Characterize the values $c$ and $k$ for the system: $x+y+z=3, x + 2y +cz = 4, 2x + 3y + 2cz = k$

Question

Characterize the values $c$ and $k$ for the system:

$$\begin{eqnarray} cx&+y&+z&=&3\\ x& + 2y& +cz& =& 4\\ 2x& + 3y& + 2cz& =& k \end{eqnarray}$$ such that the system has (i) no solution (ii) infinite solutions (iii) unique solution. I have obtained the following reduced equations using echelon reduction- $$\left(\begin{array}{c} 0 & 0 & c-1 \\ 0 & 1 & c-1 \\ 1 & 1 & 1 \end{array} \right) \left(\begin{array}{c} x \\ y \\ z \end{array} \right) = \left(\begin{array}{c} k-7 \\ 1 \\ 3 \end{array} \right)$$

(c-1)z=k-7, y+(c-1)z=1, x+y+z=3

I don't know the conditions to be imposed further for each of the cases.

Answer 1

Let the system to be $A \boldsymbol{x}= \boldsymbol{b}$.

Now, let's start with the unique solution first.

A system has an unique solution iff all pivot from the matrix $A$ are non-zero. It means $c-1 \neq 0$. So $c \neq 1$.

If $c=1$, it will turns the system to be infinitely many solution or no solution.

So, let's substitute $c=1$ back to the system.

Now, you can see that there is a nearly zero row in the system .

If you want the system to have infinitely many solution, you need to make the nearly zero row to be an all zero row.

Therefore, if $k=7$, that's the case of infinitely many solution. You need to write down $c=1,k=7$.

If $c=1$ but $k\neq 7$, there is a contradiction that $$0x+0y+0z=p,$$ where $p$ is any non zero number, that's impossible. Therefore $c=1 ,k \neq 7$ makes no solution.

Answer 2

Given the equations

cx+ y+ z= 3

x+ 2y+ cz= 4

3x+ 2y+ 2cz= k

Multiply the first equation by c and subtract the second equation from it: $(c^2-1)x+ (c- 2)y= 3c- 4$

Multiply the first equation by 2c and subtract the third equation from it: $(2c^2- 3)x+ (2c- 2)y= 6c- k$

Multiply the first of those two equations by 2c-2: $(4c^3- 4c^2- 6c+ 6)x+ (2c-2)(c-2)y= 6c^2-14c+ 8$

Multiply the second by c-2: $(2c^3- 4c^2- 3c+ 6)x+ (2c-2)(c-2)y= 6c^2- (k+12)c+ 2k$

And subtract, eliminating y, $(3c^3- 3c)x= (k-2)c+ (8-2k)

We can solve for x, and there will be a unique solution, as long as the coefficient of x, $3c(c^2- 1)$ is not 0. That is, as long as c is not 0, 1, or -1.

There will be no solution if the coefficient of x is 0 but the right side is not. If c= 0 the right side is 8- 2k so "if c= 0 and k is not 2". If c= 1 the right side is k- 2+ 8- 2k= 6- k so "if c= 1 and k is not 6". If c= -1 the right side is -k+ 2+ 8- 2k= 10- 3k so "if c= -1 and k is not 10/3.

There will be infinitely many solutions if both the coefficient of x and the right side are 0 so "if c= 0 and k= 2", "if c= 1 and k= 6", and "if c= -1 and k= 10/3".