SOLUTION: Proof of a singular matrix:
I need to show (in general) that either matrix A is "singular" or "A^2 = A^(-1)". A is an n*n square matrix such that A^4 = A. I want to know how to do
Algebra ->
College
-> Linear Algebra
-> SOLUTION: Proof of a singular matrix:
I need to show (in general) that either matrix A is "singular" or "A^2 = A^(-1)". A is an n*n square matrix such that A^4 = A. I want to know how to do
Log On
Question 39572: Proof of a singular matrix:
I need to show (in general) that either matrix A is "singular" or "A^2 = A^(-1)". A is an n*n square matrix such that A^4 = A. I want to know how to do this problem and what happens if the exponent values are different.(for example, odd number exponents?) Found 2 solutions by venugopalramana, robertb:Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! I need to show (in general) that either matrix A is "singular" or "A^2 = A^(-1)". A is an n*n square matrix such that A^4 = A. I want to know how to do this problem and what happens if the exponent values are different.(for example, odd number exponents?)
PROCEDURE IS TO,TRANSPOSITION TO LEFT OR RIGHT SIDE OF THE EQUATION, RIGHT OR LEFT MULTIPLY BOTH SIDES OF THE EQUATION WITH A OR A^(-1),TAKING COMMON FACTORS ACCORDINGLY....NOTING THAT A^N=A*A*A*...N TIMES AND
A*A^(-1)=A^(-1)*A=I...AND A*B=0 IMPLIES A=0 OR B=0.....
IN THIS CASE WE HAVE
A^4=A
(A^4)-A=0
A(A^3-I)=0...HENCE A=0...OR.....A^3-I=0
A^3=I...MULTIPLY WITH A^-1 BOTH SIDES
A^2*A*(A^-1)=I*(A^-1)=A^-1
A^2=A^(-1)
You can put this solution on YOUR website! Let . Then . (The zero matrix!)Taking the determinants of both sides, , , ,or det(A) = 0, or .
If A is nonsingular, then det(A)is NOT equal to zero, but ,(because ) so . Hence .
If A is singular , then det(A) = 0 (since a square matrix is singular if and only if its determinant is 0). In which case the value of won't matter anymore.
