document.write( "Question 570370: Using the Gauss/Jordan method to solve the system, writing all row operations between the affected matrices, using the format: rn + k x rm = Rn The system is: x-2y+z=6
\n" ); document.write( " 2x+y-3z=-3
\n" ); document.write( " x-3y+3z=10
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #367684 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "\"system%28x-2y%2Bz=6%2C%0D%0A2x%2By-3z=-3%2C%0D%0Ax-3y%2B3z=10%29\"\r\n" );
document.write( "\r\n" );
document.write( "\"system%281x-2y%2B1z=6%2C%0D%0A2x%2B1y-3z=-3%2C%0D%0A1x-3y%2B3z=10%29\"\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We want to try to end up with a matrix \r\n" );
document.write( "that looks like this:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "with 0's in the three lower left hand positions:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We can get a 0 where the 2 is by multiplying Row 1 by -2\r\n" );
document.write( "and adding it to row 2:\r\n" );
document.write( "\r\n" );
document.write( "That instruction is written as\r\n" );
document.write( "\r\n" );
document.write( "-2·R1+1·R2->R2\r\n" );
document.write( "\r\n" );
document.write( "\"matrix%283%2C1%2C-2%2C1%2C%22%22%29\"\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Notice that Row 2 will be simpler if we divide it through by 5,\r\n" );
document.write( "\r\n" );
document.write( "That instruction is written \"1%2F5\"R2->R2\r\n" );
document.write( "\r\n" );
document.write( "\"matrix%283%2C1%2C%22%22%2C1%2F5%2C%22%22%29\"\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We can get a 0 where the 1 is in the lower left corner by \r\n" );
document.write( "multiplying Row 1 by -1 and adding it to row 3:\r\n" );
document.write( "\r\n" );
document.write( "That instruction is written as\r\n" );
document.write( "\r\n" );
document.write( "-1·R1+1·R3->R3\r\n" );
document.write( "\r\n" );
document.write( "\"matrix%283%2C1%2C-1%2C%22%22%2C1%29\"\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We can get a 0 where the -1 is in the bottom row by \r\n" );
document.write( "multiplying Row 2 by 1 and adding it to row 3:\r\n" );
document.write( "\r\n" );
document.write( "That instruction is written as 1R2+1R3->R3\r\n" );
document.write( "\r\n" );
document.write( "\"matrix%283%2C1%2C%22%22%2C1%2C1%29\"\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now that we have 0's in the lower lefthand corner,\r\n" );
document.write( "we convert the matrix back to a system of equations\r\n" );
document.write( "in x, y and z:\r\n" );
document.write( "\r\n" );
document.write( "\"system%28%0D%0A%0D%0A1x-2y%2B1z=6%2C%0D%0A0x%2B1y-1z=-3%2C%0D%0A0x%2B0y%2B1z=1%29\"\r\n" );
document.write( "\r\n" );
document.write( "or just\r\n" );
document.write( "\r\n" );
document.write( "\"system%28%0D%0A%0D%0Ax-2y%2Bz=6%2C%0D%0Ay-z=-3%2C%0D%0Az=1%29\"\r\n" );
document.write( "\r\n" );
document.write( "Now we use back substitution.\r\n" );
document.write( "\r\n" );
document.write( "From the third equation, z=1, we substitute\r\n" );
document.write( "that into the middle equation, getting:\r\n" );
document.write( "\r\n" );
document.write( "y-z = -3\r\n" );
document.write( "y-1 = -3\r\n" );
document.write( "  y = -2\r\n" );
document.write( "\r\n" );
document.write( "Then substitute y=-2 and z=1 in the 1st equation:\r\n" );
document.write( "\r\n" );
document.write( "x-2(-2)+(1) = 6\r\n" );
document.write( "      x+4+1 = 6\r\n" );
document.write( "        x+5 = 6\r\n" );
document.write( "          x = 1\r\n" );
document.write( "\r\n" );
document.write( "Solution (x,y,z) = (1,-2,1)\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );