Linear Algebra - Inverse of Matrices

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SUMMARY

The discussion centers on finding the inverse of a 2-by-2 matrix A (0,8 / 5,7) as a product of elementary matrices. The user initially performed row operations to convert A into reduced row echelon form, resulting in incorrect elementary matrices. The correct elementary matrices derived from the row operations are (0,1 / 1,0), (1/5,0 / 0,1), (1,0 / 1/8,0), and (1,-7/5 / 0,1). The inverse of A is obtained by taking the inverse of these matrices in the correct order, with the last elementary matrix corrected to (1, -7/5 / 0, 1).

PREREQUISITES
  • Understanding of elementary matrices and their role in matrix operations
  • Familiarity with row operations in linear algebra
  • Knowledge of reduced row echelon form (RREF)
  • Ability to compute matrix inverses
NEXT STEPS
  • Study the process of converting matrices to reduced row echelon form using elementary row operations
  • Learn how to derive elementary matrices from specific row operations
  • Explore the concept of matrix inverses and how to compute them for 2-by-2 matrices
  • Practice problems involving the decomposition of matrices into products of elementary matrices
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators teaching matrix theory and operations.

Lorelyn
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I have a problem in which I am given a 2-by-2 matrix A (0,8 / 5,7) which I am asked to write as a product of elementary matrices. Then I am asked write the inverse of A as a product of elementary matrices. So I turned A into reduced row echlon form using the following steps:

-Switch R1 and R2
-1/5 * R1
-1/8 * R2
- (-7/5) * R2 + R1 into R1

So the elementary matrices are:
(0,1 / 1,0) * (1/5,0 / 0,1) * (1,0 / 1/8,0) and (1,-7/5 / 0,-7/5)

Right?

Then the inverse of A as a product of elementary matrices is just the inverse of each of (1,-7/5 / 0,-7/5) * (1,0 / 1/8,0) * (1/5,0 / 0,1) and (0,1 / 1,0).

Does this make sense? Because I keep getting the wrong answer and I don't know where I've gone wrong...
 
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Lorelyn said:
I have a problem in which I am given a 2-by-2 matrix A (0,8 / 5,7) which I am asked to write as a product of elementary matrices. Then I am asked write the inverse of A as a product of elementary matrices. So I turned A into reduced row echlon form using the following steps:

-Switch R1 and R2
-1/5 * R1
-1/8 * R2
- (-7/5) * R2 + R1 into R1
So the elementary matrices are:
(0,1 / 1,0) * (1/5,0 / 0,1) * (1,0 / 1/8,0) and (1,-7/5 / 0,-7/5)

Right?
No. Since you did not change the second row in the last step, the bottom row of the last elementary matrix is "0 1" not"0 -7/5". The elementary matrix corresponding to a row-operation is the matrix derived by applying that row operation to the identity matrix. Sutracting 7/5 of the second row from the first row of the identity matrix gives (1, -7/5, 0, 1).

Then the inverse of A as a product of elementary matrices is just the inverse of each of (1,-7/5 / 0,-7/5) * (1,0 / 1/8,0) * (1/5,0 / 0,1) and (0,1 / 1,0).

Does this make sense? Because I keep getting the wrong answer and I don't know where I've gone wrong...
The last elementary matrix is (1 , -7/5, 0, 1) and its inverse is (1, +7/5, 0, 1).
 
Thanks! I actually just figured it out by myself all it took was a couple hours staring at it blankly before something clicked in my brain! :)
 

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