document.write( "Question 1105260: Solve the system of equations using matrices. Use Gaussian elimination with back-substitution. x + y + z = -5
\n" ); document.write( "x - y + 3z = -1
\n" ); document.write( "4x + y + z = -2 \n" ); document.write( "

Algebra.Com's Answer #720048 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "I'm not sure what you mean by \"Gaussian elimination with back substitution\". Gaussian elimination solves the system; there is no substitution required.

\n" ); document.write( " $\"matrix%283%2C4%2C1%2C1%2C1%2C-5%2C1%2C-1%2C3%2C-1%2C4%2C1%2C1%2C-2%29\"$

\n" ); document.write( "You have the required \"1\" in position (1,1). Use it to get 0 in positions (2,1) and (3,1):
\n" ); document.write( "R2 <-- R1-R2; R3 <-- R3-4*R1: $\"matrix%283%2C4%2C1%2C1%2C1%2C-5%2C0%2C2%2C-2%2C-4%2C0%2C-3%2C-3%2C18%29\"$

\n" ); document.write( "Take out the common factors in rows 2 and 3: $\"matrix%283%2C4%2C1%2C1%2C1%2C-5%2C0%2C1%2C-1%2C-2%2C0%2C1%2C1%2C-6%29\"$

\n" ); document.write( "Use the 1 in position (2,2) to get 0 in positions (1,2) and (3,2):
\n" ); document.write( "R1 <-- R1-R2; R3 <-- R3-R2: $\"matrix%283%2C4%2C1%2C0%2C2%2C-3%2C0%2C1%2C-1%2C-2%2C0%2C0%2C2%2C-4%29\"$

\n" ); document.write( "Take out the common factor in row 3: $\"matrix%283%2C4%2C1%2C0%2C2%2C-3%2C0%2C1%2C-1%2C-2%2C0%2C0%2C1%2C-2%29\"$

\n" ); document.write( "Use the 1 in position (3,3) to get 0 in positions (1,3) and (2,3):
\n" ); document.write( "R1 <-- R1-2*R3; R2 <-- R2+R3: $\"matrix%283%2C4%2C1%2C0%2C0%2C1%2C0%2C1%2C0%2C-4%2C0%2C0%2C1%2C-2%29\"$

\n" ); document.write( "We are done; the system has a unique solution
\n" ); document.write( "x = 1; y = -4, z = -2 \n" ); document.write( "

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