Let `A=[(0,1),(2,0)] and (A^(8)+A^(6)+A^(2)+I)V=[(32),(62)]`
where I is the `(2xx2` identity matrix). Then the product of all elements of matrix V is

Options

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

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Define the Matrix A
Let \( A = \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} \).

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} \]
Calculating this gives:
\[ A^2 = \begin{pmatrix} 0 \cdot 0 + 1 \cdot 2 & 0 \cdot 1 + 1 \cdot 0 \\ 2 \cdot 0 + 0 \cdot 2 & 2 \cdot 1 + 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

### Step 3: Calculate \( A^4 \)
Next, we find \( A^4 \) by squaring \( A^2 \):
\[ A^4 = A^2 \cdot A^2 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]
Calculating this gives:
\[ A^4 = \begin{pmatrix} 2 \cdot 2 + 0 \cdot 0 & 2 \cdot 0 + 0 \cdot 2 \\ 0 \cdot 2 + 2 \cdot 0 & 0 \cdot 0 + 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \]

### Step 4: Calculate \( A^6 \)
Now, we find \( A^6 \) by multiplying \( A^4 \) by \( A^2 \):
\[ A^6 = A^4 \cdot A^2 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]
Calculating this gives:
\[ A^6 = \begin{pmatrix} 4 \cdot 2 & 0 \\ 0 & 4 \cdot 2 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} \]

### Step 5: Calculate \( A^8 \)
Next, we find \( A^8 \) by squaring \( A^4 \):
\[ A^8 = A^4 \cdot A^4 = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} \]
Calculating this gives:
\[ A^8 = \begin{pmatrix} 4 \cdot 4 & 0 \\ 0 & 4 \cdot 4 \end{pmatrix} = \begin{pmatrix} 16 & 0 \\ 0 & 16 \end{pmatrix} \]

### Step 6: Set Up the Equation
Now we need to calculate \( A^8 + A^6 + A^2 + I \), where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus,
\[ A^8 + A^6 + A^2 + I = \begin{pmatrix} 16 & 0 \\ 0 & 16 \end{pmatrix} + \begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Calculating this gives:
\[ = \begin{pmatrix} 16 + 8 + 2 + 1 & 0 \\ 0 & 16 + 8 + 2 + 1 \end{pmatrix} = \begin{pmatrix} 27 & 0 \\ 0 & 27 \end{pmatrix} \]

### Step 7: Solve for V
We know that:
\[ (A^8 + A^6 + A^2 + I)V = \begin{pmatrix} 32 \\ 62 \end{pmatrix} \]
Substituting the matrix we calculated:
\[ \begin{pmatrix} 27 & 0 \\ 0 & 27 \end{pmatrix} V = \begin{pmatrix} 32 \\ 62 \end{pmatrix} \]
Let \( V = \begin{pmatrix} x \\ y \end{pmatrix} \). Then we have:
\[ 27x = 32 \quad \text{and} \quad 27y = 62 \]
Solving for \( x \) and \( y \):
\[ x = \frac{32}{27}, \quad y = \frac{62}{27} \]

### Step 8: Calculate the Product of Elements in V
Now, we find the product of all elements in \( V \):
\[ \text{Product} = x \cdot y = \left(\frac{32}{27}\right) \cdot \left(\frac{62}{27}\right) = \frac{32 \cdot 62}{27 \cdot 27} = \frac{1984}{729} \]

### Final Answer
The product of all elements of matrix \( V \) is \( \frac{1984}{729} \).