frac{1-10}{3}\hfill \stop{assortment}[/latex]

Now we substitute the expression for [latex]z[/latex] into the second equation to solve for [latex]y[/latex] in terms of [latex]x[/latex].

[latex]\begin{array}{l}\text{ }y - 2z=0\hfill \\ \text{ }z=\frac{1-x}{3}\hfill \\ \hfill \\ y - ii\left(\frac{1-ten}{iii}\right)=0\hfill \\ \text{ }y=\frac{2 - 2x}{3}\hfill \stop{assortment}[/latex]

The generic solution is [latex]\left(x,\frac{ii - 2x}{three},\frac{i-x}{3}\right)[/latex].

Try Information technology 5

Solve the system using matrices.

[latex]\brainstorm{array}{c}x+4y-z=4\\ 2x+5y+8z=15\\ x+3y - 3z=1\end{assortment}[/latex]

Q & A

Can any organization of linear equations exist solved by Gaussian elimination?

Yes, a system of linear equations of any size can be solved by Gaussian elimination.

How To: Given a system of equations, solve with matrices using a figurer.

1. Save the augmented matrix as a matrix variable [latex]\left[A\right],\left[B\correct],\left[C\right]\text{,} \dots [/latex].
2. Use the ref( office in the calculator, calling up each matrix variable as needed.

Case 8: Solving Systems of Equations with Matrices Using a Computer

Solve the system of equations.

[latex]\begin{array}{r}\hfill 5x+3y+9z=-1\\ \hfill -2x+3y-z=-ii\\ \hfill -x - 4y+5z=1\end{array}[/latex]

Solution

Write the augmented matrix for the organisation of equations.

[latex]\left[\begin{array}{rrr}\hfill v& \hfill 3& \hfill 9\\ \hfill -2& \hfill three& \hfill -1\\ \hfill -one& \hfill -four& \hfill 5\terminate{assortment}\text{ }|\text{ }\begin{array}{r}\hfill 5\\ \hfill -2\\ \hfill -1\finish{array}\correct][/latex]

On the matrix page of the calculator, enter the augmented matrix in a higher place as the matrix variable [latex]\left[A\right][/latex].

[latex]\left[A\correct]=\left[\begin{assortment}{rrrrrrr}\hfill five& \hfill & \hfill 3& \hfill & \hfill 9& \hfill & \hfill -1\\ \hfill -two& \hfill & \hfill iii& \hfill & \hfill -1& \hfill & \hfill -2\\ \hfill -i& \hfill & \hfill -4& \hfill & \hfill five& \hfill & \hfill ane\end{array}\correct][/latex]

Utilize the ref( function in the calculator, calling up the matrix variable [latex]\left[A\right][/latex].

[latex]\text{ref}\left(\left[A\correct]\correct)[/latex]

Evaluate.

[latex]\begin{assortment}{50}\hfill \\ \left[\begin{array}{rrrr}\hfill 1& \hfill \frac{3}{five}& \hfill \frac{9}{5}& \hfill \frac{ane}{5}\\ \hfill 0& \hfill 1& \hfill \frac{13}{21}& \hfill -\frac{4}{vii}\\ \hfill 0& \hfill 0& \hfill one& \hfill -\frac{24}{187}\cease{array}\correct]\to \begin{array}{l}x+\frac{three}{v}y+\frac{9}{v}z=-\frac{1}{5}\hfill \\ \text{ }y+\frac{13}{21}z=-\frac{4}{7}\hfill \\ \text{ }z=-\frac{24}{187}\hfill \end{array}\hfill \end{array}[/latex]

Using back-substitution, the solution is [latex]\left(\frac{61}{187},-\frac{92}{187},-\frac{24}{187}\correct)[/latex].

Example 9: Applying 2 × ii Matrices to Finance

Carolyn invests a total of $12,000 in two municipal bonds, one paying x.five% interest and the other paying 12% interest. The annual interest earned on the two investments last twelvemonth was $one,335. How much was invested at each rate?

Solution

We have a organisation of ii equations in ii variables. Let [latex]ten=[/latex] the amount invested at 10.5% involvement, and [latex]y=[/latex] the amount invested at 12% interest.

