document.write( "Question 1078150: Using Gauss-Jordan elimination, write the following matrix
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Algebra.Com's Answer #692680 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "We need 0's where the 4,8,0,-9,leftmost 3, and 7 are.\r\n" );
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document.write( "To get a 0 where the 4 is:\r\n" );
document.write( "Multiply the 1st row by -2 and add it to the 2nd row,\r\n" );
document.write( "and put the result in place of row 2\r\n" );
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document.write( "That operation is abbreviated -2R1+R2->R2\r\n" );
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document.write( "To get a 0 where the 8 is:\r\n" );
document.write( "Multiply row 1 by -4 and add it to the row 3,\r\n" );
document.write( "and put the result in place of row 3\r\n" );
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document.write( "That operation is abbreviated -4R1+R3->R3\r\n" );
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document.write( "We already have a 0 at the top of column 2.\r\n" );
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document.write( "To get a 0 where the -9 is:\r\n" );
document.write( "Multiply row 2 by -3 and add it to the row 3,\r\n" );
document.write( "and put the result in place of row 3\r\n" );
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document.write( "That operation is abbreviated -3R2+R3->R3\r\n" );
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document.write( "Now that we have our 0's, we get the 1's.\r\n" );
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document.write( "We multiply row 1 by 1/2, and row 2 by -1/3.\r\n" );
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document.write( "Those operations are abbreviated (1/2)R1->R1 and (-1/3)R2->R2\r\n" );
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document.write( "Edwin

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