Let A be a non - singular square matrix such that `A^(2)=A` satisfying `(I-0.8A)^(-1)=I-alphaA` where I is a unit matrix of the same order as that of A, then the value of `-4alpha` is equal to

Answer

To solve the problem, we need to find the value of \(-4\alpha\) given the equation:

\[ (I - 0.8A)^{-1} = I - \alpha A \]

where \(A\) is a non-singular square matrix satisfying \(A^2 = A\).

### Step 1: Rewrite the equation
We start with the equation:

\[ (I - 0.8A)^{-1} = I - \alpha A \]

### Step 2: Multiply both sides by \(I - 0.8A\)
To eliminate the inverse, we multiply both sides by \(I - 0.8A\):

\[ I = (I - \alpha A)(I - 0.8A) \]

### Step 3: Expand the right-hand side
Now, we expand the right-hand side:

\[ I = I - 0.8A - \alpha A + 0.8\alpha A^2 \]

### Step 4: Substitute \(A^2 = A\)
Since \(A^2 = A\), we can substitute \(A\) for \(A^2\):

\[ I = I - 0.8A - \alpha A + 0.8\alpha A \]

### Step 5: Combine like terms
Now, we combine the terms involving \(A\):

\[ I = I - (0.8 - 0.8\alpha - \alpha)A \]

### Step 6: Set the coefficients of \(A\) to zero
For the equation to hold, the coefficient of \(A\) must be zero:

\[ 0.8 - 0.8\alpha - \alpha = 0 \]

### Step 7: Solve for \(\alpha\)
Rearranging gives:

\[ 0.8 = (0.8 + 1)\alpha \]

\[ 0.8 = 1.8\alpha \]

\[ \alpha = \frac{0.8}{1.8} = \frac{4}{9} \]

### Step 8: Calculate \(-4\alpha\)
Now we calculate \(-4\alpha\):

\[ -4\alpha = -4 \times \frac{4}{9} = -\frac{16}{9} \]

### Final Answer
Thus, the value of \(-4\alpha\) is:

\[ \boxed{-\frac{16}{9}} \]