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In Exercises 19 and \(20,\) choose \(h\) and \(k\) such that the system has (a) no solution, (b) a unique solution, and (c) many solutions. Give separate answers for each part. \(\begin{aligned} x_{1}+3 x_{2} &=2 \\ 3 x_{1}+h x_{2} &=k \end{aligned}\)

Short Answer

Expert verified
(a) No solution: \( h = 9, k \neq 6 \). (b) Unique: \( h \neq 9 \). (c) Many solutions: \( h = 9, k = 6 \).

Step by step solution

01

Understand the System of Equations

We are given a system of equations: 1. \( x_1 + 3x_2 = 2 \) 2. \( 3x_1 + hx_2 = k \). We need to choose values for \( h \) and \( k \) to obtain different types of solutions (no solution, unique solution, many solutions).
02

Condition for No Solution

For a system to have no solution, the lines represented by the equations must be parallel but not coincident. The slopes of the lines should be equal but they should have different intercepts. The slope of the first equation is \( -\frac{1}{3} \), so the slope of the second equation must also be \( -\frac{1}{3} \), which leads to \( \frac{3}{h} = \frac{1}{3} \); hence, \( h = 9 \). Substituting this in the system, the line \( 3x_1 + 9x_2 = k \) must have a different intercept from the first, meaning \( k eq 6 \) (from \[ y = -\frac{1}{3}x + 2 \]). So possible values are \( h = 9, k eq 6 \).
03

Condition for Unique Solution

For a unique solution, the determinant of the coefficients matrix must be non-zero. The matrix of coefficients is: \[ \begin{bmatrix} 1 & 3 \ 3 & h \end{bmatrix} \] The determinant is calculated as: \[ 1 \cdot h - 3 \cdot 3 = h - 9 \] For it to be non-zero, \( h eq 9 \). So, for a unique solution, \( h eq 9 \). No specific \( k \) value is needed. Example could be \( h = 5 \).
04

Condition for Many Solutions

For the system to have many solutions, the lines must be coincident which implies they are identical in all respects. Thus, both the slope and intercept must be the same. This means \( h = 9 \) and \( k = 6 \).
05

Summarize Solution Conditions

(a) No solution: \( h = 9 \) and \( k eq 6 \). (b) Unique solution: \( h eq 9 \). Example: \( h = 5 \). (c) Many solutions: \( h = 9 \) and \( k = 6 \).

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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 collection of two or more equations with the same set of variables. These systems are an integral part of linear algebra. In the given problem, we are working with a simple system consisting of two linear equations:
  • \( x_1 + 3x_2 = 2 \)
  • \( 3x_1 + hx_2 = k \)
These equations represent lines in a two-dimensional plane. The task is to find scenarios for the system that will result in different types of solutions: no solution, a unique solution, or many solutions. Understanding these possibilities is central to analyzing and solving systems of linear equations effectively.
Unique Solution
A unique solution occurs when exactly one set of values satisfies all equations in the system simultaneously. For our system of equations, this means the two lines represented by the equations intersect at exactly one point. Mathematically, this is represented by a non-zero determinant of the matrix formed by the coefficients of the variables.For the given matrix:\[\begin{bmatrix} 1 & 3 \ 3 & h \end{bmatrix}\]The determinant is calculated as \( h - 9 \). To ensure a unique solution exists, the determinant must not equal zero, which implies \( h eq 9 \). In this case, there is no restriction on the value of \( k \), as it does not affect the condition for a unique solution. By setting \( h = 5 \), for example, a unique solution is guaranteed, as the determinant becomes \( 5 - 9 = -4 \), which is non-zero.
No Solution
A system of equations has no solution when the equations represent parallel lines that never intersect. For two lines to be parallel, they must have the same slope but different intercepts. In our system, the slope of the first equation is \( -\frac{1}{3} \).Similarly, for the second equation, \( \begin{aligned} 3x_1 + hx_2 = k \end{aligned} \), the slope must also be \( -\frac{1}{3} \) to result in parallel lines. Solving \( \frac{3}{h} = \frac{1}{3} \), we find \( h = 9 \). However, to ensure no intersection, the intercept must differ from that of the first line. Thus, the second equation's intercept \( k \) must not equal \( 6 \), making the condition \( k eq 6 \). This ensures that the lines remain parallel and do not coincide, leading to no solution.
Many Solutions
The system of equations has many solutions when the equations represent the same line, meaning they overlap completely. For this to happen, both the slope and the intercept of the lines must be identical.For our equations, this means setting \( h = 9 \) for the slopes to be equal (\( -\frac{1}{3} \)). In addition, the intercepts must also match by setting \( k = 6 \). When both these conditions are satisfied, the lines are coincident, or exactly the same, leading to infinitely many solutions. This means any point on the line is a solution to both equations, highlighting the concept that some systems can have numerous solutions if the constraints align perfectly.

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Most popular questions from this chapter

Balance the chemical equations in Exercises \(5-10\) using the vector equation approach discussed in this section. When solutions of sodium phosphate and barium nitrate are mixed, the result is barium phosphate (as a precipitate) and sodium nitrate. The unbalanced equation is \(\mathrm{Na}_{3} \mathrm{PO}_{4}+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}+\mathrm{NaNO}_{3}\) [For each compound, construct a vector that lists the numbers of atoms of sodium (Na), phosphorus, oxygen, barium, and nitrogen. For instance, barium nitrate corresponds to \((0,0,6,1,2) . ]\)

a. Fill in the blank in the following statement: "If \(A\) is an \(m \times n\) matrix, then the columns of \(A\) are linearly independent if and only if \(A\) has ______ pivot columns.鈥 b. Explain why the statement in (a) is true.

Determine by inspection whether the vectors are linearly independent. Justify each answer. \(\left[\begin{array}{r}{-8} \\ {12} \\\ {-4}\end{array}\right],\left[\begin{array}{r}{2} \\ {-3} \\\ {-1}\end{array}\right]\)

Let \(T : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be the transformation that reflects each vector \(\mathbf{x}=\left(x_{1}, x_{2}, x_{3}\right)\) through the plane \(x_{3}=0\) onto \(T(\mathbf{x})=\left(x_{1}, x_{2},-x_{3}\right) .\) Show that \(T\) is a linear transformation. [See Example 4 for ideas. \(]\)

In Exercises 29鈥32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first. $$ \left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\\ {0} & {-3} & {9} & {5}\end{array}\right],\left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\ {0} & {0} & {0} & {-1}\end{array}\right] $$

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