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Show that the given systems of equations have either an unlimited number of solutions or no solution. If there is an unlimited number of solutions, find one of them. $$\begin{aligned} &3 x+y-z=-3\\\ &x+y-3 z=-5\\\ &-5 x-2 y+3 z=-7 \end{aligned}$$

Short Answer

Expert verified
The system has a unique solution: \(x = -\frac{17}{30}\), \(y = -\frac{33}{5}\), \(z = -\frac{1}{10}\).

Step by step solution

01

Set up the system of equations matrix

Write the system of equations as an augmented matrix: \[\begin{bmatrix}3 & 1 & -1 & \vert & -3\1 & 1 & -3 & \vert & -5\-5 & -2 & 3 & \vert & -7\end{bmatrix}\]
02

Perform row operations for Row Echelon Form

Start simplifying the matrix by performing row operations. Add the first row multiplied by 5 to the third row to remove the x-term in the third row:\[R_3 = R_3 + 5R_1 \Rightarrow \begin{bmatrix}3 & 1 & -1 & \vert & -3\1 & 1 & -3 & \vert & -5\0 & 3 & -2 & \vert & -22\end{bmatrix}\]Next, subtract the first row multiplied by \( \frac{1}{3} \) from the second row:\[R_2 = R_2 - \frac{1}{3}R_1 \Rightarrow \begin{bmatrix}3 & 1 & -1 & \vert & -3\0 & \frac{2}{3} & -\frac{8}{3} & \vert & -\frac{14}{3}\0 & 3 & -2 & \vert & -22\end{bmatrix}\]
03

Further simplify the augmented matrix

Continue row operations to simplify. Multiply the second row by \( \frac{3}{2} \):\[R_2 = \frac{3}{2}R_2 \Rightarrow \begin{bmatrix}3 & 1 & -1 & \vert & -3\0 & 1 & -4 & \vert & -7\0 & 3 & -2 & \vert & -22\end{bmatrix}\]Subtract three times the second row from the third row:\[R_3 = R_3 - 3R_2 \Rightarrow \begin{bmatrix}3 & 1 & -1 & \vert & -3\0 & 1 & -4 & \vert & -7\0 & 0 & 10 & \vert & -1\end{bmatrix}\]
04

Analyze for solutions

The last row translates to the equation \(0x + 0y + 10z = -1\), which simplifies to \(10z = -1\). This gives \(z = -\frac{1}{10}\). Substituting back into the second row equation \(y - 4(-\frac{1}{10}) = -7\), we calculate \(y = -7 + \frac{4}{10} = -\frac{66}{10}\). It can be further simplified and solved to check if the first row holds, but given that we have a specific solution for \(z\) and \(y\), we determine the system has a unique solution.
05

Solve for the remaining variable

Use the values found for \(y\) and \(z\) (\(z = -\frac{1}{10}\) and \(y = -\frac{33}{5}\)) in one of the original equations (e.g., the first equation):\[3x + \left(-\frac{33}{5}\right) - \left(-\frac{1}{10}\right) = -3\]Convert to a common denominator and solve for \(x\):\[3x = -3 + \frac{33}{5} - \frac{1}{10} = -3 + \frac{65}{10}\]Solve for \(x\) resulting in \(x = -\frac{17}{30}\).
06

Conclude solution type

Given the unique values computed for each variable, the system of equations has a unique solution: \(x = -\frac{17}{30}\), \(y = -\frac{33}{5}\), and \(z = -\frac{1}{10}\). There is not an infinite number of solutions, nor is it inconsistent.

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Key Concepts

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

Matrix Row Operations
When working to solve a system of linear equations, especially in a matrix form, it is common to perform operations to simplify and solve the system. These operations are known as matrix row operations. There are three main types of row operations:
  • Swapping two rows: This is useful for reordering the matrix to achieve a desired form.
  • Multiplying a row by a non-zero scalar: Allows scaling of a row to make the numbers more manageable or to facilitate elimination.
  • Adding or subtracting a multiple of one row to another: Primarily used to eliminate variables and simplify rows based on other rows.
These operations are essential as they help convert the given matrix into a simpler form, ultimately leading to easier solutions. In the provided example, various row operations were used to eliminate variables and simplify the matrix to better find a solution.
Row Echelon Form
Row Echelon Form (REF) is a specific arrangement of a matrix that makes solving systems of equations much more manageable. In this form, each leading entry of a row is in a column to the right of the leading entry of the row above it. Moreover, any rows consisting entirely of zeroes are at the bottom of the matrix.
  • Leading non-zero entries, called pivots, lead each row.
  • All entries below a pivot are zero.
  • Rows of all zeroes, if any, appear at the bottom of the matrix.
In the solution process, transforming the given matrix into Row Echelon Form made the variables' relationships clearer, facilitating the step-by-step solving of each variable. In the example, turning the matrix into REF allowed for straightforward back-substitution to find the solution variables.
Unique Solution
A unique solution in a system of equations implies that there is only one set of values for the unknowns that satisfies all of the equations in the system. When a system of linear equations has a unique solution:
  • The matrix of coefficients forms a linearly independent set of rows.
  • Each variable corresponds to a pivot column.
  • There is no row of all zeroes that corresponds to a non-zero constant after row reduction.
When calculating the solution to the system, the process confirmed a unique outcome. Transformations ended in decimal solutions for all the variables, suggesting the system was full rank and each equation added necessary, distinct information for determining each variable value.
Augmented Matrix
An augmented matrix is derived from a system of linear equations and represents both the coefficients of the variables and the constants from the equations. It serves a dual purpose:
  • The coefficient matrix reflects the system's core linear relationships.
  • The augmented column indicates the constants derived from each equation's right-hand side.
Structure-wise, an augmented matrix looks like the following:\[\begin{bmatrix} &, & \vert & \a_{11}, & a_{12}, &, & b_1 \a_{21}, & a_{22}, &, & b_2 \... & ... & \vert & ... \a_{m1}, & a_{m2}, &, & b_m \end{bmatrix}\]It provides an efficient method for applying row operations aimed at finding solutions or determining the nature of solutions, like a unique, infinite, or no solution. In the given exercise, the initial step involved forming the augmented matrix, setting the stage for subsequent simplifications and analysis.

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Most popular questions from this chapter

Set up appropriate systems of two linear equations in two unknowns and then solve the systems by determinants. All numbers are accurate to at least two significant digits. A new development has 3 -bedroom homes and 4-bedroom homes. The developer's profit was \(\$ 25,000\) from each 3 -br home, and \(\$ 35,000\) from each 4 -br home, totaling \(\$ 6,800,000\). Total annual property taxes are \(\$ 560,000,\) with \(\$ 2000\) from each 3 -br home and \(\$ 3000\) from each 4 -br home. How many of each were built?

Solve for \(u\) and \(v\) and then solve for \(x\) and \(y\). In this way, we see how to solve systems of equations involving reciprocals. $$\begin{aligned} &\frac{1}{x}-\frac{1}{y}=\frac{1}{2}\\\ &\frac{1}{x}+\frac{1}{y}=\frac{1}{4} \end{aligned}$$

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