Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let

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

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?

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

share|improve this question
1  
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

3 Answers 3

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

$(-1,\frac{1}{2},0,\frac{1}{2})$

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

share|improve this answer
    
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
1  
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
1  
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? $$\left\{ {\matrix{ {a_1x + b_1y + c_1z + d_1w = 0} \cr {a_2x + b_2y + c_2z + d_2w = 0} \cr {a_3x + b_3y + c_3z + d_3w = 0} \cr } } \right.$$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.