Linearity and Time invariant system

Time invariance

The system shown is not time invariance system. Although there is no $ t\, $ variable as the coefficient of the output or the system, the system depends on the delay of the input.

The input is shifted before undergoing the transformation

$ X[n] = X_{k + \alpha} = \delta[n - k - \alpha] = \delta[n - (k + \alpha)] \, $

$ Y[n] = Y_{k + \alpha} = (k + \alpha + 1)^2\delta[n - (k + \alpha + 1)] \, $

If the signal is shifted at the output, $ Y[n] = Y_k[n - \alpha] = (k+1)^2\delta[n-(k+1) - \alpha] = (k+1)^2\delta[n-(k + 1 + \alpha)]\, $

Both signal doesn't match, as the coeefeicent change according to the time when the signal is inputted. Therefore, the system is not time invariant

Finding the input

The system operates by shifting the input by 1 and magnify the output by the square of the input signal's final delay.

Therefore, by just inputting $ X[n] = u[n]\, $, $ Y[n] = u[n-1]\, $ will be outputted.

$ Y[n] = (-1)^2u[n-1] = u[n-1] \, $ as $ k\, $ is equal to zero.