![\"\"](\"/images/stars/stars-13.gif\")
You can put this solution on YOUR website!
\r\n" );
document.write( "Solve the system using an augmented matrix. Show you work \r\n" );
document.write( "\r\n" );
document.write( "5x + 4y - z = 1\r\n" );
document.write( "2x - 2y + z = 1\r\n" );
document.write( "-x - y + z = 2\r\n" );
document.write( "\r\n" );
document.write( "Put in the coefficients of all\r\n" );
document.write( "the unknowns that don't show them:\r\n" );
document.write( "\r\n" );
document.write( " 5x + 4y - z = 1\r\n" );
document.write( " 2x - 2y + z = 1\r\n" );
document.write( " -x - y + z = 2\r\n" );
document.write( "\r\n" );
document.write( "Erase all the letters, plus signs,\r\n" );
document.write( "and equal signs and put a line \r\n" );
document.write( "where the equal signs were: \r\n" );
document.write( "\r\n" );
document.write( "[ 5 4 -1 | 1 ]\r\n" );
document.write( "[ 2 -2 1 | 1 ]\r\n" );
document.write( "[-1 -1 1 | 2 ]\r\n" );
document.write( "\r\n" );
document.write( "It's easiest but not really necessary\r\n" );
document.write( "to rearrange the rows so that the \r\n" );
document.write( "simplest row is at the top. So\r\n" );
document.write( "swap the 1st and 3rd rows\r\n" );
document.write( "\r\n" );
document.write( "[-1 -1 1 | 2 ]\r\n" );
document.write( "[ 2 -2 1 | 1 ]\r\n" );
document.write( "[ 5 4 -1 | 1 ]\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The plan is seven-fold, in this order:\r\n" );
document.write( "\r\n" );
document.write( "1. Use the 1st row to get a 0 where the 2 is in the 2nd row\r\n" );
document.write( "2. Use the 1st row to get a 0 where the 5 is in the 3rd row\r\n" );
document.write( "3. Use the new 2nd row to get a 0 where the 4 is in the 3rd row\r\n" );
document.write( "\r\n" );
document.write( "Your matrix will then look like this:\r\n" );
document.write( "\r\n" );
document.write( " [ # # # | #]\r\n" );
document.write( " [ 0 # # | #]\r\n" );
document.write( " [ 0 0 # | #]\r\n" );
document.write( "\r\n" );
document.write( "There will be numbers where the #'s are.\r\n" );
document.write( "\r\n" );
document.write( "4. Divide each row through by the first non-zero number\r\n" );
document.write( "in each row, and the matrix will then look like this:\r\n" );
document.write( "\r\n" );
document.write( " [ 1 # # | #]\r\n" );
document.write( " [ 0 1 # | #]\r\n" );
document.write( " [ 0 0 1 | #]\r\n" );
document.write( "\r\n" );
document.write( "This form is called the reduced echelon form.\r\n" );
document.write( "\r\n" );
document.write( "4. Rewrite the matrix as three equations.\r\n" );
document.write( "5. Solve the 3rd equation for z\r\n" );
document.write( "6. Substitute that value for z into the 2nd equation and solve for y\r\n" );
document.write( "7. Substitute both the value of z and the value for y in the 1st\r\n" );
document.write( " equation and solve for x. \r\n" );
document.write( "\r\n" );
document.write( "Here we go:\r\n" );
document.write( "\r\n" );
document.write( "[-1 -1 1 | 2 ]\r\n" );
document.write( "[ 2 -2 1 | 1 ]\r\n" );
document.write( "[ 5 4 -1 | 1 ]\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "To get a zero where the 2 is, to the side, \r\n" );
document.write( "multiply the 1st row by 2 and the \r\n" );
document.write( "2nd row by 1, and add them vertically \r\n" );
document.write( "\r\n" );
document.write( " 1st row x 2 -2 -2 2 | 4\r\n" );
document.write( " 2nd row x 1 2 -2 1 | 1\r\n" );
document.write( " ----------------\r\n" );
document.write( " new 2nd row 0 -4 3 | 5\r\n" );
document.write( "\r\n" );
document.write( "Replace only the 2nd row by that bottom line\r\n" );
document.write( "in the matrix\r\n" );
document.write( "\r\n" );
document.write( "[-1 -1 1 | 2 ]\r\n" );
document.write( "[ 0 -4 3 | 5 ]\r\n" );
document.write( "[ 5 4 -1 | 1 ]\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "To get a zero where the 5 is, to the side, \r\n" );
