If A and B are two matrices of order `3xx3` satisfying `AB=A and BA=B`, then `(A+B)^(5)`is equal to

Options

A. `5(A+B)`
B. `5I`
C. `16(A+B)`
D. `32I`

Answer

To solve the problem, we need to find \((A + B)^5\) given that \(AB = A\) and \(BA = B\). Let's break this down step by step.

### Step 1: Analyze the given equations
We have two equations:
1. \(AB = A\)
2. \(BA = B\)

These equations suggest that \(A\) and \(B\) are idempotent matrices. Specifically, \(A\) is a left eigenvector of \(B\) and \(B\) is a right eigenvector of \(A\).

### Step 2: Multiply both sides of the first equation by \(B\)
Starting with \(AB = A\), we can multiply both sides by \(B\):
\[ AB = A \implies ABB = AB \implies A = AB \]
This is consistent with our original equation.

### Step 3: Multiply both sides of the second equation by \(A\)
Now, starting with \(BA = B\), we can multiply both sides by \(A\):
\[ BA = B \implies BAA = BA \implies B = BA \]
This is also consistent with our original equation.

### Step 4: Add the two equations
Now, let’s add the two equations \(AB = A\) and \(BA = B\):
\[ AB + BA = A + B \]
This shows that \(A + B\) can be expressed in terms of \(AB\) and \(BA\).

### Step 5: Consider the expression \((A + B)^2\)
Using the binomial expansion:
\[ (A + B)^2 = A^2 + AB + BA + B^2 \]
Substituting \(AB = A\) and \(BA = B\):
\[ (A + B)^2 = A^2 + A + B + B^2 \]

### Step 6: Simplify further
Since \(A^2 = A\) and \(B^2 = B\) (because \(A\) and \(B\) are idempotent):
\[ (A + B)^2 = A + A + B + B = 2(A + B) \]

### Step 7: Generalize to \((A + B)^n\)
From the previous step, we can derive that:
\[ (A + B)^2 = 2(A + B) \]
Now, let \(X = A + B\). Then:
\[ X^2 = 2X \]
This implies that \(X\) is also idempotent. We can use this property to find higher powers:
\[ X^3 = X^2 \cdot X = 2X \cdot X = 2X^2 = 2(2X) = 4X \]
\[ X^4 = X^3 \cdot X = 4X \cdot X = 4X^2 = 4(2X) = 8X \]
\[ X^5 = X^4 \cdot X = 8X \cdot X = 8X^2 = 8(2X) = 16X \]

### Step 8: Final expression for \((A + B)^5\)
Thus, we find:
\[ (A + B)^5 = 16(A + B) \]

### Step 9: Substitute back \(A + B\)
Since we know \(A + B = I + I = 2I\) (where \(I\) is the identity matrix):
\[ (A + B)^5 = 16(2I) = 32I \]

### Conclusion
The final result is:
\[ (A + B)^5 = 32I \]