Let `Aa n dB` be two `2xx2` matrices. Consider the statements (i) `A B=O => A=OorB=O` (ii)`A B=I_2 =>A=B^(-1)` (iii) `(A+B)^2=A^2+2A B+B^2` (a)(i) and (ii) are false, (iii) is true (b)(ii) and (iii) are false, (i) is true (c)(i) is false (ii) and, (iii) are true (d)(i) and (iii) are false, (ii) is true

Options

A. (i) and (ii) are false, (iii) is true
B. (ii) and (iii) are false, (i) is true
C. (i) is false, (ii) and (iii) are true
D. (i) and (iii) are false, (ii) is true

Explanation

To solve the problem, we need to analyze each statement regarding the matrices \( A \) and \( B \) and determine their validity. Let's go through each statement step by step.

### Step 1: Analyze Statement (i)
**Statement:** \( AB = O \Rightarrow A = O \text{ or } B = O \)

**Analysis:**
This statement claims that if the product of two matrices \( A \) and \( B \) is the zero matrix \( O \), then at least one of the matrices must be the zero matrix.

**Counterexample:**
Let:
\[ A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \]
Calculating \( AB \):
\[ AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \neq O \]
Now let’s take:
\[ A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]
Calculating \( AB \):
\[ AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O \]
In this case, \( A \neq O \) and \( B = O \). However, we can also find matrices \( A \) and \( B \) such that \( AB = O \) without either being zero. Therefore, statement (i) is **false**.

### Step 2: Analyze Statement (ii)
**Statement:** \( AB = I_2 \Rightarrow A = B^{-1} \)

**Analysis:**
This statement asserts that if the product of two matrices \( A \) and \( B \) is the identity matrix \( I_2 \), then \( A \) must be the inverse of \( B \).

**Verification:**
By definition, if \( AB = I_2 \), then \( A \) is indeed the inverse of \( B \) (i.e., \( A = B^{-1} \)). Therefore, statement (ii) is **true**.

### Step 3: Analyze Statement (iii)
**Statement:** \( (A + B)^2 = A^2 + 2AB + B^2 \)

**Analysis:**
This statement resembles the algebraic identity for the square of a binomial. However, matrix multiplication is not commutative, which means \( AB \neq BA \) in general.

**Counterexample:**
Let:
\[ A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \]
Calculating \( (A + B)^2 \):
\[ A + B = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \]
\[ (A + B)^2 = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \]
Now calculating \( A^2 + 2AB + B^2 \):
\[ A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad B^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \quad AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \]
Thus,
\[ A^2 + 2AB + B^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 2\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \]
Since \( (A + B)^2 \neq A^2 + 2AB + B^2 \), statement (iii) is **false**.

### Conclusion
From the analysis:
- Statement (i) is false.
- Statement (ii) is true.
- Statement (iii) is false.

Thus, the correct option is **(d)**: (i) and (iii) are false, (ii) is true.