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 #257611 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"A%5E4+=+A\"$ . Then $\"A%5E4+-A=O\"$ . (The zero matrix!)Taking the determinants of both sides,
\n" ); document.write( " $\"det%28A%5E4+-A%29=det%28O%29+=+0\"$ ,
\n" ); document.write( " $\"det%28A%28A%5E3+-+I%29%29+=+0\"$ ,
\n" ); document.write( " $\"det%28A%29%2Adet%28A%5E3+-+I%29+=+0\"$ ,or det(A) = 0, or $\"det%28A%5E3+-+I%29+=+0\"$ .
\n" ); document.write( "If A is nonsingular, then det(A)is NOT equal to zero, but $\"A%5E3+-+I=+O\"$ ,(because $\"A%5E3+=+A%5E4%2AA%5E-1+=+A%2AA%5E-1+=+I\"$ ) so $\"det%28A%5E3+-+I%29+=+0\"$ . Hence
\n" ); document.write( " $\"A%5E2+=+A%5E3%2AA%5E-1+=+I%2AA%5E-1+=+A%5E-1\"$ .
\n" ); document.write( "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 $\"det%28A%5E3+-+I%29\"$ won't matter anymore. \n" ); document.write( "

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