6A

$ \,x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1)\, $


Proof:

$ x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1) \to Time Shift(t0) \to z(t)=y(t-t0) $

$ \, =(k+1)^{2}x(t-t0-k-1)\, $


$ x(t-k) \to Time Shift(t0) \to y(t)=x(t-t0-k) \to System \to z(t)=(k1+1)^{2}y(t-k1-1) $

$ \, =(t0+k+1)^{2}x(t-t0-k-1)\, $

It is not time invariant.


6B

If superposition works, integrating input will yield integration of output, integration of $ u[n-1] $ is $ \delta[n-1] $, thus u[n] as an input will yield the desired output.

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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