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$$\left\{ {\matrix{ {a_1x + b_1y + c_1z = d_1} \cr {a_2x + b_2y + c_2z = d_2} \cr {a_3x + b_3y + c_3z = d_3} \cr } } \right.$$

It's given that the unique solution for the system is $(2,-1,0)^T$.

Find the solutions set for the following linear system:

$$\left\{ {\matrix{ {d_1x + b_1y + c_1z = a_1} \cr {d_2x + b_2y + c_2z = a_2} \cr {d_3x + b_3y + c_3z = a_3} \cr } } \right.$$

Well, since the original system has a unique solution, I can infer all the stuff which are true for this kind ($\det(A)\ne 0$, $Ax=0$ has the trivial solution, $A$ is invertible, etc..)

I couldn't make the connection though, to the desired system.
What's the trick here?

Maybe subtracting the desired system's rows from the original system's row?

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I hope you are not intending to eliminate the person in your icon - he is much too important for that ;-) – David Jun 28 '14 at 14:10
Haha, good one :) – Elimination Jun 28 '14 at 14:11
up vote 1 down vote accepted

Think of it this way

you have solved

$$\left\{ {\matrix{ {a_1x + b_1y + c_1z + d_1 w=0} \cr {a_2x + b_2y + c_2z + d_2 w=0} \cr {a_3x + b_3y + c_3z + d_3 w=0} \cr } } \right. $$ and the solution is $(-2a,a,0,a)$ setting $a=-1$, makes the coefficients for the $d$'s into $-1$ and gives the original solution.

now you want so solve $$\left\{ {\matrix{ {-a_1 + b_1y + c_1z + d_1 w=0} \cr {-a_2 + b_2y + c_2z + d_2 w=0} \cr {-a_3 + b_3y + c_3z + d_3 w=0} \cr } } \right. $$

Here we want to chose $a$ so that the coefficient if the $a_i$'s is $-1$ that is $-2a=-1$ so $a=\frac{1}{2}$ and this gives the solution


So noting the rearrangement of the variables equation as given in the question we would have $x=\frac{1}{2}, y=\frac{1}{2}, z=0$

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I can see you turned the system equation to the form of $Ax=0$ with $4$ variables where the coefficients of $w$ are the $b$'s of the original system equation. How can you conclude immediately the solutions set is $(2a,-a,0,a)$? I'm probably missing something very fundamental here. – Elimination Jun 28 '14 at 14:21
No your confusing me,...the solution to a homogenous system depends on a arbitrary constant. – Rene Schipperus Jun 28 '14 at 14:27
Can you help me solve the second system? The signs are the same for $y,z$ but inverse for $x, w$. So... – Elimination Jun 28 '14 at 14:56
I have edited my answer I hope it answers your question. Thanks for accepting it. – Rene Schipperus Jun 28 '14 at 19:20

Hint. From the given information, can you find all solutions of the following system? $$\left\{ {\matrix{ {a_1x + b_1y + c_1z + d_1w = 0} \cr {a_2x + b_2y + c_2z + d_2w = 0} \cr {a_3x + b_3y + c_3z + d_3w = 0} \cr } } \right.$$

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Let $$ A_1 = \left( \begin{array}{ccc} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{array} \right) \\ A_2 = \left( \begin{array}{ccc} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{array} \right) \\ A_3 = \left( \begin{array}{ccc} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{array} \right) \\ $$ Using Cramer's rule, we can say that, $$ \frac{\det(A_1)}{\det(A)} = 2,\frac{\det(A_2)}{\det(A)} = -1,\frac{\det(A_3)}{\det(A)} = 0 $$

For the second system the solutions are $$ \frac{\det(A)}{\det(A_1)} = \frac{1}{2},-\frac{\det(A_2)}{\det(A_1)} = \frac{1}{2},-\frac{\det(A_3)}{\det(A_1)} = 0 $$ Remember that when we interchange two columns in a matrix, the sign of its determinant changes.

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