document.write( "Question 39572: Proof of a singular matrix:
\n" ); document.write( "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?) \n" ); document.write( "

Algebra.Com's Answer #25103 by venugopalramana(3286) $\"\"$ $\"About$

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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?)
\n" ); document.write( "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
\n" ); document.write( "A*A^(-1)=A^(-1)*A=I...AND A*B=0 IMPLIES A=0 OR B=0.....
\n" ); document.write( "IN THIS CASE WE HAVE
\n" ); document.write( "A^4=A
\n" ); document.write( "(A^4)-A=0
\n" ); document.write( "A(A^3-I)=0...HENCE A=0...OR.....A^3-I=0
\n" ); document.write( "A^3=I...MULTIPLY WITH A^-1 BOTH SIDES
\n" ); document.write( "A^2*A*(A^-1)=I*(A^-1)=A^-1
\n" ); document.write( "A^2=A^(-1) \n" ); document.write( "

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