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The values of $K$ for which the system \begin{cases} x + y = 5\\ y + z = 4\\ x + 2Kz = 1 \end{cases} has only one solution are…

Will the answer be ( All Real Numbers ) or ( the empty set )?

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5 Answers 5

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A general method is using Gaussian elimination; the matrix of the system is $$ \left[\begin{array}{ccc|c} 1&1&0&5\\ 0&1&1&4\\ 1&0&2K&1 \end{array}\right] $$ Subtract the first row from the third: $$ \left[\begin{array}{ccc|c} 1&1&0&5\\ 0&1&1&4\\ 0&-1&2K&-4 \end{array}\right] $$ Sum the third row with the second: $$ \left[\begin{array}{ccc|c} 1&1&0&5\\ 0&1&1&4\\ 0&0&1+2K&0 \end{array}\right] $$ Now, if $1+2K\ne0$, the system has unique solution. If $1+2K=0$, the system has infinitely many solutions.

You need no smart substitutions or note anything special about the equations: just work out the reduced form of the matrix and you'll get all the information in this case and all similar ones.

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Well, really just combine the first two equations by subtracting the second from the first:

$$x+y-(y+z)=5-(4)$$ $$\Rightarrow x-z=1$$

The coefficient of $z$, which is equal to $2K$, is numerically $-1$. Hence:

$$-1=2K \\ \therefore K=-\frac12$$

The answer is neither all real numbers nor an empty set.

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it's confused. the values of the other variables will be x=1 , y=4 , z=0 .. if we solve for k in the third equation and substitute .. it will be k=3/0 which is undefined ?!!!?!?!? –  Maher May 2 at 22:50
    
@Maher You're getting confused. $K$ is a constant; it will always be $-\frac12$. You can't solve it's value by plugging in numbers. –  Shahar May 2 at 22:59

$1 = 5 - 4 = (x+y) - (y+z) = x - z$. So: $x = 1 + z$, and $1 + z + 2Kz = 1$. So: $z(1+2K) = 0$. This equation has only $1$ solution if $1 + 2K \neq 0$ or $K \neq -\dfrac{1}{2}$. So then $z = 0$, and $y = 4$, and $x = 1$. So $K \neq -\dfrac{1}{2}$

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So, what's the value of K ? –  Maher May 2 at 22:22
    
If we take the third equation and solve for k , then k = (1-x)/(2z) ... and if we substitute by the numbers you gave for x and z .. then the result will be k=-3/0 .. and that is undefined ?!!! –  Maher May 2 at 22:33

Just try to solve the equations. You get: $$x-z=1$$ $$x+2Kz=1$$ And the latter equation is useless when $2K=-1$, because then you have 3 variables and 2 equations - that's not enough. Otherwise you have 3 unique equations and 3 variables. That means there's only one solution.

Other way of showing this is finding when is the determinant zero (the matrix of the array is irregular). The matrix of this array is: $$\left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 2K \end{array} \right)$$ The determinant is zero (matrix is irregular, meaning it's rows are lineary codependent) when $$2K+1=0$$ and you get the same solution $$K=-\frac{1}{2}$$

So the answer to your question is all reals except for $-\frac{1}{2}$, because for $K=-\frac{1}{2}$ the three equations are codependent (meaning you get the third one with clever combination of the other two)

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about k you mean equal or not equal ??? –  Maher May 2 at 22:39
    
If $K=-\frac{1}{2}$ you only have 2 independent equations. When you have 2 independent linear equations and 3 variables, there's infinite number of solutions. –  Tom83B May 2 at 22:42
    
ok there's something that i still don't understand. the values of variables are x=1 , y=4 , z=0 .. if we solve for k in 3rd equation and substitute we will get .. k=3/0 which is undefined !! how's that? –  Maher May 2 at 23:03

Wolfram

$ y = 5 - x \\ 5 - x + z = 4 \\ z = 4 + x - 5 \\ z = x - 1 \\ x + 2k(x - 1) = 1 \\ 2k(x - 1) = (-x - 1) \\ -2k(x-1) = (x - 1) \\ -2k = 1 \\ k = -\frac{1}{2}, 2k + 1 \neq{} 0 \\ x = 1, y = 4, z = 0 $

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