If `A = [[0, 1],[3,0]]and (A^(8) + A^(6) + A^(4) + A^(2) + I) V= [[0],[11]],`
where `V` is a vertical vector and `I` is the `2xx2` identity
matrix and if `lambda` is sum of all elements of vertical vector
`V`, the value of `11 lambda` is

Explanation

To solve the problem step by step, we will calculate the powers of the matrix \( A \) and then substitute these into the equation given in the problem.

### Step 1: Define the Matrix
Let \( A = \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix} \).

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} (0 \cdot 0 + 1 \cdot 3) & (0 \cdot 1 + 1 \cdot 0) \\ (3 \cdot 0 + 0 \cdot 3) & (3 \cdot 1 + 0 \cdot 0) \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I \]
where \( I \) is the identity matrix.

### Step 3: Calculate \( A^4 \)
Now, we calculate \( A^4 \) as \( A^2 \cdot A^2 \):
\[ A^4 = A^2 \cdot A^2 = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} 3 \cdot 3 & 0 \\ 0 & 3 \cdot 3 \end{pmatrix} = \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} = 9I \]

### Step 4: Calculate \( A^6 \)
Next, we calculate \( A^6 \) as \( A^4 \cdot A^2 \):
\[ A^6 = A^4 \cdot A^2 = \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} \cdot \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} 9 \cdot 3 & 0 \\ 0 & 9 \cdot 3 \end{pmatrix} = \begin{pmatrix} 27 & 0 \\ 0 & 27 \end{pmatrix} = 27I \]

### Step 5: Calculate \( A^8 \)
Now, we calculate \( A^8 \) as \( A^4 \cdot A^4 \):
\[ A^8 = A^4 \cdot A^4 = \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} \cdot \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} 9 \cdot 9 & 0 \\ 0 & 9 \cdot 9 \end{pmatrix} = \begin{pmatrix} 81 & 0 \\ 0 & 81 \end{pmatrix} = 81I \]

### Step 6: Substitute into the Given Equation
Now we substitute \( A^2, A^4, A^6, A^8 \) into the equation:
\[ A^8 + A^6 + A^4 + A^2 + I = 81I + 27I + 9I + 3I + I \]
Calculating the sum:
\[ = (81 + 27 + 9 + 3 + 1)I = 121I \]

### Step 7: Set Up the Equation with \( V \)
We have:
\[ 121I \cdot V = \begin{pmatrix} 0 \\ 11 \end{pmatrix} \]
This implies:
\[ V = \frac{1}{121} \begin{pmatrix} 0 \\ 11 \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{11}{121} \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{1}{11} \end{pmatrix} \]

### Step 8: Calculate \( \lambda \)
The sum of the elements of vector \( V \) is:
\[ \lambda = 0 + \frac{1}{11} = \frac{1}{11} \]

### Step 9: Calculate \( 11\lambda \)
Finally, we calculate:
\[ 11\lambda = 11 \cdot \frac{1}{11} = 1 \]

### Final Answer
The value of \( 11\lambda \) is \( \boxed{1} \).