91Ó°ÊÓ

Fill in the missing entries by performing the indicated row operations to obtain the rowreduced matrices. $$ \begin{array}{l} \text { }\left[\begin{array}{rrr|r} 0 & 1 & 3 & -4 \\ 1 & 2 & 1 & 7 \\ 1 & -2 & 0 & 1 \end{array}\right] \stackrel{R_{1} \leftrightarrow R_{2}}{\longrightarrow}\left[\begin{array}{rrr|r} \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ 1 & -2 & 0 & 1 \end{array}\right]\\\ \stackrel{R_{3}-R_{1}}{\longrightarrow}\left[\begin{array}{ccc|r} 1 & 2 & 1 & 7 \\ 0 & 1 & 3 & -4 \\ \cdot & \cdot & \cdot & \cdot \end{array}\right] \frac{R_{1}+\frac{1}{2} R_{3}}{R_{3}+4 R_{2}}\left[\begin{array}{ccc|c} \cdot & \cdot & \cdot & \cdot \\ 0 & 1 & 3 & -4 \\ \cdot & \cdot & \cdot & \cdot \end{array}\right]\\\ \stackrel{\frac{1}{11} R_{3}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & \frac{1}{2} & 4 \\ 0 & 1 & 3 & -4 \\ . & \cdot & . & . \end{array}\right] \frac{R_{1}-\frac{1}{2} R_{3}}{R_{2}-3 R_{3}}\left[\begin{array}{ccc|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right] \end{array} $$

Step by step solution

01

Swap Rows 1 and 2

Perform the operation \(R_1 \leftrightarrow R_2\): $$ \left[\begin{array}{rrr|r} 0 & 1 & 3 & -4 \\ 1 & 2 & 1 & 7 \\ 1 & -2 & 0 & 1 \end{array}\right] \stackrel{R_{1} \leftrightarrow R_{2}}{\longrightarrow} \left[\begin{array}{rrr|r} 1 & 2 & 1 & 7 \\ 0 & 1 & 3 & -4 \\ 1 & -2 & 0 & 1 \end{array}\right] $$

02

Subtract Row 1 from Row 3

Perform the operation \(R_3 - R_1\), $$ \stackrel{R_{3}-R_{1}}{\longrightarrow}\left[\begin{array}{ccc|r} 1 & 2 & 1 & 7 \\ 0 & 1 & 3 & -4 \\ 0 & -4 & -1 & -6 \end{array}\right] $$

03

Add 4 times Row 2 to Row 3

Perform the operation \(R_3+4R_2\), $$ \frac{R_{1}+\frac{1}{2} R_{3}}{R_{3}+4 R_{2}}\left[\begin{array}{rrr|r} 1 & 2 & 1 & 7 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 11 & -22 \end{array}\right] $$

04

Multiply Row 3 by 1/11

Perform the operation \(\frac{1}{11}R_3\), $$ \stackrel{\frac{1}{11} R_{3}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 1 & -2 \end{array}\right] $$

05

Subtract 3 times Row 3 from Row 2 to eliminate the entry above pivot element

Perform the operation \(R_2 - 3R_3\), $$ \frac{R_{1}-\frac{1}{2} R_{3}}{R_{2}-3 R_{3}}\left[\begin{array}{ccc|r} 1 & 2 & 1 & 7 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right] $$

06

Subtract 1/2 times Row 3 from Row 1

Perform the operation \(R_1 - \frac{1}{2}R_3\), $$ \frac{R_{1}-\frac{1}{2} R_{3}}{R_{2}-3 R_{3}}\left[\begin{array}{ccc|r} 1 & 2 & 0 & 8 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right] $$

07

Subtract 2 times Row 2 from Row 1

Perform the operation \(R_1 - 2R_2\), $$ \frac{R_{1}-\frac{1}{2} R_{3}}{R_{2}-3 R_{3}}\left[\begin{array}{ccc|r} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right] $$

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Key Concepts

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

Row Swapping

Row swapping is an operation where two rows in a matrix are exchanged with each other. This is often used in matrix reduction to easily place a desired row in the correct position. The primary purpose of row swapping is to simplify further operations by setting the matrix up in a more favorable order.
For example, in the given exercise, the first step involves swapping Row 1 and Row 2. Originally, the first row was \([0, 1, 3, -4]\) and the second row was \([1, 2, 1, 7]\). By swapping them, the matrix becomes:

• First Row: \([1, 2, 1, 7]\)
• Second Row: \([0, 1, 3, -4]\)

This helps position the row that begins with '1' at the start, simplifying the row reduction process. Remember, row swapping is a reversible operation, meaning you can always switch back the rows if needed.

Matrix Reduction

Matrix reduction is a method used to simplify matrices into simpler forms such as Row Echelon Form (REF) or Reduced Row Echelon Form (RREF). The goal is often to make solving systems of equations easier.
In our exercise, after performing row operations and achieving the RREF, we simplify the system to solve it efficiently.
Matrix reduction involves making leading coefficients (pivots) equal to one, and clearing out other numbers in the column.

• For instance, to simplify step 4 in the original solution: multiply Row 3 by \(1/11\), turning\([0, 0, 11, -22]\) into \([0, 0, 1, -2]\). This converts the last row into a pivotal form, i.e., ideal for achieving RREF.

Thus, achieving a form of the matrix where solutions can be directly interpreted or further manipulated easily.

Augmented Matrix

An augmented matrix is a representation of a system of linear equations, combining the coefficient matrix and the constants into a single matrix. This form helps visualize and perform row operations efficiently.
In the original problem, the matrix given was augmented by the constants from the equations:

• The "|" line separates coefficients of variables from the constants.
• For example, the initial augmented matrix is: \(\begin{array}{ccc|r}0 & 1 & 3 & -4 \1 & 2 & 1 & 7 \1 & -2 & 0 & 1\end{array}\)

Augmented matrices are useful because they enable simultaneous handling of multiple linear equations, thereby simplifying the algebraic manipulations needed to solve the systems.

Elementary Row Operations

Elementary Row Operations are fundamental matrix operations that transform a matrix into a simpler form without changing its solutions. There are three types of elementary row operations:

• Row swapping: Exchanging two rows.
• Row multiplication: Multiplying a row by a non-zero scalar.
• Row addition: Replacing a row by adding to it a multiple of another row.

In the exercise, elementary row operations were used to achieve the row-reduced matrix:

• "Subtract Row 1 from Row 3" is a row addition operation, simplifying Row 3.
• By "adding 4 times Row 2 to Row 3," you eliminate values to simplify Row 3.

Each operation fundamentally transforms the matrix while maintaining the solution of its system of equations. These steps are critical in iterative processes like matrix reduction to solve linear equations effectively.