5. For the system below find A^k using the inverse Z...

5. For the system below find $A^k$ using the inverse Z transform of $(zl-A)^{-1}z$. Find the zero-input $x_{zi}(k)$ from $A^k x(0)$. Write an expression for the full solution assuming the input is $(-1)^k$, for $k > 0$. For the full solution use the time domain formula (with the summation formula) and plug in the zero-input part and plug in all available values into the sum part and simplify as much as possible. $x(k+1) = A(k)x(k) + B(k)u(k)$ $A = \begin{bmatrix} 0 & 1 \ -1/6 & -5/6 \end{bmatrix}$, $u(k) = (-1)^k$ for $k >= 0$ ($u(k) = 0$ for $k < 0$), $x(0) = \begin{bmatrix} 1 \ 0 \end{bmatrix}$

Transcript

00:01 In this question, we want solution of this system.

00:05 That means we want value of x, y and z.

00:09 Here, we can write this as matrix a multiplied by matrix x equal to matrix b.

00:21 Here, matrix a is matrix of coefficients of x, y and z.

00:28 That means in first row we can write 1, 1 and 1, coefficients of x, y, z.

00:37 And in second row 1, 1 and minus 4, 1, 1 minus 4, and in 3rd row, minus 4, 1 1, 1, 1.

00:46 1, minus 4, 1, 1.

00:51 This is matrix a.

00:54 Now, matrix x is column matrix of elli, demands x y z x y z and matrix b is column matrix of 510 0 510 0 now to find out x this a will come this side and x will be a inverse b now to find out x we have to find out x we have to solve for a inverse.

01:35 Now we know that a inverse equal to 1 divided by determinant a multiplied by adjoint a.

01:45 Now here we find out we will find out determinant a.

01:53 Here we can see that determinant a will be we solve with respect to first row.

02:00 Now with respect to this one we can 1 minus 4 that means 1 plus 4 now with respect to this 1 minus 1 we get 1 minus 4 that means minus 16 1 minus 16 now with respect to this 1 minus 4 that means plus 4 1 plus 4 now we get 1 plus 4 4 that means 5 multiplied by 1 that means 5 1 minus 16 minus 15 multiplied by minus 1 that means plus 15 now 1 plus 4 5 5 multiplied by 1 will be 5 we get 25 this is determinant a now adjoint of matrix a will be we can write here adjoint of the matrix a will be transpose of co -factors of elements of matrix a that means co -factor of 1 will be 1 minus minus 4 that means 1 plus 4 that means 5 now co -factor of this 1 will be 1 minus 16 that means minus 15 but here we have to multiply it with minus therefore we get plus 15.

03:41 Now, co -factor of this one will be 1 plus 4, that means 5.

03:48 Now, co -factor of this one will be 1 minus 1, that means 0.

03:54 Now, co -factor of this one will be 1 plus 4, that means 5.

04:02 Now co -factor of this minus 4 will be 1 plus 4 that means 5 but here we have to multiply with minus now co -factor of this minus 4 will be minus 4 minus 1 that means minus 5 now co -factor of this 1 will be minus 4 minus 1 that means minus 5 but here we have to multiply with minus therefore we get plus 5...

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