[latex]\begin{array}{fifty}\text{ }x+y=12,000\hfill \\ 0.105x+0.12y=1,335\hfill \stop{array}[/latex]

Every bit a matrix, we have

[latex]\left[\begin{array}{rr}\hfill one& \hfill one\\ \hfill 0.105& \hfill 0.12\stop{array}\text{ }|\text{ }\begin{array}{r}\hfill 12,000\\ \hfill 1,335\finish{assortment}\right][/latex]

Multiply row 1 by [latex]-0.105[/latex] and add the consequence to row 2.

[latex]\left[\begin{assortment}{rr}\hfill 1& \hfill 1\\ \hfill 0& \hfill 0.015\end{array}\text{ }|\text{ }\brainstorm{array}{r}\hfill 12,000\\ \hfill 75\terminate{array}\right][/latex]

And so,

[latex]\begin{array}{50}0.015y=75\hfill \\ \text{ }y=five,000\hfill \end{assortment}[/latex]

So [latex]12,000 - five,000=7,000[/latex].

Thus, $5,000 was invested at 12% involvement and $7,000 at 10.5% interest.

Example x: Applying iii × three Matrices to Finance

Ava invests a full of $10,000 in three accounts, ane paying 5% interest, some other paying 8% interest, and the tertiary paying 9% interest. The annual interest earned on the three investments last year was $770. The corporeality invested at 9% was twice the corporeality invested at 5%. How much was invested at each charge per unit?

Solution

We take a organisation of three equations in three variables. Let [latex]x[/latex] be the amount invested at 5% interest, let [latex]y[/latex] be the amount invested at 8% interest, and let [latex]z[/latex] be the amount invested at 9% interest. Thus,

[latex]\begin{array}{fifty}\text{ }x+y+z=10,000\hfill \\ 0.05x+0.08y+0.09z=770\hfill \\ \text{ }2x-z=0\hfill \stop{array}[/latex]

As a matrix, we take

[latex]\left[\begin{array}{rrr}\hfill 1& \hfill i& \hfill 1\\ \hfill 0.05& \hfill 0.08& \hfill 0.09\\ \hfill 2& \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{assortment}{r}\hfill x,000\\ \hfill 770\\ \hfill 0\end{assortment}\right][/latex]

Now, we perform Gaussian elimination to achieve row-echelon form.

[latex]\begin{array}{l}\begin{assortment}{50}\hfill \\ -0.05{R}_{ane}+{R}_{2}={R}_{ii}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill one& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 2& \hfill & \hfill 0& \hfill & \hfill -1& \hfill \stop{array}|\begin{assortment}{rr}\hfill & \hfill ten,000\\ \hfill & \hfill 270\\ \hfill & \hfill 0\end{array}\right]\hfill \end{array}\hfill \\ -two{R}_{one}+{R}_{three}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill one& \hfill & \hfill one& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 0& \hfill & \hfill -two& \hfill & \hfill -three& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill -twenty,000\end{assortment}\right]\hfill \\ \frac{i}{0.03}{R}_{two}={R}_{ii}\to \left[\begin{array}{rrrrrr}\hfill 0& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill ane& \hfill & \hfill \frac{four}{three}& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill nine,000\\ \hfill & \hfill -20,000\end{assortment}\right]\hfill \\ ii{R}_{ii}+{R}_{iii}={R}_{3}\to \left[\brainstorm{assortment}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill one& \hfill \\ \hfill 0& \hfill & \hfill one& \hfill & \hfill \frac{4}{iii}& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill -\frac{1}{iii}& \hfill \finish{array}|\brainstorm{array}{rr}\hfill & \hfill x,000\\ \hfill & \hfill nine,000\\ \hfill & \hfill -2,000\end{array}\correct]\hfill \end{array}[/latex]

The third row tells united states [latex]-\frac{one}{3}z=-ii,000[/latex]; thus [latex]z=6,000[/latex].

The 2d row tells u.s.a. [latex]y+\frac{iv}{iii}z=9,000[/latex]. Substituting [latex]z=six,000[/latex], nosotros go

[latex]\begin{array}{r}\hfill y+\frac{four}{iii}\left(6,000\correct)=nine,000\\ \hfill y+8,000=9,000\\ \hfill y=i,000\end{array}[/latex]

The offset row tells us [latex]10+y+z=10,000[/latex]. Substituting [latex]y=i,000[/latex] and [latex]z=6,000[/latex], nosotros get

[latex]\begin{array}{l}x+1,000+6,000=10,000\hfill \\ \text{ }x=3,000\text{ }\hfill \terminate{array}[/latex]

The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $six,000 invested at ix% interest.

