document.write( "Question 87113: solve homogeneous system of equations\r
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\n" ); document.write( "4x - 4y + z = 0\r
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Algebra.Com's Answer #63078 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
solve homogeneous system of equations
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document.write( "Every homogeneous system of 3 equations in\r\n" );
document.write( "3 unknowns has solution (x,y,z) = (0,0,0).\r\n" );
document.write( "But sometimes homogeneous systems have other\r\n" );
document.write( "solutions.  So we go through Gauss elimination:\r\n" );
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document.write( " x - 2y + 3z = 0\r\n" );
document.write( "3x - 7y - 4z = 0\r\n" );
document.write( "4x - 4y +  z = 0\r\n" );
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document.write( "Make this augmented matrix of coefficients:\r\n" );
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document.write( "[1 -2  3 | 0]\r\n" );
document.write( "[3 -7 -4 | 0]\r\n" );
document.write( "[4 -4  1 | 0]\r\n" );
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document.write( "The idea is to get 0's where the three red\r\n" );
document.write( "numbers are.\r\n" );
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document.write( "[1 -2  3 | 0]\r\n" );
document.write( "[3 -7 -4 | 0]\r\n" );
document.write( "[4 -4  1 | 0]\r\n" );
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document.write( "[1 -2  3 | 0]\r\n" );
document.write( "[3 -7 -4 | 0]\r\n" );
document.write( "[4 -2  1 | 0]\r\n" );
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document.write( "Get a 0 where the red 3 is by multiplying row 1 by\r\n" );
document.write( "-3 and adding it to 1 times row 2, but restore row 1:\r\n" );
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document.write( "-3×[1 -2  3 | 0]\r\n" );
document.write( " 1×[3 -7 -4 | 0]\r\n" );
document.write( "   [4 -2  1 | 0]\r\n" );
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document.write( "[1  -2   3 | 0]\r\n" );
document.write( "[0  -1 -13 | 0]\r\n" );
document.write( "[4  -2   1 | 0] \r\n" );
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document.write( "Get a 0 where the red 4 is by multiplying row 1 by\r\n" );
document.write( "-4 and adding it to 1 times row 3, but restore row 1:\r\n" );
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document.write( "-4×[1  -2   3 | 0]\r\n" );
document.write( "   [0  -1 -13 | 0]\r\n" );
document.write( " 1×[4  -2   1 | 0]\r\n" );
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document.write( "[1  -2   3 | 0]\r\n" );
document.write( "[0  -1 -13 | 0]\r\n" );
document.write( "[0   6 -11 | 0]\r\n" );
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document.write( "Get a 0 where the red 6 is by multiplying row 2 by\r\n" );
document.write( "6 and adding it to 1 times row 3, but restore row 2:\r\n" );
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document.write( "  [1  -2   3 | 0]\r\n" );
document.write( "6×[0  -1 -13 | 0]\r\n" );
document.write( "1×[0   6 -11 | 0] \r\n" );
document.write( "  \r\n" );
document.write( "[1  -2   3 | 0]\r\n" );
document.write( "[0  -1 -13 | 0]\r\n" );
document.write( "[0   0 -89 | 0]\r\n" );
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document.write( "Get a 1 where the red -1 is by dividing row 2 thru by -1\r\n" );
document.write( "Get a 1 where the red -89 is by dividing row 3 thru by -89\r\n" );
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document.write( "[1  -2   3 | 0]\r\n" );
document.write( "[0   1  13 | 0]\r\n" );
document.write( "[0   0   1 | 0]\r\n" );
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document.write( "Convert back into a system of equations:\r\n" );
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document.write( "1x - 2y +  3z = 0\r\n" );
document.write( "0x + 1y + 13z = 0\r\n" );
document.write( "0x + 0y +  1z = 0\r\n" );
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document.write( "Erase the first 0 term of the 2nd equation,\r\n" );
document.write( "Erase the first two 0 terms of the third equation:\r\n" );
document.write( "Erase the 1 coefficients\r\n" );
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document.write( " x + 2y +  3z = 0\r\n" );
document.write( "      y + 13z = 0\r\n" );
document.write( "            z = 0\r\n" );
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document.write( "The last equation tells us the only value of z is 0\r\n" );
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document.write( "Substitute in the 2nd equation:\r\n" );
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document.write( "     y + 13(0) = 0\r\n" );
document.write( "             y = 0\r\n" );
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document.write( "Substitute z = 0 and y = 0 in the first equation:\r\n" );
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document.write( "  x + 2(0) + 3(0) = 0\r\n" );
document.write( "        x + 0 + 0 = 0\r\n" );
document.write( "                x = 0\r\n" );
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document.write( "So there is but one solution,\r\n" );
document.write( "the obvious one.\r\n" );
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document.write( "(x, y, z) = (0, 0, 0)\r\n" );
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document.write( "Edwin
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