Id `[1//25 0x1//25]=[5 0-a5]^(-2)` , then the value of `x` is `a//125` b. `2a//125` c. `2a//25` d. none of these

Options

A. `a//125`
B. `2a//125`
C. `2a//25`
D. none of these

Explanation

To solve the equation \( \begin{bmatrix} \frac{1}{25} & 0 \\ x & \frac{1}{25} \end{bmatrix} = \left( \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix} \right)^{-2} \), we will follow these steps:

### Step 1: Find the inverse of the matrix \( A \)
Let \( A = \begin{bmatrix} 5 & 0 \\ -a & 5 \end{bmatrix} \).

The formula for the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
For our matrix \( A \):
- \( a = 5 \)
- \( b = 0 \)
- \( c = -a \)
- \( d = 5 \)

Calculating \( ad - bc \):
\[ ad - bc = 5 \cdot 5 - 0 \cdot (-a) = 25 \]

Thus, the inverse of \( A \) is:
\[ A^{-1} = \frac{1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \]

### Step 2: Calculate \( A^{-2} \)
To find \( A^{-2} \), we need to square \( A^{-1} \):
\[ A^{-2} = (A^{-1})^2 = A^{-1} \cdot A^{-1} \]
Calculating:
\[ A^{-1} = \frac{1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \]
So,
\[ A^{-2} = \left( \frac{1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \right) \cdot \left( \frac{1}{25} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \right) \]
\[ = \frac{1}{625} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \]

### Step 3: Perform the matrix multiplication
Calculating the product:
\[ \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ a & 5 \end{bmatrix} = \begin{bmatrix} 5 \cdot 5 + 0 \cdot a & 5 \cdot 0 + 0 \cdot 5 \\ a \cdot 5 + 5 \cdot a & a \cdot 0 + 5 \cdot 5 \end{bmatrix} \]
\[ = \begin{bmatrix} 25 & 0 \\ 10a & 25 \end{bmatrix} \]

Thus,
\[ A^{-2} = \frac{1}{625} \begin{bmatrix} 25 & 0 \\ 10a & 25 \end{bmatrix} = \begin{bmatrix} \frac{25}{625} & 0 \\ \frac{10a}{625} & \frac{25}{625} \end{bmatrix} = \begin{bmatrix} \frac{1}{25} & 0 \\ \frac{2a}{125} & \frac{1}{25} \end{bmatrix} \]

### Step 4: Compare both sides
Now we have:
\[ \begin{bmatrix} \frac{1}{25} & 0 \\ x & \frac{1}{25} \end{bmatrix} = \begin{bmatrix} \frac{1}{25} & 0 \\ \frac{2a}{125} & \frac{1}{25} \end{bmatrix} \]

### Step 5: Solve for \( x \)
From the comparison, we can see that:
\[ x = \frac{2a}{125} \]

### Conclusion
Thus, the value of \( x \) is \( \frac{2a}{125} \).