Help with simultaneous equations
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From: Fred Nurk on 4 Jul 2010 19:11 Solve the equations 2x + 2y - z = 1 and x - y + z = 0 for x, y and z in terms of a parameter lamda. I've three equations: 4y - 3z = 1 4x + z = 1 3x + y = 1 What does 'for x, y and z in terms of a parameter lamda' mean?
From: OwlHoot on 4 Jul 2010 19:37 On Jul 5, 12:11 am, Fred Nurk <albert.xtheunkno...(a)gmail.com> wrote: > > Solve the equations 2x + 2y - z = 1 and x - y + z = 0 for x, y and z > in terms of a parameter lamda. > > I've three equations: > 4y - 3z = 1 > 4x + z = 1 > 3x + y = 1 > > What does 'for x, y and z in terms of a parameter lamda' mean? You don't have three equations, because those three are linearly dependent. Their determinant is zero: | 0 4 -3 | | | | 4 0 1 | | | | 3 1 0 | You need to throw another equation into the pot, to add to the original pair, and make three equations with non-zero determinant. The easiest way to do that is to think of the original pair as a couple of dot products of vectors, one of which is the "unknown vector": (2, 2, -1).(x, y, z) = 1 and (1, -1, 1).(x, y, z) = 0 If your third equation uses the vector product of (2, 2, 1) and (1, -1, 1), which is perpendicular to both these, then the determinant of the resulting three equations cannot be zero: ((2, 2, -1) ^ (1, -1, 1)).(x, y, z) = \lambda That's where the \lambda comes in - You don't know the numeric value of the resulting dot product. So it represents the "wiggle room" in the system. (If you had more variables, e.g. 4 equations in five variables, the same trick can be done by making each coefficient on the left hand side of your extra "dummy" equation the cofactor of the resulting determinant. That way when you expand the resulting determinant along the bottom row, it will be a sum of squares, and hence non-zero (unless the system is little or no more than multiples of the same equation). Cheers John Ramsden
From: Alois Steindl on 4 Jul 2010 19:39 Am 2010-07-05 01:11, schrieb Fred Nurk: > Solve the equations 2x + 2y - z = 1 and x - y + z = 0 for x, y and z in > terms of a parameter lamda. > > I've three equations: > 4y - 3z = 1 > 4x + z = 1 > 3x + y = 1 > > What does 'for x, y and z in terms of a parameter lamda' mean? Hello, homework? How do you arrive at these equations? The first system of equations consists of 2 equations, the second one of 3. How comes? What have you tried so far? Alois
From: PiperAlpha167 on 4 Jul 2010 19:07 > Solve the equations 2x + 2y - z = 1 and x - y + z = 0 > for x, y and z in > terms of a parameter lamda. > > I've three equations: > 4y - 3z = 1 > 4x + z = 1 > 3x + y = 1 > > What does 'for x, y and z in terms of a parameter > lamda' mean? Write the first two equations in matrix form. Take the LHS as the standard (e-frame) matrix representation of the underlying linear transformation (call it T). So T is a map on R^3 onto R^2. Consider the one-dimensional linear manifold residing in the domain of T whose image under T is (1,0). I suggest that you've been asked to parametrize this manifold. Call the map s, and consider s: R -> R^3 defined by the rule, s(t) = (1/4, 1/4, 0) + (-1/4, 3/4, 1)t. (My 't' is your 'lambda'.) Now relate the triples on the manifold given above to your other second system of equations. As already noted, that system's coeffiecient matrix has zero determinant. But in this case, there's an infinity of solutions to the system. All the solutions are on the manifold given above, i.e., it constitutes the solution set for the second system.
From: Fred Nurk on 5 Jul 2010 00:09
Fred Nurk wrote: > <snip> > What does 'for x, y and z in terms of a parameter lamda' mean? The answer is: x = (1 - lamda) / 4, y = (3lamda + 1) / 4, z = lamda. My reasoning: Take z to be lamda and write equations with x and y as subject. All that was needed was to rearrange the second and first equations in that order. What do you think of my reasoning? |