document.write( "Question 661767: Use matrices to solve the system. (If the system has infinitely many solutions, express your answer in terms of c, where x = x(c), y = y(c), and z = c. If the system has no solution, enter NONE for each answer.)
\n" ); document.write( " 2x-3y+z=-3
\n" ); document.write( "3x+2y-z=1
\n" ); document.write( "5x-2y+z=5\r
\n" ); document.write( "\n" ); document.write( "(x,y,z)=? \n" ); document.write( "
Algebra.Com's Answer #411815 by math-vortex(648)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "Hi, there--\r\n" );
document.write( "\r\n" );
document.write( "Your three equations are: \r\n" );
document.write( "2x-3y+z=-3\r\n" );
document.write( "3x+2y-z=1\r\n" );
document.write( "5x-2y+z=5\r\n" );
document.write( "\r\n" );
document.write( "The 3x4 augmented matrix is\r\n" );
document.write( "\r\n" );
document.write( "2, -3, 1, -3\r\n" );
document.write( "3, 2, -1, 1\r\n" );
document.write( "5, -2, 1, 5\r\n" );
document.write( "\r\n" );
document.write( "Now we perform row operations to attempt to transform the matrix into reduce row echelon \r\n" );
document.write( "form (rref).\r\n" );
document.write( "\r\n" );
document.write( "Step one. We want the first element in Row 1 to be a 1. We multiply the entire row by 1/2 because 1/2 times 2 is 1.\r\n" );
document.write( "\r\n" );
document.write( "Multiply (1/2)*R1.\r\n" );
document.write( "\r\n" );
document.write( "1, -3/2, 1/2, -3/2\r\n" );
document.write( "3, 2, -1, 1\r\n" );
document.write( "5, -2, 1, 5\r\n" );
document.write( "\r\n" );
document.write( "Now we want the first element in Row 2 to be 0. We add -3 times Row 1 to Row 2.\r\n" );
document.write( "\r\n" );
document.write( "Add -3*R1 to R2.\r\n" );
document.write( "\r\n" );
document.write( "1, -3/2, 1/2, -3/2\r\n" );
document.write( "0, 13/2, -5/2\r\n" );
document.write( "5,-2, 1, 5\r\n" );
document.write( "\r\n" );
document.write( "Next, we want the first element in Row 3 to be 0. We add -5 times Row 1 to Row 3.\r\n" );
document.write( "\r\n" );
document.write( "Add -5*R1 to R3.\r\n" );
document.write( "\r\n" );
document.write( "1, -3/2, 1/2, -3/2\r\n" );
document.write( "0, 13/2, -5/2, 11/2\r\n" );
document.write( "0, 11/2, -3/2, 25/2\r\n" );
document.write( "\r\n" );
document.write( "Notice that Column 1 looks like 1 0 0. We Want the 2nd Column to look like 0 1 0 so we repeat \r\n" );
document.write( "the process we just performed on the first column.\r\n" );
document.write( "\r\n" );
document.write( "Multiply (2/13)*R2.\r\n" );
document.write( "\r\n" );
document.write( "1, -3/2, 1/2, -3/2\r\n" );
document.write( "0, 1, -5/13, 11/13\r\n" );
document.write( "0, 11/2, -3/2, 25/2\r\n" );
document.write( "\r\n" );
document.write( "Add (3/2)*R2 to R1.\r\n" );
document.write( "\r\n" );
document.write( "1, 0, -1/13, -3/13\r\n" );
document.write( "0, 1, -5/13, 11/13\r\n" );
document.write( "0, 11/2, -3/2, 25/2\r\n" );
document.write( "\r\n" );
document.write( "Add (-11/2)*R2 to R3.\r\n" );
document.write( "\r\n" );
document.write( "1, 0, -1/13, -3/13\r\n" );
document.write( "0, 1, -5/13, 11/13\r\n" );
document.write( "0, 0, 8/13, 102/13\r\n" );
document.write( "\r\n" );
document.write( "We want the third column to look like 0 0 1. We multiply every element in Row 3 by 13/8 \r\n" );
document.write( "because (8/13)*(13/8)=1. Then repeat the above process to get zeroes in the other column 3 \r\n" );
document.write( "positions.\r\n" );
document.write( "\r\n" );
document.write( "Multiply (13/8)*R3\r\n" );
document.write( "\r\n" );
document.write( "1, 0 -1/13, -3/13\r\n" );
document.write( "0, 1, -5/13, 11/13\r\n" );
document.write( "0, 0, 1, 51/4\r\n" );
document.write( "\r\n" );
document.write( "Add (5/13)*R3 to R2.\r\n" );
document.write( "\r\n" );
document.write( "1, 0, -1/13, -3/13\r\n" );
document.write( "0, 1, 0, 23/4\r\n" );
document.write( "0, 0, 1, 51/4\r\n" );
document.write( "\r\n" );
document.write( "Add (1/13)*R3 to R1.\r\n" );
document.write( "\r\n" );
document.write( "1, 0, 0, 3/4\r\n" );
document.write( "0, 1, 0, 23/4\r\n" );
document.write( "0, 0, 1, 51/4\r\n" );
document.write( "\r\n" );
document.write( "The matrix is now in rref. (x, y, z) = (3/4, 23/4, 51/4)\r\n" );
document.write( "\r\n" );
document.write( "Were is a link to a row operation calculator website where you can enter your matrix and \r\n" );
document.write( "experiment with row operations to find the solution. This is excellent practice. You might \r\n" );
document.write( "considering trying it with the two matrix problems you submitted here so you can see how \r\n" );
document.write( "nicely it works.\r\n" );
document.write( "\r\n" );
document.write( "http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=roc\r\n" );
document.write( "\r\n" );
document.write( "Best,\r\n" );
document.write( "Mrs. Figgy\r\n" );
document.write( "
\n" ); document.write( "
\n" );