Endeavour It vi

A small shoe visitor took out a loan of $one,500,000 to expand their inventory. Part of the money was borrowed at 7%, function was borrowed at viii%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the almanac interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.

Solution

Key Concepts

• An augmented matrix is i that contains the coefficients and constants of a system of equations.
• A matrix augmented with the constant cavalcade can be represented every bit the original system of equations.
• Row operations include multiplying a row past a constant, calculation one row to another row, and interchanging rows.
• We can utilize Gaussian elimination to solve a system of equations.
• Row operations are performed on matrices to obtain row-echelon course.
• To solve a arrangement of equations, write information technology in augmented matrix form. Perform row operations to obtain row-echelon grade. Back-substitute to find the solutions.
• A reckoner can be used to solve systems of equations using matrices.
• Many real-world problems can be solved using augmented matrices.

Glossary

augmented matrix

a coefficient matrix adjoined with the constant cavalcade separated by a vertical line inside the matrix brackets

coefficient matrix

a matrix that contains only the coefficients from a system of equations

Gaussian elimination

using elementary row operations to obtain a matrix in row-echelon form

main diagonal

entries from the upper left corner diagonally to the lower right corner of a square matrix

row-echelon class

afterwards performing row operations, the matrix form that contains ones down the master diagonal and zeros at every space beneath the diagonal

row-equivalent

2 matrices [latex]A[/latex] and [latex]B[/latex] are row-equivalent if one can exist obtained from the other by performing basic row operations

row operations

adding ane row to another row, multiplying a row by a constant, interchanging rows, and and then on, with the goal of achieving row-echelon form

Section Exercises

ane. Can any organization of linear equations be written as an augmented matrix? Explain why or why non. Explain how to write that augmented matrix.

2. Tin can any matrix exist written as a arrangement of linear equations? Explain why or why not. Explain how to write that system of equations.

3. Is at that place only one correct method of using row operations on a matrix? Endeavor to explain ii different row operations possible to solve the augmented matrix [latex]\left[\begin{array}{rr}\hfill 9& \hfill three\\ \hfill 1& \hfill -2\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 0\\ \hfill vi\terminate{array}\right][/latex].

4. Tin can a matrix whose entry is 0 on the diagonal be solved? Explain why or why non. What would you do to remedy the state of affairs?

five. Can a matrix that has 0 entries for an entire row have ane solution? Explain why or why not.

For the following exercises, write the augmented matrix for the linear organisation.

six. [latex]\begin{array}{50}8x - 37y=viii\\ 2x+12y=iii\end{array}[/latex]

7. [latex]\begin{array}{50}\text{ }16y=four\hfill \\ 9x-y=2\hfill \cease{array}[/latex]

8. [latex]\brainstorm{assortment}{l}\text{ }3x+2y+10z=three\hfill \\ -6x+2y+5z=thirteen\hfill \\ \text{ }4x+z=xviii\hfill \end{array}[/latex]

9. [latex]\begin{array}{l}\hfill \\ \text{ }x+5y+8z=19\hfill \\ \text{ }12x+3y=4\hfill \\ 3x+4y+9z=-vii\hfill \end{array}[/latex]

10. [latex]\begin{array}{l}6x+12y+16z=4\hfill \\ \text{ }19x - 5y+3z=-9\hfill \\ \text{ }x+2y=-eight\hfill \end{array}[/latex]

For the post-obit exercises, write the linear system from the augmented matrix.

11. [latex]\left[\begin{array}{rr}\hfill -2& \hfill 5\\ \hfill half-dozen& \hfill -xviii\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 5\\ \hfill 26\stop{array}\right][/latex]

12. [latex]\left[\begin{assortment}{rr}\hfill 3& \hfill 4\\ \hfill x& \hfill 17\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 10\\ \hfill 439\end{array}\right][/latex]

13. [latex]\left[\begin{assortment}{rrr}\hfill 3& \hfill 2& \hfill 0\\ \hfill -1& \hfill -9& \hfill 4\\ \hfill 8& \hfill 5& \hfill vii\end{assortment}\text{ }|\text{ }\begin{array}{r}\hfill 3\\ \hfill -i\\ \hfill viii\terminate{assortment}\correct][/latex]

