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Chapter 2: Problem 8

Show that this pair of augmented matrices are row equivalent, assuming \(a d-b c \neq 0\) $$\left(\begin{array}{ll|l}a & b & e \\ c & d & f\end{array}\right) \sim\left(\begin{array}{cc|c} 1 & 0 & \frac{d e-b f}{a d-b c} \\ 0 & 1 & \frac{a f-c e}{a d-b c}\end{array}\right)$$

Short Answer

Expert verified
The matrices are row equivalent through a series of row operations that transform the initial matrix into the target matrix.

Step by step solution

01

Identify the Matrices and Goal

We are given an initial augmented matrix \[\begin{pmatrix} a & b & e \ c & d & f \end{pmatrix}\] and a target matrix \[\begin{pmatrix} 1 & 0 & \frac{d e-b f}{a d-b c} \ 0 & 1 & \frac{a f-c e}{a d-b c} \end{pmatrix}\]. Our goal is to use row operations to transform the initial matrix into the target matrix, showing they are row equivalent.
02

Apply Row Operations to Form RREF

Perform row operations on the initial matrix to reach the reduced row-echelon form (RREF):1. Make the element in the first row, first column equal to 1 by dividing the entire first row by \(a\). Now, we have: \[\begin{pmatrix} 1 & \frac{b}{a} & \frac{e}{a} \ c & d & f \end{pmatrix} \]2. Eliminate the first element in the second row by transforming it with the first row: \( R_2 := R_2 - c \times R_1 \). This gives: \[\begin{pmatrix} 1 & \frac{b}{a} & \frac{e}{a} \ 0 & d - c\frac{b}{a} & f - c\frac{e}{a} \end{pmatrix} \]
03

Normalize Second Row

3. Normalize the second row by making the leading coefficient (second element) equal to 1: Divide the second row by \(d - c\frac{b}{a}\): \[ R_2 := \frac{1}{d - c\frac{b}{a}}R_2 \] This results in: \[\begin{pmatrix} 1 & \frac{b}{a} & \frac{e}{a} \ 0 & 1 & \frac{f - c\frac{e}{a}}{d - c\frac{b}{a}} \end{pmatrix} \]
04

Simplify and Restructure

4. Simplify the expression for the second row, third column: \[ \frac{f - c\frac{e}{a}}{d - c\frac{b}{a}} = \frac{af - ce}{ad - bc} \]5. Eliminate the second column鈥檚 first row entry by using the second row: \( R_1 := R_1 - \frac{b}{a} \times R_2 \), obtaining: \[\begin{pmatrix} 1 & 0 & \frac{de - bf}{ad - bc} \ 0 & 1 & \frac{af - ce}{ad - bc} \end{pmatrix} \]This matches the target matrix given in the problem.

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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 an essential concept when solving systems of linear equations. It combines the coefficient matrix of a system with its constants.
The augmented matrix is usually separated by a vertical line, which helps to distinguish the coefficients from the constants.
For example, given the system:
  • Equation 1: \[ ax + by = e \]
  • Equation 2: \[ cx + dy = f \]

It can be represented as an augmented matrix like this:\[ \left( \begin{array}{cc|c} a & b & e \ c & d & f \end{array} \right) \]
This structure allows us to use matrix operations to solve the system efficiently and keeps track of the equations' solutions as they are altered via row operations.
Row Operations
Row operations are a set of methods used to manipulate matrices in a way that doesn't change the solution of the associated system of equations. These operations are fundamental in transforming a matrix into its reduced row-echelon form (RREF).
There are three types of row operations:
  • Swap two rows.
  • Multiply a row by a nonzero scalar.
  • Add or subtract a multiple of one row to another row.
These operations are powerful because they allow us to simplify matrices, revealing solutions to systems of equations. In our case, row operations were used to transform the initial augmented matrix to achieve the desired RREF, indicating row equivalency.
Reduced Row-Echelon Form (RREF)
The Reduced Row-Echelon Form (RREF) of a matrix is an advanced concept in linear algebra that simplifies solving systems by presenting a clear path to the solution.
In RREF, the matrix follows certain conditions:
  • Each leading entry (first non-zero number from the left, in each row) is 1.
  • Each leading 1 is the only non-zero entry in its column.
  • The leading 1 of each row appears to the right of the leading 1 in the previous row.
  • Any row containing only zeros is at the bottom of the matrix.
These properties make it easier to read off the solutions to a system of equations directly from the matrix. In our problem, RREF was used to demonstrate row equivalency between two augmented matrices, showing the transformations needed to move from the initial matrix to the solution form.

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Most popular questions from this chapter

Show that the RREF of a matrix is unique.

Check that the system of equations corresponding to the augmented matrix $$\left(\begin{array}{rr|r}1 & 4 & 10 \\\3 & 13 & 9 \\\4 & 17 & 20\end{array}\right)$$ has no solutions. If you remove one of the rows of this matrix, does the new matrix have any solutions? In general, can row equivalence be affected by removing rows? Explain why or why not.

Solve the following linear system: $$\begin{array}{l}2 x_{1}+5 x_{2}-8 x_{3}+2 x_{4}+2 x_{5}=0 \\ 6 x_{1}+2 x_{2}-10 x_{3}+6 x_{4}+8 x_{5}=6 \\ 3 x_{1}+6 x_{2}+2 x_{3}+3 x_{4}+5 x_{5}=6 \\ 3 x_{1}+1 x_{2}-5 x_{3}+3 x_{4}+4 x_{5}=3 \\ 6 x_{1}+7 x_{2}-3 x_{3}+6 x_{4}+9 x_{5}=9\end{array}$$ Be sure to set your work out carefully with equivalence signs \(\sim\) between each step, labeled by the row operations you performed.

Consider the augmented matrix: $$\left(\begin{array}{rr|r}2 & -1 & 3 \\\\-6 & 3 & 1\end{array}\right)$$ Give a geometric reason why the associated system of equations has no solution. (Hint, plot the three vectors given by the columns of this augmented matrix in the plane.) Given a general augmented matrix $$\left(\begin{array}{ll|l}a & b & e \\\c & d & f\end{array}\right)$$ can you find a condition on the numbers \(a, b, c\) and \(d\) that corresponds to the geometric condition you found?

Another method for solving linear systems is to use row operations to bring the augmented matrix to Row Echelon Form (REF as opposed to RREF). In REF, the pivots are not necessarily set to one, and we only require that all entries left of the pivots are zero, not necessarily entries above a pivot. Provide a counterexample to show that row echelon form is not unique. Once a system is in row echelon form, it can be solved by "back substitution." Write the following row echelon matrix as a system of equations, then solve the system using back-substitution. $$\left(\begin{array}{lll|l}2 & 3 & 1 & 6 \\\0 & 1 & 1 & 2 \\\0 & 0 & 3 & 3\end{array}\right)$$
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