Simple Impulse Response
as the comments to your question mention you have not done the first step: system classification


Is the system linear? We have to input three different signal in the system and compare the outputs: the two signals $x_1$ and $x_2$ and the sum signal $x_{1+2} = x_1 + x_2$


\begin{align}
 y_1 &amp;= (-1)^t x_1 \\ 
 y_2 &amp;= (-1)^t x_2 \\ 
 y_{1+2} &amp;= (-1)^t (x_1 + x_2) \\ 
\end{align}

and even at this point it is easy to see that $y_1 + y_2 = y_{1+2}$ that means the system is linear


Is the system time variant? it's usually very easy to say yes here when your system is a function of the time (or n in your case) $z(t) = (-1)^t $ Mathematically you have to input a single signal $x(t)$ at two points in time and check if the system gives you different outputs at different times:


\begin{align}
 y_1 &amp;= (-1)^t x(t) \\ 
 y_2 &amp;= (-1)^t x(t-\Delta t) \\ 
\end{align}

to compare 1 and 2 you have to time shift the output of the first system to make sure they overlap the way you want.

\begin{align}
 y_1 &amp;= (-1)^t x(t) \\ 
 y_{1-\Delta t} &amp;= (-1)^{t - \Delta t} x(t - \Delta t) \\ 
 y_2 &amp;= (-1)^t x(t-\Delta t) \\ 
\end{align}

the comparison shows that those two are not the same, which means your system changes over time

\begin{align}
(-1)^{t - \Delta t} x(t - \Delta t) &amp;= (-1)^t x(t-\Delta t) \\ 
(-1)^{t - \Delta t} &amp;= (-1)^t  \\ 
(-1)^{ - \Delta t} &amp;= 1  \\ 
\end{align}

this is not true in general, that means your system is linear and time variant and you can not use the description of a LTI System. But there's a definition of the impulse response of a time variant system that you could use:

Time-varying &quot;impulse response&quot;

\begin{align}
y(t) = \int h(\tau, t) x(t - \tau) d\tau 
\end{align}

since your system is memory less and kausal (i don't think there's a trivial mathematical proof for those two properties but it is easy to see), your time variant impulse response is rather simple: 

\begin{align}
h(\tau, t) = \delta(\tau) (-1)^t
\end{align}

which leads to the collapse of the integral in the time variant convolution (just check the rules for integration over the dirac function if you don't remember them): 

\begin{align}
y(t) = \int \delta(\tau) (-1)^t x(t - \tau) d\tau \\
y(t) = (-1)^t x(t)
\end{align}

And the last line ( and basically your first line) is the best description for linear memory less time variant systems - a simple multiplication with a time variant function or in general just:

\begin{align}
y(t) = x(t) z(t) \\
\end{align}

bye bye stackexchange you managed to annoy me enough in my first two answers for me to decide not to come back.