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Solve the systems in Exercises \(11-14\) $$ \begin{aligned} x_{2}+4 x_{3} &=-5 \\ x_{1}+3 x_{2}+5 x_{3} &=-2 \\ 3 x_{1}+7 x_{2}+7 x_{3} &=6 \end{aligned} $$

Short Answer

Expert verified
The system has no solution (it's inconsistent).

Step by step solution

01

Write the system in matrix form

The given system of equations:\[\begin{aligned} x_{2}+4x_{3} &=-5 \x_{1}+3x_{2}+5x_{3} &=-2 \3x_{1}+7x_{2}+7x_{3} &=6\end{aligned}\]can be represented as a matrix equation \( A\mathbf{x} = \mathbf{b} \) where:\[A = \begin{bmatrix} 0 & 1 & 4 \1 & 3 & 5 \3 & 7 & 7\end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -5 \ -2 \ 6 \end{bmatrix}\]
02

Use row reduction to find row-echelon form

We aim to convert the matrix \( A \) into row-echelon form using elementary row operations. 1. Start with the original augmented matrix:\[\begin{bmatrix} 0 & 1 & 4 & | & -5 \1 & 3 & 5 & | & -2 \3 & 7 & 7 & | & 6\end{bmatrix}\]2. Swap Row 1 and Row 2:\[\begin{bmatrix} 1 & 3 & 5 & | & -2 \0 & 1 & 4 & | & -5 \3 & 7 & 7 & | & 6\end{bmatrix}\]3. Eliminate the \( x_1 \) term in Row 3 by replacing Row 3: Row 3 = Row 3 - 3*Row 1:\[\begin{bmatrix} 1 & 3 & 5 & | & -2 \0 & 1 & 4 & | & -5 \0 & -2 & -8 & | & 12\end{bmatrix}\]4. Replace Row 3: Row 3 = Row 3 + 2*Row 2:\[\begin{bmatrix} 1 & 3 & 5 & | & -2 \0 & 1 & 4 & | & -5 \0 & 0 & 0 & | & 2\end{bmatrix}\]Notice that the last row indicates contradictory information, implying no solution.
03

Interpret the row-echelon form

The resulting row-echelon matrix:\[\begin{bmatrix} 1 & 3 & 5 & | & -2 \0 & 1 & 4 & | & -5 \0 & 0 & 0 & | & 2\end{bmatrix}\]This shows that the last equation "0 = 2" is a contradiction. Thus, the system does not have a solution.

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Key Concepts

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

Row Reduction: Simplifying Matrices
Row reduction is a method used to simplify matrices, allowing us to solve a system of equations more easily. When you want to solve a system, you first translate it into a matrix form. Then, row reduction helps transform this matrix into a simpler form—often row-echelon form—making it more comprehensible. This simplification is achieved through a series of steps called elementary row operations.

Imagine adjusting a set of equations so that solving them step-by-step becomes as straightforward as possible. It's like tidying up a messy room, clearing space for clear, logical thinking. Each operation intends to clear terms, reduce complexity, and reveal the simplicity beneath the mess.

When studying row reduction, focus on becoming familiar with the process. It involves regularly swapping and scaling rows, or adding multiples of one row to another. These basic moves form the core of almost any matrix-related problem solving in mathematics. Understanding this concept deeply helps in managing equations efficiently, leading to insights into the potential solutions—or lack thereof—as seen in equations with contradictory elements.
Elementary Row Operations: The Building Blocks
Elementary row operations are the key actions you perform during row reduction in a matrix. These operations allow you to change the equations without altering the solutions, essentially reshaping your set of equations into a more workable form. There are three primary types of elementary row operations:

  • Swapping two rows: Like swapping places with someone else in line, making it easier to manage who comes first.
  • Multiplying a row by a non-zero scalar: This helps balance equations, similar to stretching or shrinking to fit a new size.
  • Adding or subtracting a multiple of one row to another row: This adjusts one equation using another to cancel terms, aiding clarity and simplification.

Understanding these operations is crucial because they transform a complex multivariable setup into a much clearer image. They help us focus by eliminating distractions brought by complex terms. The beauty of these operations is in their simplicity and power; they allow for equations to be tailored perfectly to reveal their underlying truths.
Row-Echelon Form: Reaching Simplified Clarity
Row-echelon form is a structured, simplified version of a matrix achieved via row reduction. This form is neat and predictable, making it easier to extract solutions or understand the nature of the system of equations.

To achieve row-echelon form, a matrix must satisfy some conditions:
  • All non-zero rows (i.e., rows with at least one non-zero element) are above any rows of all zeros.
  • The leading entry or the first non-zero number from the left in a non-zero row is always to the right of the leading entry of the row above it.
  • All entries in a column below a leading entry are zero.

Think of row-echelon form as a staircase; it steps down row by row, each step clearly defined. When you see this form, you're witnessing the clean, structured outcome of various row operations. In solving a system of equations, this form facilitates seeing solutions directly—or spotting contradictions, as in cases where a system has no solution.

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Most popular questions from this chapter

In Exercises 29–32, 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. $$ \left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\\ {0} & {-3} & {9} & {5}\end{array}\right],\left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\ {0} & {0} & {0} & {-1}\end{array}\right] $$

[M] Use as many columns of \(A\) as possible to construct a matrix \(B\) with the property that the equation \(B \mathbf{x}=\mathbf{0}\) has only the trivial solution. Solve \(B \mathbf{x}=\mathbf{0}\) to verify your work. \(A=\left[\begin{array}{rrrrr}{8} & {-3} & {0} & {-7} & {2} \\ {-9} & {4} & {5} & {11} & {-7} \\ {6} & {-2} & {2} & {-4} & {4} \\ {5} & {-1} & {7} & {0} & {10}\end{array}\right]\)

Use the vectors \(\mathbf{u}=\left(u_{1}, \ldots, u_{n}\right), \mathbf{v}=\left(v_{1}, \ldots, v_{n}\right),\) and \(\mathbf{w}=\left(w_{1}, \ldots, w_{n}\right)\) to verify the following algebraic properties of \(\mathbb{R}^{n}\) . a. \((\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})\) b. \(c(\mathbf{u}+\mathbf{v})=c \mathbf{u}+c \mathbf{v}\) for each scalar \(c\)

Suppose \(A\) is an \(m \times n\) matrix with the property that for all \(\mathbf{b}\) in \(\mathbb{R}^{m}\) the equation \(A \mathbf{x}=\mathbf{b}\) has at most one solution. Use the definition of linear independence to explain why the columns of \(A\) must be linearly independent.

Each statement in Exercises 33–38 is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is true, give a justification. (One specific example cannot explain why a statement is always true. You will have to do more work here than in Exercises 21 and 22.) If \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\) are linearly independent vectors in \(\mathbb{R}^{4},\) then \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is also linearly independent. [Hint: Think about \(x_{1} \mathbf{v}_{1}+x_{2} \mathbf{v}_{2}+x_{3} \mathbf{v}_{3}+0 \cdot \mathbf{v}_{4}=\mathbf{0} . ]\)

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