14. [latex]\left[\begin{array}{rrr}\hfill 8& \hfill 29& \hfill 1\\ \hfill -i& \hfill 7& \hfill five\\ \hfill 0& \hfill 0& \hfill 3\finish{array}\text{ }|\text{ }\begin{array}{r}\hfill 43\\ \hfill 38\\ \hfill 10\end{array}\right][/latex]

fifteen. [latex]\left[\begin{array}{rrr}\hfill four& \hfill 5& \hfill -2\\ \hfill 0& \hfill 1& \hfill 58\\ \hfill 8& \hfill vii& \hfill -three\end{assortment}\text{ }|\text{ }\begin{array}{r}\hfill 12\\ \hfill 2\\ \hfill -5\cease{assortment}\right][/latex]

For the following exercises, solve the organization past Gaussian emptying.

xvi. [latex]\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 0\stop{array}\text{ }|\text{ }\begin{array}{r}\hfill 3\\ \hfill 0\finish{array}\right][/latex]

17. [latex]\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 1& \hfill 0\end{assortment}\text{ }|\text{ }\begin{assortment}{r}\hfill one\\ \hfill ii\stop{array}\right][/latex]

xviii. [latex]\left[\begin{array}{rr}\hfill i& \hfill 2\\ \hfill iv& \hfill 5\finish{array}\text{ }|\text{ }\begin{array}{r}\hfill 3\\ \hfill 6\end{array}\correct][/latex]

19. [latex]\left[\begin{array}{rr}\hfill -one& \hfill ii\\ \hfill 4& \hfill -5\stop{array}\text{ }|\text{ }\brainstorm{array}{r}\hfill -3\\ \hfill six\terminate{array}\right][/latex]

twenty. [latex]\left[\begin{array}{rr}\hfill -2& \hfill 0\\ \hfill 0& \hfill ii\end{assortment}\text{ }|\text{ }\brainstorm{array}{r}\hfill 1\\ \hfill -i\end{array}\right][/latex]

21. [latex]\begin{assortment}{l}\text{ }2x - 3y=-9\hfill \\ 5x+4y=58\hfill \stop{array}[/latex]

22. [latex]\brainstorm{array}{50}6x+2y=-4\\ 3x+4y=-17\end{assortment}[/latex]

23. [latex]\begin{array}{50}2x+3y=12\hfill \\ \text{ }4x+y=fourteen\hfill \end{array}[/latex]

24. [latex]\begin{array}{fifty}-4x - 3y=-2\hfill \\ \text{ }3x - 5y=-thirteen\hfill \finish{array}[/latex]

25. [latex]\begin{array}{fifty}-5x+8y=3\hfill \\ 10x+6y=v\hfill \end{array}[/latex]

26. [latex]\begin{assortment}{l}\text{ }3x+4y=12\hfill \\ -6x - 8y=-24\hfill \end{array}[/latex]

27. [latex]\begin{assortment}{l}-60x+45y=12\hfill \\ \text{ }20x - 15y=-4\hfill \stop{array}[/latex]

28. [latex]\begin{assortment}{l}11x+10y=43\\ 15x+20y=65\end{assortment}[/latex]

29. [latex]\begin{assortment}{fifty}\text{ }2x-y=2\hfill \\ 3x+2y=17\hfill \terminate{array}[/latex]

30. [latex]\brainstorm{array}{l}\begin{assortment}{l}\\ -i.06x - 2.25y=five.51\cease{array}\hfill \\ -5.03x - 1.08y=5.40\hfill \end{array}[/latex]

31. [latex]\begin{array}{l}\frac{three}{4}x-\frac{three}{five}y=4\\ \frac{1}{4}x+\frac{2}{3}y=1\stop{assortment}[/latex]

32. [latex]\begin{array}{50}\frac{1}{four}10-\frac{2}{3}y=-1\\ \frac{one}{2}x+\frac{1}{3}y=iii\end{array}[/latex]

33. [latex]\left[\brainstorm{array}{rrr}\hfill one& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 1\\ \hfill 0& \hfill 0& \hfill ane\stop{assortment}\text{ }|\text{ }\begin{array}{r}\hfill 31\\ \hfill 45\\ \hfill 87\end{array}\right][/latex]

