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 conditions on the matrix \(A\). Let's go through the steps systematically.

### Step 1: Understand the given equation
We have the equation:
\[ (I - 0.8A)^{-1} = I - \alpha A \]
where \(I\) is the identity matrix and \(A\) is a non-singular square matrix such that \(A^2 = A\).

### Step 2: Multiply both sides by \((I - 0.8A)\)
To eliminate the inverse on the left side, 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:
\[ I = I - 0.8A - \alpha A + 0.8\alpha A^2 \]
Since \(A^2 = A\), we can substitute \(A\) for \(A^2\):
\[ I = I - 0.8A - \alpha A + 0.8\alpha A \]

### Step 4: Simplify the equation
Now, we can simplify the equation:
\[ I = I - (0.8 + \alpha - 0.8\alpha)A \]
This simplifies to:
\[ I = I - (0.8 + \alpha(1 - 0.8))A \]
\[ I = I - (0.8 + 0.2\alpha)A \]

### Step 5: Set the coefficients of \(A\) to zero
Since the identity matrix \(I\) must equal itself, the coefficient of \(A\) must be zero:
\[ 0.8 + 0.2\alpha = 0 \]

### Step 6: Solve for \(\alpha\)
Now we solve for \(\alpha\):
\[ 0.2\alpha = -0.8 \]
\[ \alpha = -\frac{0.8}{0.2} = -4 \]

### Step 7: Calculate \(-4\alpha\)
Now we need to find \(-4\alpha\):
\[ -4\alpha = -4(-4) = 16 \]

### Final Answer
Thus, the value of \(-4\alpha\) is:
\[ \boxed{16} \]