If `[a b c1-a]` is an idempotent matrix and `f(x)=x-^2=b c=1//4` , then the value of `1//f(a)` is ______.

Explanation

To solve the problem step by step, we need to analyze the given information and derive the required result.

### Step 1: Understanding Idempotent Matrix
An idempotent matrix \( M \) satisfies the property \( M^2 = M \). Given the matrix:
\[ M = \begin{bmatrix} a & b \\ c & 1-a \end{bmatrix} \]
we need to compute \( M^2 \) and set it equal to \( M \).

### Step 2: Calculate \( M^2 \)
Calculating \( M^2 \):
\[ M^2 = \begin{bmatrix} a & b \\ c & 1-a \end{bmatrix} \begin{bmatrix} a & b \\ c & 1-a \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + b(1-a) \\ ac + c(1-a) & bc + (1-a)^2 \end{bmatrix} \]

### Step 3: Set \( M^2 = M \)
Now, equate \( M^2 \) to \( M \):
\[ \begin{bmatrix} a^2 + bc & ab + b(1-a) \\ ac + c(1-a) & bc + (1-a)^2 \end{bmatrix} = \begin{bmatrix} a & b \\ c & 1-a \end{bmatrix} \]
From this, we can derive the following equations:
1. \( a^2 + bc = a \)
2. \( ab + b(1-a) = b \)
3. \( ac + c(1-a) = c \)
4. \( bc + (1-a)^2 = 1-a \)

### Step 4: Simplifying the Equations
From equation 1:
\[ a^2 + bc = a \implies a^2 - a + bc = 0 \]
From equation 2:
\[ ab + b - ab = b \implies b = b \quad \text{(this is always true)} \]
From equation 3:
\[ ac + c - ac = c \implies c = c \quad \text{(this is always true)} \]
From equation 4:
\[ bc + (1-a)^2 = 1-a \implies (1-a)^2 = 1-a - bc \]

### Step 5: Substitute \( bc = \frac{1}{4} \)
We know from the problem statement that \( bc = \frac{1}{4} \). Substitute this into the equation:
\[ (1-a)^2 = 1-a - \frac{1}{4} \]
This simplifies to:
\[ (1-a)^2 = \frac{3}{4} - a \]

### Step 6: Solve for \( f(a) \)
Now, we need to find \( f(a) = a - a^2 \). From the equation \( a^2 - a + \frac{1}{4} = 0 \), we can find \( f(a) \):
\[ f(a) = \frac{1}{4} \]

### Step 7: Find \( \frac{1}{f(a)} \)
Now, we need to calculate \( \frac{1}{f(a)} \):
\[ \frac{1}{f(a)} = \frac{1}{\frac{1}{4}} = 4 \]

### Final Answer
Thus, the value of \( \frac{1}{f(a)} \) is:
\[ \boxed{4} \]