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$$2x+6y-z=10$$ $$3x+2y+2z=1$$ $$5x+4y+3z=4$$ I do not find any solution of this three equations.I know that if two equations or planes are parallel then we do't find any solution.But I can't understand what is the problem here.Are they parallel.and if they are parallel How can I understand that.please help me.

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I was gonna suggest Cramer's Rule, but you beat me to it. Good suggestion. –  recursive recursion Jan 12 at 16:09
1  
Why should there be a solution? The line of intersection of the first two planes could be parallel to the third plane. –  André Nicolas Jan 12 at 16:12
    
@fazla Rabbi Mashur : I would recommend Gauss-Jordan elimination, avoiding fractions as much as possible. –  Stefan Smith Jan 12 at 16:18

5 Answers 5

up vote 2 down vote accepted

Note that $$ \det\left(\begin{array}{ccc} 2&6&-1 \\ 3&2&2 \\ 5 & 4 & 3 \end{array}\right)=0$$

So there is either no so solution or an infinte amount of solutions.

Now substract the second line from the first line three times, and substract it from the third line twice, tends to

$$\begin{array}{ccc} -7x-5z &=& 7\\ -x-z &=& 2 \end{array}$$

Can you continue from this point?

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According to Wolfram Alpha,

$$z = 2x + 6y - 10, z = -\dfrac{3x}{2} - y + \dfrac{1}{2}, z = -\dfrac{5x}{3} - \dfrac{4y}{3} + \dfrac{4}{3}$$

So there seems to be no solution for this problem.

If we augment 3-by-3 matrix with 3-by-1 matrix and determine reduced row echelon form of 3-by-4 matrix, then we obtain

$$\left[\begin{array}{rrr|r} 1 & 0 & 1 & 0\\ 0 & 1 & -\frac{1}{2} & 0\\ 0 & 0 & 0 & 1 \end{array}\right]$$

(Unfortunately, MathJax won't let me set the augmented matrix like this.)

So indeed, there is no solution for this problem.

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There is no solution. Maybe try your luck with $$\tag{(2)+2(1)}7x+14y=21$$ $$\tag{(3)+3(1)}11x+22y=34$$ first.

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@user127001 Right, but only after correcting the mistake André hinted at. (Who works with numbers anyway?) –  Hagen von Eitzen Jan 12 at 16:20

If your three equations have solutions, they have to satisfy the followings :

$(2)+2\times (1)$ will give you $$x+2y=3.$$ $(3)+3\times (1)$ will give you $$x+2y=34/11.$$

However, these does not have any solution (these represent two parallel lines).

So, there is no solution for the given three equations.

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Note that $$\ det\begin{pmatrix} 2 & 6 & -1 \\ 3 & 2 & 2 \\ 5 & 4 & 3 \end{pmatrix}= 0$$ while $$rank \begin{pmatrix} 2 & 6 & -1 & 10\\ 3 & 2 & 2 & 1\\ 5 & 4 & 3 & 4 \end{pmatrix}=3$$ because $$\begin{vmatrix} 6 & -1 & 10 \\ 2 & 2 & 1\\ 4 & 3 & 4 \end{vmatrix} \neq 0$$ so the system has no solutions.

On the other hand reducing the augmented matrix $$ \begin{pmatrix} 2 & 6 & -1 & 10\\ 3 & 2 & 2 & 1\\ 5 & 4 & 3 & 4 \end{pmatrix} $$ $$ \longrightarrow \begin{pmatrix} 2 & 6 & -1 & 10\\ 0 & -2 & 1 & -7\\ 0 & 0 & 0 & 35 \end{pmatrix} $$. Seeing the last row you realize that can't be possible.

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