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Systems of Linear EquationsThe general methods by which we solve a set of simultaneous linear equations can lead us to a pattern such as:
![]() This is the situation of no solution. The last row gives us the equation
0 = 7
which is clearly an impossible situation. If you attempted to solve the system of equations that led to the above augmented matrix using substitution, you might obtain along the way something like:
If you solve the second of these for x2 and substitute it into the third, the x3's subtract out, and you get:
x2 = −7/2 + 3/2•x3 and
4(−7/2 + 3/2•x3) − 6x3 = −7 and
−14 + 6x3 − 6x3 = −7 therefore
−14 = −7 which is an indication of an impossibility and hence no solution!
![]() is not impossible. The last row produces an equation:
0 = 0
This situation indicates an infinite number of solutions. We can set the last variable, say x3, to any Real Number, t:
x3 = t
Then we can solve for the other two variables by substituting this into the equation for the second row to get x2 and finally substituting into the equation for the first row to obtain x1:
Another way of indicating this would be:
{(x1,x2,x3) | x1 = 9 − 8t, x2 = −7/2 + 3/2•t, and x3 = t; t ∈ (−&infin,&infin)}
which is read: "the set of all triplets, (x1,x2,x3), such that ....; where t is any Real value"
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© 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008 by Lawrence Turner |
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