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Find the general solutions of the systems whose augmented matrices are given in Exercises \(7-14\) . \(\left[\begin{array}{llll}{1} & {3} & {4} & {7} \\ {3} & {9} & {7} & {6}\end{array}\right]\)

Short Answer

Expert verified
General solution: \(x_1 = -5 - 3t\), \(x_2 = t\), \(x_3 = 3\), where \(t\) is any real number.

Step by step solution

01

Write the Augmented Matrix

Start by identifying the given augmented matrix from the problem statement:\[\left[\begin{array}{cccc}1 & 3 & 4 & 7 \ 3 & 9 & 7 & 6\end{array}\right]\] This matrix represents the system of equations.
02

Transform to Row Echelon Form

Use row operations to transform the matrix to Row Echelon Form (REF). Focus on making zeros below the leading 1 in the first row: Subtract 3 times the first row from the second row:\[R_2 = R_2 - 3R_1\]The new form of the matrix is:\[\left[\begin{array}{cccc}1 & 3 & 4 & 7 \ 0 & 0 & -5 & -15\end{array}\right]\]
03

Find Parameters for Free Variables

Notice the second row of the matrix indicates a dependency among the variables. Write the system from the matrix:1. \(x_1 + 3x_2 + 4x_3 = 7\)2. \(-5x_3 = -15\) Solve the second equation for \(x_3\):\[ x_3 = 3 \]
04

Express Variables in Terms of Free Variable

Since \(x_3 = 3\), substitute \(x_3\) back into the original equations to find \(x_1\) and \(x_2\).Using equation 1:\[x_1 + 3x_2 + 4(3) = 7\]\[x_1 + 3x_2 + 12 = 7\]\[x_1 + 3x_2 = -5\]
05

Solve for Specific Variables

Solve the equation \(x_1 + 3x_2 = -5\) for \(x_1\) in terms of \(x_2\):\[x_1 = -5 - 3x_2\]Since \(x_2\) is a free variable, the general solution can be expressed in terms of \(x_2\).
06

Write the General Solution

Combining our expressions, the general solution of the system is:\[x_1 = -5 - 3x_2, \quad x_2 = t, \quad x_3 = 3\]where \(t\) is any real number, indicating that \(x_2\) is a free variable.

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Key Concepts

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

Augmented Matrix
An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single matrix. This simplifies the process of solving the system, especially when using techniques like Gaussian elimination. Each row in the augmented matrix corresponds to an equation, and each column represents either a variable's coefficient or the constant term.

For instance, the augmented matrix \[\left[\begin{array}{cccc}1 & 3 & 4 & 7 \3 & 9 & 7 & 6 \end{array}\right]\]represents a system of two equations in three variables \(x_1, x_2,\) and \(x_3\). The last column in this matrix is the constants column that follows the equal sign in the equations. Understanding how to form and interpret an augmented matrix is fundamental to solving systems of equations efficiently.
Row Echelon Form
Transforming an augmented matrix into Row Echelon Form (REF) is a critical step in solving a system of equations. REF is an organized structure that makes it easier to solve equations through back-substitution. The key features of REF include having leading ones in each row, and zeros below these leading ones.

Achieving this form involves using row operations that are similar to performing arithmetic operations on equations. For example, given the matrix \[\left[\begin{array}{cccc}1 & 3 & 4 & 7 \3 & 9 & 7 & 6 \end{array}\right]\]a common step would be to eliminate the value in the first position of the second row to create a zero. This results in:\[\left[\begin{array}{cccc}1 & 3 & 4 & 7 \0 & 0 & -5 & -15 \end{array}\right]\]
This systematic elimination allows us to focus on each variable starting from the top, making the problem much more manageable.
System of Equations
A system of equations consists of multiple equations that are solved together, usually because they share common variables. The goal is to find values for those variables that satisfy all equations simultaneously. Systems can be linear or non-linear, with linear systems being the focus of linear algebra.

