Matrix and inverse matrix question?

Click For Summary
SUMMARY

The discussion focuses on solving the equation A² = λA - 2I to find the inverse of a specific matrix A. The user initially attempted to manipulate the equation by multiplying both sides by A⁻¹ but became confused due to the presence of two unknowns: A⁻¹ and λ. A participant clarified that λ can be deduced from the specific matrix A provided, emphasizing the importance of isolating I and factoring out A to find the multiplicative inverse correctly.

PREREQUISITES
  • Understanding of matrix operations, specifically multiplication and inversion.
  • Familiarity with eigenvalues and eigenvectors, particularly the concept of λ in matrix equations.
  • Knowledge of algebraic manipulation techniques in linear algebra.
  • Experience with identity matrices and their properties.
NEXT STEPS
  • Study the process of finding the inverse of a matrix using algebraic methods.
  • Learn how to deduce eigenvalues from matrix equations, focusing on the relationship between A and λ.
  • Explore the implications of the identity matrix in matrix equations and its role in finding inverses.
  • Practice solving similar matrix equations to reinforce understanding of linear algebra concepts.
USEFUL FOR

Students studying linear algebra, mathematicians working with matrix equations, and educators teaching matrix theory will benefit from this discussion.

nishantve1
Messages
74
Reaction score
1

Homework Statement


how do I solve this question
https://www.dropbox.com/s/bn958xnm5483s6u/photo%20%281%29.JPG
(This link if the image is not visible : https://www.dropbox.com/s/bn958xnm5483s6u/photo (1).JPG)
just so the equation is not clear it says A2 = λA - 2I

The inverse should be found through the equation in the question and not through the adjoint method

Homework Equations


The Attempt at a Solution



The equation says
A2 = λA - 2I
So, I mulitplied by A-1
This gives
A2 A-1 = λA A-1 - 2IA-1
{A-1 . A = I}

now the equation becomes
A = λI - 2A-1

I am stuck here since I cannot simply find A-1 by the adjoint method this equation has two unknowns A-1 and λ . Or maybe I am misinterpreting the question IDK
 
Physics news on Phys.org
But \lambda is not an unkown.

You are given a very specific matrix A. From your specific matrix A, and from your equation A^2= \lambda A- 2I, you can deduce what \lambda is.
 
How did I miss that . Thanks
 
I see no point in looking for specific values of \lambda. To find the multiplicative inverse of A:
1) algebraically manipulate the equation, A^2= \lambda A- 2I, to get "I" alone on the right.
2) factor out an "A".
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 69 ·
3
Replies
69
Views
10K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K