34. [latex]\left[\begin{assortment}{rrr}\hfill 1& \hfill 0& \hfill one\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 1\finish{array}\text{ }|\text{ }\begin{assortment}{r}\hfill 50\\ \hfill 20\\ \hfill -xc\stop{assortment}\right][/latex]

35. [latex]\left[\brainstorm{array}{rrr}\hfill 1& \hfill two& \hfill three\\ \hfill 0& \hfill 5& \hfill half-dozen\\ \hfill 0& \hfill 0& \hfill 8\end{array}\text{ }|\text{ }\begin{assortment}{r}\hfill 4\\ \hfill 7\\ \hfill 9\end{array}\right][/latex]

36. [latex]\left[\begin{array}{rrr}\hfill -0.i& \hfill 0.iii& \hfill -0.one\\ \hfill -0.4& \hfill 0.2& \hfill 0.1\\ \hfill 0.6& \hfill 0.1& \hfill 0.7\end{array}\text{ }|\text{ }\begin{assortment}{r}\hfill 0.2\\ \hfill 0.eight\\ \hfill -0.8\end{assortment}\right][/latex]

37. [latex]\begin{assortment}{l}\text{ }-2x+3y - 2z=3\hfill \\ \text{ }4x+2y-z=nine\hfill \\ \text{ }4x - 8y+2z=-6\hfill \terminate{array}[/latex]

38. [latex]\brainstorm{array}{l}\text{ }x+y - 4z=-4\hfill \\ \text{ }5x - 3y - 2z=0\hfill \\ \text{ }2x+6y+7z=30\hfill \end{array}[/latex]

39. [latex]\begin{array}{l}\text{ }2x+3y+2z=1\hfill \\ \text{ }-4x - 6y - 4z=-2\hfill \\ \text{ }10x+15y+10z=five\hfill \end{array}[/latex]

40. [latex]\begin{array}{l}\text{ }x+2y-z=i\hfill \\ -x - 2y+2z=-2\hfill \\ 3x+6y - 3z=5\hfill \finish{array}[/latex]

41. [latex]\begin{array}{l}\text{ }x+2y-z=one\hfill \\ -10 - 2y+2z=-2\hfill \\ \text{ }3x+6y - 3z=iii\hfill \finish{array}[/latex]

42. [latex]\begin{array}{l}\text{ }\text{ }x+y=ii\hfill \\ \text{ }x+z=1\hfill \\ -y-z=-three\hfill \cease{array}[/latex]

43. [latex]\begin{array}{fifty}x+y+z=100\hfill \\ \text{ }x+2z=125\hfill \\ -y+2z=25\hfill \terminate{array}[/latex]

44. [latex]\begin{array}{50}\frac{1}{4}ten-\frac{ii}{iii}z=-\frac{1}{2}\\ \frac{i}{5}x+\frac{one}{3}y=\frac{iv}{7}\\ \frac{one}{5}y-\frac{1}{3}z=\frac{2}{9}\stop{array}[/latex]

45. [latex]\begin{array}{l}-\frac{1}{two}x+\frac{1}{ii}y+\frac{1}{7}z=-\frac{53}{14}\hfill \\ \text{ }\frac{1}{2}x-\frac{1}{2}y+\frac{ane}{4}z=3\hfill \\ \text{ }\frac{one}{4}10+\frac{one}{5}y+\frac{one}{3}z=\frac{23}{xv}\hfill \end{array}[/latex]

46. [latex]\begin{array}{l}-\frac{1}{2}x-\frac{1}{3}y+\frac{1}{iv}z=-\frac{29}{half dozen}\hfill \\ \text{ }\frac{1}{5}x+\frac{i}{6}y-\frac{1}{7}z=\frac{431}{210}\hfill \\ -\frac{ane}{8}ten+\frac{1}{9}y+\frac{ane}{10}z=-\frac{49}{45}\hfill \end{array}[/latex]

For the post-obit exercises, use Gaussian elimination to solve the arrangement.

