Are These Row Equivalent Matrices? Why Am I Getting Different Results?

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SUMMARY

The discussion centers on the confusion regarding row equivalent matrices derived from a given matrix A. The user calculated matrix B using a TI-83 Plus calculator, resulting in a matrix that does not match the book's provided answer. The book's matrix is correctly identified as not being in Reduced Row Echelon Form (RREF), while the user's matrix also fails to meet RREF criteria. The unique RREF of matrix A can be verified using Wolfram|Alpha, emphasizing the importance of accurate row-reduction techniques.

PREREQUISITES
  • Understanding of matrix operations and row reduction techniques
  • Familiarity with Reduced Row Echelon Form (RREF) and its properties
  • Experience using graphing calculators, specifically the TI-83 Plus
  • Basic knowledge of linear algebra concepts, including row equivalency theorem
NEXT STEPS
  • Learn how to perform row reduction manually to ensure accuracy
  • Explore the properties of RREF and its uniqueness in linear algebra
  • Utilize Wolfram|Alpha for matrix calculations and verifications
  • Study common pitfalls in matrix operations to avoid errors in calculations
USEFUL FOR

Students and educators in linear algebra, mathematicians working with matrices, and anyone seeking to understand row equivalency and RREF in matrix theory.

Pull and Twist
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I am having trouble with the following problem;

a.) Find a matrix B in reduced echelon form such that B is row equivalent to the given matrix A.

A=$$\left[\begin{array}{c}1 & 2 & 3 & -1 \\ 3 & 5 & 8 & -2 \\ 1 & 1 & 2 & 0 \end{array}\right]$$

So using my calculator I am able to get,

B=$$\left[\begin{array}{c}1 & 5/8 & 8/3 & -2/3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

The problem is that the book claims that B should be;

B=$$\left[\begin{array}{c}1 & 2 & 3 & -1 \\ 0 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

Which is weird cause according to the row equivalency theorem, any B of A in REF should be the same unique row equivalent matrix. Why am I not getting the same result?? I am using a TI-83 Plus.
 
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PullandTwist said:
I am having trouble with the following problem;

a.) Find a matrix B in reduced echelon form such that B is row equivalent to the given matrix A.

A=$$\left[\begin{array}{c}1 & 2 & 3 & -1 \\ 3 & 5 & 8 & -2 \\ 1 & 1 & 2 & 0 \end{array}\right]$$

So using my calculator I am able to get,

Unfortunately, we cannot verify your calculator input, nor the accuracy of said calculator. Sorry about that.

B=$$\left[\begin{array}{c}1 & 5/8 & 8/3 & -2/3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

The problem is that the book claims that B should be;

B=$$\left[\begin{array}{c}1 & 2 & 3 & -1 \\ 0 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

Which is weird cause according to the row equivalency theorem, any B of A in REF should be the same unique row equivalent matrix. Why am I not getting the same result?? I am using a TI-83 Plus.

only RREF is unique, the matrix in your book appears to be properly calculated, but it is NOT in RREF, nor is your matrix $B$ (you have non-zero entries in row 1 above the leading one in row 2, as does the matrix in your book). The RREF can be found here:

Wolfram|Alpha: Computational Knowledge Engine

I verified your book's answer, using the following steps:

a) Subtract 3 times row 1 from row 2
b) Multiply row 2 by -1
c) Subtract row 1 from row 3
d) Add row 2 to row 3

Row-reduction is a path fraught with peril, simple mistakes can ruin everything.
 

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