
Patrick D. answered 04/26/17
Tutor
5
(10)
Patrick the Math Doctor
X Y Z C
1 3 4 1
3 4 5 3
1 8 11 2
First, do -3*row1 + row2 and -Row1 + row3
X Y Z C
1 3 4 1
0 -5 -7 0
0 5 7 1
1 3 4 1
0 -5 -7 0
0 5 7 1
Finally, add rows 2 and 3 and store the result in row 2
1 3 4 1
0 -5 -7 0
0 0 0 1
The system has no solution as the last equation reads 0 = 1.
It is an inconsistent system.
To see this more clearly, do this without the matrix.
First pair the top two equations and eliminate x:
x + 3y + 4z = 1 <--- mutiplies by -3, then adds
3x + 4y + 5z = 3
-3x - 9y -12z = -3
3x + 4y + 5z = 3
3x + 4y + 5z = 3
-5y - 7z = 0 <--- remember this equation as equation A ***
Pairs up the last two equations and eliminates x.
3x + 4y + 5z = 3
x + 8y + 11z = 2 <-- multiplies by -3, then adds
3x + 4y + 5z = 3
-3x + -24y + -33z = -6
-20y + - 28z = -3 <--- remember this equation as equation B ***
putting equations A and B together:
-5y - 7z = 0
-20y + -28z = -3
Notice that equation B is exactly 4 times equation A. So multiplying equation A by -4 and the adding:
20y + 28z = 0
-20y + -28z = -3
0y + 0z = -3
0 + 0 = -3
0 = -3 <-- game over. system has no solution