47. [latex]\begin{array}{l}\frac{ten - 1}{vii}+\frac{y - 2}{viii}+\frac{z - 3}{4}=0\hfill \\ \text{ }x+y+z=six\hfill \\ \text{ }\frac{ten+two}{3}+2y+\frac{z - 3}{3}=five\hfill \finish{array}[/latex]

48. [latex]\begin{array}{fifty}\frac{x - i}{four}-\frac{y+1}{4}+3z=-i\hfill \\ \text{ }\frac{x+5}{ii}+\frac{y+7}{4}-z=4\hfill \\ \text{ }x+y-\frac{z - ii}{ii}=1\hfill \end{array}[/latex]

49. [latex]\begin{assortment}{l}\text{ }\frac{x - three}{4}-\frac{y - 1}{3}+2z=-1\hfill \\ \frac{x+5}{2}+\frac{y+5}{2}+\frac{z+5}{2}=viii\hfill \\ \text{ }ten+y+z=i\hfill \finish{array}[/latex]

fifty. [latex]\begin{array}{l}\frac{ten - three}{10}+\frac{y+iii}{two}-2z=three\hfill \\ \text{ }\frac{x+five}{4}-\frac{y - 1}{8}+z=\frac{three}{2}\hfill \\ \frac{x - 1}{four}+\frac{y+four}{2}+3z=\frac{iii}{2}\hfill \end{array}[/latex]

51. [latex]\begin{array}{50}\text{ }\frac{x - 3}{four}-\frac{y - 1}{three}+2z=-one\hfill \\ \frac{x+5}{ii}+\frac{y+5}{two}+\frac{z+v}{2}=7\hfill \\ \text{ }ten+y+z=1\hfill \cease{assortment}[/latex]

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution.

52. Every day, a cupcake store sells five,000 cupcakes in chocolate and vanilla flavors. If the chocolate flavour is three times as popular equally the vanilla flavor, how many of each cupcake sell per 24-hour interval?

53. At a competing cupcake shop, $4,520 worth of cupcakes are sold daily. The chocolate cupcakes price $ii.25 and the ruby-red velvet cupcakes cost $1.75. If the total number of cupcakes sold per solar day is 2,200, how many of each flavor are sold each day?

54. You invested $10,000 into two accounts: ane that has unproblematic iii% interest, the other with ii.5% interest. If your full interest payment after 1 yr was $283.50, how much was in each account later on the year passed?

55. You invested $2,300 into account 1, and $2,700 into account 2. If the total amount of interest after one year is $254, and account 2 has ane.5 times the interest rate of business relationship ane, what are the involvement rates? Presume simple involvement rates.

56. Bikes'R'Us articles bikes, which sell for $250. It costs the manufacturer $180 per bike, plus a startup fee of $iii,500. Afterwards how many bikes sold will the manufacturer interruption fifty-fifty?

57. A major appliance store is considering purchasing vacuums from a small manufacturer. The shop would be able to purchase the vacuums for $86 each, with a delivery fee of $9,200, regardless of how many vacuums are sold. If the shop needs to start seeing a profit after 230 units are sold, how much should they accuse for the vacuums?

58. The three almost popular ice foam flavors are chocolate, strawberry, and vanilla, comprising 83% of the flavors sold at an ice cream shop. If vanilla sells ane% more twice strawberry, and chocolate sells 11% more than vanilla, how much of the full water ice cream consumption are the vanilla, chocolate, and strawberry flavors?

59. At an ice cream store, three flavors are increasing in demand. Last twelvemonth, banana, pumpkin, and rocky route water ice cream made up 12% of total ice cream sales. This yr, the same 3 ice creams made upward 16.ix% of water ice foam sales. The rocky road sales doubled, the banana sales increased by 50%, and the pumpkin sales increased by 20%. If the rocky road water ice cream had i less percent of sales than the banana ice cream, find out the per centum of ice cream sales each individual ice cream made last twelvemonth.

lx. A bag of mixed nuts contains cashews, pistachios, and almonds. There are 1,000 total nuts in the bag, and in that location are 100 less almonds than pistachios. The cashews weigh iii g, pistachios weigh iv thousand, and almonds weigh 5 g. If the purse weighs iii.7 kg, detect out how many of each type of nut is in the bag.

61. A pocketbook of mixed nuts contains cashews, pistachios, and almonds. Originally at that place were 900 basics in the bag. thirty% of the almonds, twenty% of the cashews, and 10% of the pistachios were eaten, and at present there are 770 nuts left in the pocketbook. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to brainstorm with.

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Source: https://courses.lumenlearning.com/precalctwo/chapter/solving-systems-with-gaussian-elimination/