document.write( "multiply the 1st row by 5 and the \r\n" );
document.write( "3rd row by 1, and add them vertically \r\n" );
document.write( "\r\n" );
document.write( " 1st row x 5 -5 -5 5 | 10\r\n" );
document.write( " 3rd row x 1 5 4 -1 | 1\r\n" );
document.write( " ---------------\r\n" );
document.write( " new 3rd row 0 -1 4 | 11\r\n" );
document.write( "\r\n" );
document.write( "Replace only the 3rd row by that bottom line\r\n" );
document.write( "in the matrix\r\n" );
document.write( "\r\n" );
document.write( "[12 -4 -7 | 8]\r\n" );
document.write( "[ 0 -26 13 | 37]\r\n" );
document.write( "[ 0 104 107 | -106]\r\n" );
document.write( "\r\n" );
document.write( "[-1 -1 1 | 2 ]\r\n" );
document.write( "[ 0 -4 3 | 5 ]\r\n" );
document.write( "[ 0 -1 4 | 11 ]\r\n" );
document.write( "\r\n" );
document.write( "To get a zero where the -1 is, to the side, \r\n" );
document.write( "multiply the 2nd row by 1 and the \r\n" );
document.write( "3rd row by -4, and add them vertically \r\n" );
document.write( "\r\n" );
document.write( " 2nd row x 1 [ 0 -4 3 | 5 ]\r\n" );
document.write( " 3rd row x -4 [ 0 4 -16 | -44 ] \r\n" );
document.write( " ----------------------\r\n" );
document.write( " new 3rd row 0 0 -13 | -39 \r\n" );
document.write( "\r\n" );
document.write( "Replace only the 3rd row by that bottom line\r\n" );
document.write( "in the matrix\r\n" );
document.write( "\r\n" );
document.write( "[-1 -1 1 | 2 ]\r\n" );
document.write( "[ 0 -4 3 | 5 ]\r\n" );
document.write( "[ 0 0 -13 | -39 ]\r\n" );
document.write( "\r\n" );
document.write( "Now we want to get the first non-zero number\r\n" );
document.write( "in each row to be 1. so we divide the 1st row\r\n" );
document.write( "through by -1, the 2nd row through by -4 and\r\n" );
document.write( "the third row through by -13:\r\n" );
document.write( "\r\n" );
document.write( "[ 1 1 -1 | -22]\r\n" );
document.write( "[ 0 1 -3/4 | -5/4]\r\n" );
document.write( "[ 0 0 1 | 3]\r\n" );
document.write( "\r\n" );
document.write( "This is reduced echelon form. If your teacher\r\n" );
document.write( "wants you to get it all the way to row-reduced\r\n" );
document.write( "echelon form, you'll have to get 0's where the\r\n" );
document.write( "1 and the -1 are in the 1st row and the -3/4\r\n" );
document.write( "in the 2nd row. But it can be \r\n" );
document.write( "solved from the reduced echelon form as\r\n" );
document.write( "follows. \r\n" );
document.write( "\r\n" );
document.write( "Erase the brackets and the line, and put the \r\n" );
document.write( "letters, plus signs, and equal signs back in:\r\n" );
document.write( "\r\n" );
document.write( " 1x + 1y - 1z = -2\r\n" );
document.write( " 0x + 1y - 3/4z = -5/4\r\n" );
document.write( " 0x + 0y + 1z = 3\r\n" );
document.write( "\r\n" );
document.write( "Erase the 0 terms and the 1 coefficients:\r\n" );
document.write( "\r\n" );
document.write( " x + y - z = -2\r\n" );
document.write( " y - 3/4z = -5/4\r\n" );
document.write( " z = 3\r\n" );
document.write( "\r\n" );
document.write( "The third equation tells us that z = 3.\r\n" );
document.write( "\r\n" );
document.write( "Subtitute 3 for z in the 2nd equation\r\n" );
document.write( "\r\n" );
document.write( " y - {3/4}(3) = -5/4\r\n" );
document.write( " y - 9/4 = -5/4\r\n" );
document.write( " y = -5/4 + 9/4\r\n" );
document.write( " y = 4/4\r\n" );
document.write( " y = 1\r\n" );
document.write( "\r\n" );
document.write( "Now substitute y = 1 and z = 3 in\r\n" );
document.write( "the 1st equation:\r\n" );
document.write( "\r\n" );
document.write( " x + y - z = -2\r\n" );
document.write( " x + 1 - 3 = -2\r\n" );
document.write( " x - 2 = -2\r\n" );
document.write( " x = 0\r\n" );
document.write( "\r\n" );
document.write( "So the solution is\r\n" );
document.write( "\r\n" );
document.write( "(x, y, z) = (0, 1, 3)\r\n" );
document.write( "\r\n" );
document.write( "Edwin \r\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "