document.write( "Question 1093175: x+y+z=-2 2x+8y+5z=5 x+2y-z=2 solve by row equivalent \n" ); document.write( "
Algebra.Com's Answer #707784 by Edwin McCravy(20056) $\"\"$ $\"About$
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$\"system%28x+%2B+y+%2B+z=+-2%2C%0D%0A2x+%2B+8y+%2B+5z=+5%2C%0D%0Ax+%2B+2y+-+z=+2%29\"$ \r
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document.write( "Write that as a matrix by dropping the letters\r\n" );
document.write( "and putting vertical line instead of equal signs:\r\n" );
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document.write( "The idea is to get three zeros in the three positions\r\n" );
document.write( "in the lower left corner of the matrix, where the elements\r\n" );
document.write( "I've colored red are:\r\n" );
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document.write( "To get a 0 where the red 2 on the left of the middle row is,\r\n" );
document.write( "multiply R1 by -2 and add it to 1 times R2, and put it in place \r\n" );
document.write( "of the present R2.  That's written as\r\n" );
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document.write( "-2R1+1R2->R2\r\n" );
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document.write( "To make it easy, write the multipliers to the left of the two\r\n" );
document.write( "rows you're working with; that is, put a -2 by R1 and a 1 by R2\r\n" );
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document.write( "We are going to change only R2.  Although R1 gets multiplied\r\n" );
document.write( "by -2 we are going to just do that mentally and add it to R2, but\r\n" );
document.write( "not really change R1.\r\n" );
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document.write( "Since it just happens that row 2 can be divided through by 3,\r\n" );
document.write( "without getting fractions, let's do it.  We multiply row 2 by\r\n" );
document.write( "1/3.  That operation is written\r\n" );
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document.write( "R2->R2\r\n" );
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document.write( "To get a 0 where the lower left red 1 is, multiply R1\r\n" );
document.write( "by -1 and add it to 1 times R3.  That's written as\r\n" );
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document.write( "-1R1+1R3->R3\r\n" );
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document.write( "Write the multipliers to the left of the two rows you're \r\n" );
document.write( "working with; that is, put a -1 by R1 and a 1 by R3\r\n" );
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document.write( "We are going to change only R3. \r\n" );
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document.write( "It will be more convenient to get a 1 where the 2 is,\r\n" );
document.write( "so we swap rows 2 and 3.  This is written\r\n" );
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document.write( "R2<->R3\r\n" );
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document.write( "To get a 0 where the red 2 is, multiply R2\r\n" );
document.write( "by -2 and add it to R3.  That's written as\r\n" );
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document.write( "-2R2+R3->R3\r\n" );
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document.write( "Write the multipliers to the left of the two\r\n" );
document.write( "rows you're working with; that is, put a -2 by R2 and a 1 by R3\r\n" );
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document.write( "We are going to change only R3. \r\n" );
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document.write( "Since it just happens that row 3 can be divided through by 5,\r\n" );
document.write( "without getting fractions, let's do it.  We multiply row 3 by\r\n" );
document.write( "1/5.  That operation is written\r\n" );
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document.write( "R3<->R3\r\n" );
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document.write( "Now that we have 0's in the three positions in the\r\n" );
document.write( "lower left corner of the matrix, we change the matrix\r\n" );
document.write( "back to equations:\r\n" );
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document.write( "he third equation gives us f\r\n" );
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document.write( "Substitute -1 for z in the middle equation:\r\n" );
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document.write( "Substitute -1 for z and 2 for y in the top equation:\r\n" );
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document.write( "So the solution is \r\n" );
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document.write( "Edwin
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