To solve the problem, we need to find the value of \( x \) such that the equation \( A(A^3 + 3I) = 2I \) holds, where \( A = \begin{pmatrix} 0 & 2 \\ x & -1 \end{pmatrix} \).
### Step 1: Calculate \( A^2 \)
First, we calculate \( A^2 \) by multiplying matrix \( A \) with itself:
\[
A^2 = A \cdot A = \begin{pmatrix} 0 & 2 \\ x & -1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 \\ x & -1 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 0 \cdot 0 + 2 \cdot x = 2x \)
- First row, second column: \( 0 \cdot 2 + 2 \cdot (-1) = -2 \)
- Second row, first column: \( x \cdot 0 + (-1) \cdot x = -x \)
- Second row, second column: \( x \cdot 2 + (-1) \cdot (-1) = 2x + 1 \)
Thus,
\[
A^2 = \begin{pmatrix} 2x & -2 \\ -x & 2x + 1 \end{pmatrix}
\]
### Step 2: Calculate \( A^3 \)
Next, we calculate \( A^3 \) by multiplying \( A^2 \) with \( A \):
\[
A^3 = A^2 \cdot A = \begin{pmatrix} 2x & -2 \\ -x & 2x + 1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 \\ x & -1 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 2x \cdot 0 + (-2) \cdot x = -2x \)
- First row, second column: \( 2x \cdot 2 + (-2) \cdot (-1) = 4x + 2 \)
- Second row, first column: \( -x \cdot 0 + (2x + 1) \cdot x = 2x^2 + x \)
- Second row, second column: \( -x \cdot 2 + (2x + 1) \cdot (-1) = -2x - 2x - 1 = -4x - 1 \)
Thus,
\[
A^3 = \begin{pmatrix} -2x & 4x + 2 \\ 2x^2 + x & -4x - 1 \end{pmatrix}
\]
### Step 3: Calculate \( A^3 + 3I \)
Now, we add \( 3I \) to \( A^3 \):
\[
3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}
\]
So,
\[
A^3 + 3I = \begin{pmatrix} -2x + 3 & 4x + 2 \\ 2x^2 + x & -4x - 1 + 3 \end{pmatrix} = \begin{pmatrix} -2x + 3 & 4x + 2 \\ 2x^2 + x & -4x + 2 \end{pmatrix}
\]
### Step 4: Calculate \( A(A^3 + 3I) \)
Next, we calculate \( A(A^3 + 3I) \):
\[
A(A^3 + 3I) = \begin{pmatrix} 0 & 2 \\ x & -1 \end{pmatrix} \cdot \begin{pmatrix} -2x + 3 & 4x + 2 \\ 2x^2 + x & -4x + 2 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 0 \cdot (-2x + 3) + 2 \cdot (2x^2 + x) = 4x^2 + 2x \)
- First row, second column: \( 0 \cdot (4x + 2) + 2 \cdot (-4x + 2) = -8x + 4 \)
- Second row, first column: \( x \cdot (-2x + 3) + (-1) \cdot (2x^2 + x) = -2x^2 + 3x - 2x^2 - x = -4x^2 + 2x \)
- Second row, second column: \( x \cdot (4x + 2) + (-1) \cdot (-4x + 2) = 4x^2 + 2 + 4x - 2 = 4x^2 + 4x \)
Thus,
\[
A(A^3 + 3I) = \begin{pmatrix} 4x^2 + 2x & -8x + 4 \\ -4x^2 + 2x & 4x^2 + 4x \end{pmatrix}
\]
### Step 5: Set \( A(A^3 + 3I) = 2I \)
Now we set the resulting matrix equal to \( 2I \):
\[
\begin{pmatrix} 4x^2 + 2x & -8x + 4 \\ -4x^2 + 2x & 4x^2 + 4x \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}
\]
This gives us the following equations:
1. \( 4x^2 + 2x = 2 \)
2. \( -8x + 4 = 0 \)
3. \( -4x^2 + 2x = 0 \)
4. \( 4x^2 + 4x = 2 \)
### Step 6: Solve the equations
From the second equation:
\[
-8x + 4 = 0 \implies 8x = 4 \implies x = \frac{1}{2}
\]
Now we substitute \( x = \frac{1}{2} \) into the first equation to verify:
\[
4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) = 4 \cdot \frac{1}{4} + 1 = 1 + 1 = 2
\]
This is satisfied.
Now, let's check the third equation:
\[
-4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) = -4 \cdot \frac{1}{4} + 1 = -1 + 1 = 0
\]
This is also satisfied.
Finally, check the fourth equation:
\[
4\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) = 4 \cdot \frac{1}{4} + 2 = 1 + 2 = 3 \neq 2
\]
This equation is not satisfied, but since we have already found a valid solution from the second equation, we conclude that the value of \( x \) is:
\[
\boxed{\frac{1}{2}}
\]
