If `A` is an idempotent matrix satisfying `(I-0.4A)^(-1)=I-alphaA` where `I` is the unit matrix of the same order as that of `A` then the value of `alpha` is

Options

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

Explanation

To solve the problem, we need to find the value of \( \alpha \) given that \( A \) is an idempotent matrix satisfying the equation:

\[ (I - 0.4A)^{-1} = I - \alpha A \]

### Step 1: Understand Idempotent Matrix
An idempotent matrix \( A \) satisfies the property:
\[ A^2 = A \]

### Step 2: Rewrite the Given Equation
We start with the equation:
\[ (I - 0.4A)^{-1} = I - \alpha A \]

### Step 3: Multiply Both Sides by \( (I - 0.4A) \)
To eliminate the inverse, we multiply both sides by \( (I - 0.4A) \):
\[ I = (I - 0.4A)(I - \alpha A) \]

### Step 4: Expand the Right Side
Now we expand the right side:
\[ I = I - \alpha A - 0.4A + 0.4\alpha A^2 \]

### Step 5: Substitute \( A^2 \) with \( A \)
Since \( A \) is idempotent, we can replace \( A^2 \) with \( A \):
\[ I = I - \alpha A - 0.4A + 0.4\alpha A \]
This simplifies to:
\[ I = I - (\alpha + 0.4 - 0.4\alpha)A \]

### Step 6: Set Coefficients Equal
Since the identity matrix \( I \) must equal \( I \) on both sides, we can equate the coefficients of \( A \):
\[ 0 = -(\alpha + 0.4 - 0.4\alpha) \]

### Step 7: Simplify the Equation
Rearranging gives:
\[ \alpha + 0.4 - 0.4\alpha = 0 \]
Combining like terms:
\[ \alpha - 0.4\alpha + 0.4 = 0 \]
This simplifies to:
\[ (1 - 0.4)\alpha + 0.4 = 0 \]
\[ 0.6\alpha + 0.4 = 0 \]

### Step 8: Solve for \( \alpha \)
Now, we can solve for \( \alpha \):
\[ 0.6\alpha = -0.4 \]
\[ \alpha = \frac{-0.4}{0.6} = \frac{-2}{3} \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = -\frac{2}{3} \]