For example, using the system represented by the matrix \[\left[\begin{array}{cccc}1 & 3 & 4 & 7 \3 & 9 & 7 & 6 \end{array}\right]\]the corresponding equations are:
  • \(x_1 + 3x_2 + 4x_3 = 7\)
  • \(3x_1 + 9x_2 + 7x_3 = 6\)
Solving these equations typically involves strategies like substitution or elimination, and often involves representing them using matrices to streamline the process with matrix operations. Each solution to the system corresponds to a point in space where the equations intersect.
Free Variable
In the context of solving systems of linear equations, a free variable is one that is not bound by a unique solution but can take any real number value. This typically occurs in underdetermined systems, where there are more variables than equations. Free variables lead to a range of possible solutions, often represented as a parameterized expression.

For example, in the system arising from the augmented matrix we worked with, there are three variables but fewer independent equations. This results in \(x_2\) being a free variable because it does not have a specific solution from the system. Hence, the general solution can be expressed in terms of \(x_2\):
  • \(x_1 = -5 - 3x_2\)
  • \(x_2 = t\)
  • \(x_3 = 3\)
where \(t\) is any real number. Free variables enable the formation of a general solution, giving insights into the behavior and solution space of the system.

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Most popular questions from this chapter

Exercises \(17-20\) refer to the matrices \(A\) and \(B\) below. Make appropriate calculations that justify your answers and mention an appropriate theorem. $$ A=\left[\begin{array}{rrrr}{1} & {3} & {0} & {3} \\ {-1} & {-1} & {-1} & {1} \\\ {0} & {-4} & {2} & {-8} \\ {2} & {0} & {3} & {-1}\end{array}\right] \quad B=\left[\begin{array}{rrrr}{1} & {3} & {-2} & {2} \\ {0} & {1} & {1} & {-5} \\\ {1} & {2} & {-3} & {7} \\ {-2} & {-8} & {2} & {-1}\end{array}\right] $$ How many rows of \(A\) contain a pivot position? Does the equation \(A \mathbf{x}=\mathbf{b}\) have a solution for each \(\mathbf{b}\) in \(\mathbb{R}^{4} ?\)

Determine by inspection whether the vectors are linearly independent. Justify each answer. \(\left[\begin{array}{l}{5} \\ {1}\end{array}\right],\left[\begin{array}{l}{2} \\\ {8}\end{array}\right],\left[\begin{array}{l}{1} \\\ {3}\end{array}\right],\left[\begin{array}{r}{-1} \\ {7}\end{array}\right]\)

\([\mathbf{M}]\) In Exercises \(37-40,\) let \(T\) be the linear transformation whose standard matrix is given. In Exercises 37 and \(38,\) decide if \(T\) is a one-to- one mapping. In Exercises 39 and \(40,\) decide if \(T\) maps \(\mathbb{R}^{5}\) onto \(\mathbb{R}^{5} .\) Justify your answers. $$ \left[\begin{array}{rrrrr}{4} & {-7} & {3} & {7} & {5} \\ {6} & {-8} & {5} & {12} & {-8} \\ {-7} & {10} & {-8} & {-9} & {14} \\ {3} & {-5} & {4} & {2} & {-6} \\ {-5} & {6} & {-6} & {-7} & {3}\end{array}\right] $$

Suppose an \(m \times n\) matrix \(A\) has \(n\) pivot columns. Explain why for each \(\mathbf{b}\) in \(\mathbb{R}^{m}\) the equation \(A \mathbf{x}=\mathbf{b}\) has at most one solution. [Hint: Explain why \(A \mathbf{x}=\mathbf{b}\) cannot have infinitely many solutions.]

Let \(\mathbf{u}=\left[\begin{array}{l}{7} \\ {2} \\ {5}\end{array}\right], \mathbf{v}=\left[\begin{array}{l}{3} \\ {1} \\ {3}\end{array}\right],\) and \(\mathbf{w}=\left[\begin{array}{l}{6} \\ {1} \\ {0}\end{array}\right]\) It can be shown that \(3 \mathbf{u}-5 \mathbf{v}-\mathbf{w}=\mathbf{0} .\) Use this fact (and no row operations) to find \(x_{1}\) and \(x_{2}\) that satisfy the equation $$ \left[\begin{array}{ll}{7} & {3} \\ {2} & {1} \\ {5} & {3}\end{array}\right]\left[\begin{array}{l}{x_{1}} \\\ {x_{2}}\end{array}\right]=\left[\begin{array}{l}{6} \\ {1} \\\ {0}\end{array}\right] $$

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