what value of K does the system have a unique solution

Question

$\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\ \end{cases}$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$ and reduced it

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\0&-1-2K&K+2&1\\0&10-K&-5&-1\end{array}\right]$

But now I am not sure how i could remove $10-K$ with $-1-2K$ any help would be appreciated

Answer 1

If $k\neq -\frac12$ $$\left( \begin{matrix} 1 & k & -1 & | & 2\\ 2 & -1 & k & | & 5 \\ 1 & 10 & -6 & | & 1 \\ \end{matrix} \right) \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left( \begin{matrix} 1 & k & -1 & | & 2\\ 0 & -1-2k & k+2 & | & 1 \\ 0 & 10-k & -5 & | & -1 \\ \end{matrix} \right) \xrightarrow[\text{$(2k+1\neq0)$}]{R_3=R_3-(\frac{10-k}{2k+1})R_2} $$ $$\left( \begin{matrix} 1 & k & -1 & | & 2\\ 0 & -1-2k & k+2 & | & 1 \\ 0 & 0 & \frac{-k^2-2k+15}{2k+1} & | & \frac{-3k+9}{2k+1} \\ \end{matrix} \right) $$ Which has a unique solution $\iff \frac{-3k+9}{2k+1} \neq 0 \iff -3k+9\neq 0 \iff k\neq3$

If $k=-\frac12$ it is easy to show that the system has a unique solution.

Answer 2

Without solving you the problem, you can ALWAYS proceed with the following way if you don't spot any easy factorisation.

Suppose that, in your matrix, you have two values $a_{i,j}$ and $a_{i+1,j}$ you want to reduce in that way. You can ALWAYS exploit the property of linearity, and proceed with the substitution $R_{i+1} = a_{i,j}R_{i+1} - a_{i+1,j}R_{i}$, providing that $a_{i,j} \neq 0$.

In your case, proceed with the substitution: $R_3 = (-1-2K)R_3 - (10-K)R_2$, providing (i.e. check at the end if the solution allows it) that $-1-2K \neq 0$.

Answer 3

Let's solve this system:
$x_1+kx_2-x_3=2 \implies 2x_1+2kx_2-2x_3=4$
$2x_1-x_2+kx_3=5$
$x_1+10x_2-6x_3=1 \implies 2x_1+20x_2-12x_3=2$

Subtracting each pair of equations we have:
$(k+2)x_3-(2k+1)x_2=1$
$(k-10)x_2+5x_3=1$
$-21x_2+(k+12)x_3=3$

Finally :
$(3k+6)x_3-(6k+3)x_2=(3k-30)x_2+15x_3=(k+12)x_3-21x_2=3$
Which is easy to solve since from the two first ones:
$(3k+6)x_3-(6k+3)x_2=(3k-30)x_2+15x_3 \implies (3k-9)x_3=(9k-27)x_2 \implies x_3=3x_2$ (unless $k=3$)
By the second and the third ones:
$(3k-30)x_2+15x_3=(k+12)x_3-21x_2 \implies (3k-9)x_2=(k-3)x_3 \implies x_3=3x_2$ (again unless $k=3$)
And by the first and the last ones:
$(3k+6)x_3-(6k+3)x_2=(k+12)x_3-21x_2 \implies (2k-6)x_3=(6k-18)x_2 \implies x_3=3x_2$
So all pairs $(x,3x)$ will satisfy these equations and hence there are infinitely many answers. But then there is only a unique $x$ such that these three creatures are exactly equal to $3$ so there is a unique answer. (to be more exact, you shall put $(x_2,x_3)=(\frac{1}{k+5},\frac{3}{k+5})$ The case $k=3$ can also be easily checked. (hint: it does not satisfy)