document.write( "Question 40518This question is from textbook
\n" ); document.write( ": I am struggling with this proof, please help!\r
\n" ); document.write( "\n" ); document.write( "Prove that the row vectos of an n x n intertible matrix A form a basis for R^n
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\n" ); document.write( "Thank you!
\n" ); document.write( "Glenna \n" ); document.write( "

Algebra.Com's Answer #26035 by kev82(151) $\"\"$ $\"About$

You can put this solution on YOUR website!
Hi,\r
\n" ); document.write( "\n" ); document.write( "For the vectors to form a basis of

, there must be n of them and they must be linearly independent. Clearly there are n of them as the matrix is an nxn matrix. So the proof reduces to showing that the rows of an invertable matrix are linearly independent.\r
\n" ); document.write( "\n" ); document.write( "Lets assume that an invertable matrix exists which has linearly dependent rows. Because the rows are linearly dependent we can perform elementary row operations to make a row entirely of zeroes. (Note elementary row operations don't affect the determinant) Now evaluating the determinant along this row of zeroes gives zero.\r
\n" ); document.write( "\n" ); document.write( "An invertable matrix must have non-zero determinant, so by contradiction there can be no invertable matrix with linearly dependent rows. Hence every invertable matrix has linearly independent rows which is what we needed.\r
\n" ); document.write( "\n" ); document.write( "Hope that helps,
\n" ); document.write( "Kev \n" ); document.write( "

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