document.write( "Question 87122: Prove or disprove whether the following matrices are equivalent or not. Explain how you came up with your answer.\r
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Algebra.Com's Answer #63331 by Edwin McCravy(20067)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Prove or disprove whether the following matrices are \r\n" );
document.write( "equivalent or not. Explain how you came up with your answer. \r\n" );
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document.write( "[1 2 3]\r\n" );
document.write( "[0 1 1] \r\n" );
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document.write( "[3 2 5]\r\n" );
document.write( "[5 4 9]\r\n" );
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document.write( "We must get both of these into \"row reduced echelon form\"\r\n" );
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document.write( "Let's get the first one to row reduced echelon form:\r\n" );
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document.write( "[1 2 3]\r\n" );
document.write( "[0 1 1]\r\n" );
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document.write( "Get a 0 where the 2 is by multiplying the bottom row by -2\r\n" );
document.write( "and adding it to 1 times the top row, then restoring \r\n" );
document.write( "the bottom row:\r\n" );
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document.write( " 1[1 2 3]\r\n" );
document.write( "-2[0 1 1]\r\n" );
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document.write( "[1 0 1]\r\n" );
document.write( "[0 1 1]\r\n" );
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document.write( "That is in row reduced echelon form because:\r\n" );
document.write( "1. the leading 1 in the bottom row is farther\r\n" );
document.write( "   to the right than the leading 1 in the top\r\n" );
document.write( "   row.\r\n" );
document.write( "2. Each of the leading 1's are the only non-zero\r\n" );
document.write( "   numbers in the columns they are in. \r\n" );
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document.write( "Now, let's get the second one to row reduced echelon form:\r\n" );
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document.write( "[3 2 5]\r\n" );
document.write( "[5 4 9]\r\n" );
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document.write( "Get a 0 where the 5 on the bottom row is by multiplying the \r\n" );
document.write( "top row by -5 and adding it to 3 times the bottom row, then \r\n" );
document.write( "restoring the top row:\r\n" );
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document.write( "-5[3 2 5]\r\n" );
document.write( " 3[5 4 9]\r\n" );
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document.write( "[3 2 5]\r\n" );
document.write( "[0 2 2]\r\n" );
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document.write( "Get a 0 where the 2 on the top row is by multiplying the \r\n" );
document.write( "bottom row by -1 and adding it to 1 times the top row, then \r\n" );
document.write( "restoring the bottom row:\r\n" );
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document.write( " 1[3 2 5]\r\n" );
document.write( "-1[0 2 2]\r\n" );
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document.write( "[3 0 3]\r\n" );
document.write( "[0 2 2]\r\n" );
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document.write( "Get a 1 where the first 3 in the top row is by dividing the\r\n" );
document.write( "top row through by 3.\r\n" );
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document.write( "Get a 1 where the first 2 in the bottom row is by dividing the\r\n" );
document.write( "bottom row through by 2.\r\n" );
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document.write( "[3 0 3]÷3\r\n" );
document.write( "[0 2 2]÷2\r\n" );
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document.write( "[1 0 1]\r\n" );
document.write( "[0 1 1]\r\n" );
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document.write( "So the given matrices are equivalent because the have the\r\n" );
document.write( "same row reduced echelon form, namely\r\n" );
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document.write( "[1 0 1]\r\n" );
document.write( "[0 1 1] \r\n" );
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document.write( "Note: It didn't apply here but in some matrices, to have \r\n" );
document.write( "row reduced echelon form, all all-zero rows must appear\r\n" );
document.write( "at the bottom of the matrix.  This can be accomplished\r\n" );
document.write( "by swapping rows.\r\n" );
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document.write( "Edwin
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