Let A and B are two non - singular matrices of order 3 such that `A+B=2I` and `A^(-1)+B^(-1)=3I`, then AB is equal to (where, I is the identity matrix of order 3)

Options

A. A
B. B
C. `(2I)/(3)`
D. 2I

Answer

To solve the problem step by step, we will use the given equations involving the matrices \( A \) and \( B \).

### Step 1: Write down the given equations
We are given two equations:
1. \( A + B = 2I \) (Equation 1)
2. \( A^{-1} + B^{-1} = 3I \) (Equation 2)

### Step 2: Rewrite Equation 2 using the property of inverses
Using the property of matrix inverses, we can rewrite Equation 2:
\[ A^{-1} + B^{-1} = A^{-1}B^{-1}(B + A) = 3I \]
This implies:
\[ A^{-1} + B^{-1} = (A + B)^{-1}(A + B) \]
However, we will directly manipulate Equation 2 without this transformation for simplicity.

### Step 3: Multiply Equation 2 by \( A \)
We multiply Equation 2 by \( A \) from the left:
\[ A(A^{-1} + B^{-1}) = A(3I) \]
This simplifies to:
\[ I + AB^{-1} = 3A \]
Rearranging gives us:
\[ AB^{-1} = 3A - I \quad \text{(Equation 3)} \]

### Step 4: Multiply Equation 3 by \( B \)
Now, we will post-multiply Equation 3 by \( B \):
\[ AB^{-1}B = (3A - I)B \]
This simplifies to:
\[ A = 3AB - B \]
Rearranging gives:
\[ 3AB = A + B \]

### Step 5: Substitute \( A + B \) from Equation 1
From Equation 1, we know \( A + B = 2I \). Substituting this into the equation gives:
\[ 3AB = 2I \]

### Step 6: Solve for \( AB \)
Dividing both sides by 3, we find:
\[ AB = \frac{2I}{3} \]

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