How to find values of a matrix to make it consistent

Question

So I have this system of equations

$2x+ky+2z=0$

$x-y+z=1$

$y-z=k$

I augment it as to get:

$\begin{pmatrix}2 & k & 2 & 0\\ 1 & -1 & 1 & 1 \\ 0 & 1 & -1 & k \end{pmatrix}$

If i reduced the latter in reduced echelon form I get:

$\begin{pmatrix} 1 & 0 &0 &k+1\\ 0 &1 &-1& k\\ 0 &0 &-k-2 & k²+2(k+1) \end{pmatrix}$

I need to find the k values for which the system is consistent. It seems to me that the system is consistent when $k\neq -2$.

However, when I plug this matrix in the ti nspire cx cas calculator, I get that the reduced echelon form is

$\begin{pmatrix} 1 &k/2 &1& 0\\ 0 &1 & -1 &k\\ 0 &0& 1 &-(k²+2k+2)/(k+2) \end{pmatrix}$

The system for me is still consistent when $k$ is not equal to $-2$, because if it were equal to $-2$ then $z= a$ divided by 0, which makes it undefined.

However, when I try to compute the solutions, meaning $x$, $y$ and $z$, I get different solutions for the matrices. How come? Which one do you think is correct?

E.g with MY REF I get that $x= k+1$. With REF on the calculator I get that $x=-y(k/2)-z$.

Can somebody please explain the differences with the ccalculator's REF and my REF , or if maybe the soltuions are the same and i just need to expan better? Thanks!

Answer 1

From the augmented matrix we obtain

$$\begin{bmatrix} 2 & k & 2 & 0\\ 1 &-1 &1 &1 \\ 0 &1 &-1 &k \\ \end{bmatrix} \to \begin{bmatrix} 2 & k & 2 & 0\\ 0&-2-k&0 &2 \\ 0 &1 &-1 &k \\ \end{bmatrix} \to \begin{bmatrix} 2 & k & 2 & 0\\ 0&-2-k&0 &2 \\ 0 &0&-2-k &k(k+2)+2\\ \end{bmatrix}$$

and thus the system is consistent and has an unique solution for $k\neq -2$ otherwise for $k=-2$ for the second and third equation we obtain $0=2$.

Note also that for $k=-2$ the first 2 equations ofbthe original system become

• $2x-2y+2x=0 \implies x-y+z=0$
• $x-y+z=1$

which are clearly not consistent.

Answer 2

Why don't you try to find all x, y , z in terms of k Proceed to plug them in the equation generated by your calculator and check if they match. Hopefully it provides a relationship between all variables by default...

Answer 3

Note that: $$\begin{bmatrix} 2 & k & 2 & 0\\ 1 &-1 &1 &1 \\ 0 &1 &-1 &k \\ \end{bmatrix} \to \begin{bmatrix} 1 & \frac k2 & 1 & 0\\ 1 &-1 &1 &1 \\ 0 &1 &-1 &k \\ \end{bmatrix} \to \begin{bmatrix} 1 & \frac k2 & 1 & 0\\ 0 &-\frac k2-1 &0 &1 \\ 0 &1 &-1 &k \\ \end{bmatrix} \to \\ \begin{bmatrix} 1 & \frac k2 & 1 & 0\\ 0 &1 &0 &-\frac{2}{k+2} \\ 0 &1 &-1 &k \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & \frac{2}{k+2}\\ 0 &1 &0 &-\frac{2}{k+2} \\ 0 &0 &-1 &k+\frac{2}{k+2} \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 0 & k+1\\ 0 &1 &0 &-\frac{2}{k+2} \\ 0 &0 &1 &-k-\frac{2}{k+2} \\ \end{bmatrix}.$$ Hence, it is consistent only when $k\ne-2$.

Also, if you plug in $y$ and $z$: $$x=-y(k/2)-z=-\left(-\frac{2}{k+2}\right)\cdot \frac k2-\left(-k-\frac{2}{k+2}\right)=k+1.$$