document.write( "Question 1118807: Solve using gauss elimination method\r
\n" ); document.write( "\n" ); document.write( "x+2y+2z=1; 2x+y+z=2; 3x+2y+2z=3; and y+z=0 \n" ); document.write( "

Algebra.Com's Answer #734275 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "With 4 equations and only 3 variables, there is a possibility that the system is over-constrained; there might not be a solution. Fortunately, it turns out that is not the case.

\n" ); document.write( "The beginning matrix for Gaussian elimination is

\n" ); document.write( " $\"matrix%284%2C4%2C1%2C2%2C2%2C1%2C2%2C1%2C1%2C2%2C3%2C2%2C2%2C3%2C0%2C1%2C1%2C0%29\"$

\n" ); document.write( "With the standard procedure for Gaussian elimination, the first step is to get a 1 in row 1 column 1 and 0s in the rest of column 1. The 1 in row 1 column 1 is already there; we can use it to get 0's in rows 2 and 3 of column 1.

\n" ); document.write( "Replace row 2 with (row 2 - 2*row 1); replace row 3 with (row3 - 3*row1):
\n" ); document.write( " $\"matrix%284%2C4%2C1%2C2%2C2%2C1%2C0%2C-3%2C-3%2C0%2C0%2C-4%2C-4%2C0%2C0%2C1%2C1%2C0%29\"$

\n" ); document.write( "I won't show the details of the next few steps. Clearly rows 2 and 3 are multiples of row 4. So if we move row 4 to be row 2 and use row 2 to simplify the last two rows, we get

\n" ); document.write( " $\"matrix%284%2C4%2C1%2C2%2C2%2C1%2C0%2C1%2C1%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%29\"$

\n" ); document.write( "Last, replace row 1 with (row1 - 2*row2), giving

\n" ); document.write( " $\"matrix%284%2C4%2C1%2C0%2C0%2C1%2C0%2C1%2C1%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%2C0%29\"$

\n" ); document.write( "The two non-zero rows then give us the solution:

\n" ); document.write( "x = 1; y+z = 0

\n" ); document.write( "This of course represents an infinite family of solutions. It is common to represent that family using a parameter:

\n" ); document.write( "x = 1; y = t, x = -t \n" ); document.write( "

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