Let `A=[(0,1),(2,0)] and (A^(8)+A^(6)+A^(2)+I)V=[(32),(62)]`
where V is the `(2xx1` identity matrix). Then the product of all elements of matrix V is

Options

A. 2.72
B. 1
C. 3
D. `-2`

Answer

To solve the problem step by step, we will first compute the powers of the matrix \( A \), then we will evaluate the expression \( A^8 + A^6 + A^2 + I \), and finally, we will find the matrix \( V \) and calculate the product of its elements.

### Step 1: Compute \( A^2 \)

Given the matrix:
\[ A = \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} \]

We calculate \( A^2 \) as follows:
\[ A^2 = A \cdot A = \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 0 \cdot 0 + 1 \cdot 2 = 2 \)
- First row, second column: \( 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Second row, first column: \( 2 \cdot 0 + 0 \cdot 2 = 0 \)
- Second row, second column: \( 2 \cdot 1 + 0 \cdot 0 = 2 \)

Thus,
\[ A^2 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2I \]

### Step 2: Compute \( A^4 \)

Now, we compute \( A^4 \):
\[ A^4 = A^2 \cdot A^2 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 2 \cdot 2 + 0 \cdot 0 = 4 \)
- First row, second column: \( 2 \cdot 0 + 0 \cdot 2 = 0 \)
- Second row, first column: \( 0 \cdot 2 + 2 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + 2 \cdot 2 = 4 \)

Thus,
\[ A^4 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} = 4I \]

### Step 3: Compute \( A^8 \)

Next, we compute \( A^8 \):
\[ A^8 = A^4 \cdot A^4 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 4 \cdot 4 + 0 \cdot 0 = 16 \)
- First row, second column: \( 4 \cdot 0 + 0 \cdot 4 = 0 \)
- Second row, first column: \( 0 \cdot 4 + 4 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + 4 \cdot 4 = 16 \)

Thus,
\[ A^8 = \begin{pmatrix} 16 & 0 \\ 0 & 16 \end{pmatrix} = 16I \]

### Step 4: Compute \( A^6 \)

Now, we compute \( A^6 \):
\[ A^6 = A^4 \cdot A^2 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 4 \cdot 2 + 0 \cdot 0 = 8 \)
- First row, second column: \( 4 \cdot 0 + 0 \cdot 2 = 0 \)
- Second row, first column: \( 0 \cdot 2 + 4 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + 4 \cdot 2 = 8 \)

Thus,
\[ A^6 = \begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} = 8I \]

### Step 5: Compute \( A^8 + A^6 + A^2 + I \)

Now, we can compute:
\[ A^8 + A^6 + A^2 + I = 16I + 8I + 2I + I \]

Calculating the sum:
\[ = (16 + 8 + 2 + 1)I = 27I \]

### Step 6: Solve for \( V \)

We have the equation:
\[ 27I \cdot V = \begin{pmatrix} 32 \\ 62 \end{pmatrix} \]

This implies:
\[ 27V = \begin{pmatrix} 32 \\ 62 \end{pmatrix} \]

Thus,
\[ V = \begin{pmatrix} \frac{32}{27} \\ \frac{62}{27} \end{pmatrix} \]

### Step 7: Calculate the product of elements of \( V \)

The product of the elements of \( V \) is:
\[ \text{Product} = \left(\frac{32}{27}\right) \cdot \left(\frac{62}{27}\right) = \frac{32 \cdot 62}{27^2} = \frac{1984}{729} \]

### Final Answer

The product of all elements of matrix \( V \) is:
\[ \frac{1984}{729} \]