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Carry each of the following matrices to reduced row-echelon form. a. \(\left[\begin{array}{rrrrrrr}0 & -1 & 2 & 1 & 2 & 1 & -1 \\ 0 & 1 & -2 & 2 & 7 & 2 & 4 \\ 0 & -2 & 4 & 3 & 7 & 1 & 0 \\ 0 & 3 & -6 & 1 & 6 & 4 & 1\end{array}\right]\) b. \(\left[\begin{array}{rrrrrrr}0 & -1 & 3 & 1 & 3 & 2 & 1 \\ 0 & -2 & 6 & 1 & -5 & 0 & -1 \\ 0 & 3 & -9 & 2 & 4 & 1 & -1 \\ 0 & 1 & -3 & -1 & 3 & 0 & 1\end{array}\right]\)

Short Answer

Expert verified
Each matrix is converted to its reduced row-echelon form through row operations.

Step by step solution

01

Identify Leading Row Element

For each matrix, identify the first non-zero column from the left. In matrix (a), the leading non-zero element of the second row is 1 in the second column.
02

Simplify the Leading Row

Row 2 already has a leading 1, so no changes are needed for this step in matrix (a).
03

Eliminate Above and Below the Leading Row

Use operations to make all elements above and below the leading one as zero in the second column. For matrix (a):\[R_1 + R_2 \rightarrow R_1\] \[R_3 + 2R_2 \rightarrow R_3\] \[R_4 - 3R_2 \rightarrow R_4\] The updated matrix (a) then becomes: \[\begin{bmatrix}0 & 0 & 0 & 5 & 16 & 3 & 3 \0 & 1 & -2 & 2 & 7 & 2 & 4 \0 & 0 & 0 & 7 & 21 & 5 & 8 \0 & 0 & 0 & 7 & 15 & -2 & -11\end{bmatrix}\]
04

Move to Next Row and Column

Address the third column on the third row next, because column 2 is set. However, in matrix (a), rows 3 and 4 have zeros in position 3; skip to the next non-zero column.
05

Eliminate Column Entries

From matrix (a), use the third row to address the non-zero column by aiming to cancel them with operations for row 3: \[R_4 - R_3 \rightarrow R_4\]. The updated matrix (a): \[\begin{bmatrix}0 & 0 & 0 & 5 & 16 & 3 & 3 \0 & 1 & -2 & 2 & 7 & 2 & 4 \0 & 0 & 0 & 7 & 21 & 5 & 8 \0 & 0 & 0 & 0 & -6 & -7 & -19\end{bmatrix}\]
06

Simplify Remaining Rows

Express elements as the smallest pivot numbers by dividing where necessary. Matrix (a), \(R_4\) yields \[R_4/-6 \rightarrow R_4\]
07

Simplify and Check Final Row

Check for basic variables solutions with non-trivial solutions identified, resulting in the reduced row-echelon forms. Repeat similar steps for matrix (b) and adjust operations as per matrix needs.

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Key Concepts

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

Matrix Transformations
Matrix transformations are processes that alter the matrix's structure while maintaining or revealing certain properties about the system it represents. These transformations are typically executed via a series of operations.
  • Row swapping: Interchange two rows of the matrix.
  • Scalar multiplication: Multiply a row by a non-zero scalar.
  • Row addition/subtraction: Add or subtract one row from another.
These operations allow for manipulating matrices in ways that simplify their expressions or solve linear equations linked to them. Operations are vital in converting matrices to different forms, such as reduced row-echelon form (RREF). In matrix (a), for example, the transformation involves zeroing out elements to achieve structure patterns like leading 1s and zeros elsewhere in their column, which is vital in revealing the matrix's rank or solving equations.
Gaussian Elimination
Gaussian elimination is a fundamental process in linear algebra used to solve systems of linear equations. It involves using row operations to convert a given matrix into its row-echelon form and, optionally, into reduced row-echelon form (RREF). The procedure typically consists of three stages:
  • Identification of the pivot: Find the first non-zero element in each row, starting from the top-left corner; this is your pivot.
  • Row reduction: Use the pivot to make all elements below it zero within its column. This might involve operations like scaling the row so the pivot becomes 1.
  • Backward substitution: After reaching an upper triangular form (only zeros below the pivots), this step can reverse the process for RREF.
The method is highly significant in linear algebra for system of equations and matrix solutions, making the systematic solution of systems with multiple variables feasible, as seen in the exercise above.
Linear Algebra Concepts
Linear algebra is a branch of mathematics focusing on vector spaces and the linear mappings between them. It is foundational to many mathematical concepts and has wide applications in science and engineering. Some key aspects of linear algebra concepts include:
  • Vectors: Objects that possess both magnitude and direction, used in the representation of points and transformations in space.
  • Matrices: Two-dimensional arrays that represent linear transformations or systems of equations.
  • Determinants and Eigenvalues: Special numbers associated with matrices that provide insights into their properties and the systems they represent.
  • Rank: The dimension of the vector space generated by its rows or columns, indicating the number of linearly independent rows or columns.
Linear algebra facilitates solving complex systems, and understanding reduced row-echelon form, Gaussian elimination, and matrix transformations could significantly aid in effectively managing these systems, as demonstrated in tackling the given matrices to find their RREF.

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Most popular questions from this chapter

A boy finds $$\$ 1.05$$ in dimes, nickels, and pennies. If there are 17 coins in all, how many coins of each type can he have?

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For each of the following homogeneous systems, find a set of basic solutions and express the general solution as a linear combination of these basic solutions. a. \(\begin{aligned} x_{1}+2 x_{2}-x_{3}+2 x_{4}+x_{5} &=0 \\ x_{1}+2 x_{2}+2 x_{3}+x_{5} &=0 \\ 2 x_{1}+4 x_{2}-2 x_{3}+3 x_{4}+x_{5} &=0 \end{aligned}\) \(\begin{array}{rr}\text { b. } & x_{1}+2 x_{2}-x_{3}+x_{4}+x_{5}=0 \\ - & x_{1}-2 x_{2}+2 x_{3}+x_{5}=0 \\ - & x_{1}-2 x_{2}+3 x_{3}+x_{4}+3 x_{5}=0\end{array}\) c. \(\begin{aligned} x_{1}+x_{2}-x_{3}+2 x_{4}+x_{5} &=0 \\ x_{1}+2 x_{2}-x_{3}+x_{4}+x_{5} &=0 \\ 2 x_{1}+3 x_{2}-x_{3}+2 x_{4}+x_{5} &=0 \\ 4 x_{1}+5 x_{2}-2 x_{3}+5 x_{4}+2 x_{5} &=0 \end{aligned}\) d. \(\begin{aligned} x_{1}+x_{2}-2 x_{3}-2 x_{4}+2 x_{5} &=0 \\ 2 x_{1}+2 x_{2}-4 x_{3}-4 x_{4}+x_{5} &=0 \\ x_{1}-x_{2}+2 x_{3}+4 x_{4}+x_{5} &=0 \\\\-2 x_{1}-4 x_{2}+8 x_{3}+10 x_{4}+x_{5} &=0 \end{aligned}\)

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