Consider the following in respect of two non-singular matrices A and B of same order :
1. `det(A + B) = det A + det B`
2. `(A+B)^(-1)=A^(-1)+B^(-1)`
Which of the above is/ar correct ?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the given problem, we need to analyze the two statements regarding two non-singular matrices \( A \) and \( B \) of the same order.

### Step 1: Verify the first statement \( \text{det}(A + B) = \text{det}(A) + \text{det}(B) \)

1. **Define Matrices A and B**:
Let \( A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \) and \( B = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \).

2. **Calculate \( A + B \)**:
\[ A + B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 2 & 4 \end{pmatrix} \]

3. **Calculate \( \text{det}(A + B) \)**:
\[ \text{det}(A + B) = \text{det}\begin{pmatrix} 3 & 2 \\ 2 & 4 \end{pmatrix} = (3)(4) - (2)(2) = 12 - 4 = 8 \]

4. **Calculate \( \text{det}(A) \)**:
\[ \text{det}(A) = \text{det}\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1 \]

5. **Calculate \( \text{det}(B) \)**:
\[ \text{det}(B) = \text{det}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3 \]

6. **Verify the statement**:
\[ \text{det}(A) + \text{det}(B) = 1 + 3 = 4 \]
Since \( \text{det}(A + B) = 8 \) and \( \text{det}(A) + \text{det}(B) = 4 \), the first statement is **false**.

### Step 2: Verify the second statement \( (A + B)^{-1} = A^{-1} + B^{-1} \)

1. **Calculate \( A^{-1} \)**:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} \]

2. **Calculate \( B^{-1} \)**:
\[ B^{-1} = \frac{1}{\text{det}(B)} \text{adj}(B) = \frac{1}{3} \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} \end{pmatrix} \]

3. **Calculate \( A^{-1} + B^{-1} \)**:
\[ A^{-1} + B^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} + \begin{pmatrix} \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} \end{pmatrix} = \begin{pmatrix} 2 + \frac{2}{3} & -1 - \frac{1}{3} \\ -1 - \frac{1}{3} & 1 + \frac{2}{3} \end{pmatrix} \]
\[ = \begin{pmatrix} \frac{6}{3} + \frac{2}{3} & -\frac{3}{3} - \frac{1}{3} \\ -\frac{3}{3} - \frac{1}{3} & \frac{3}{3} + \frac{2}{3} \end{pmatrix} = \begin{pmatrix} \frac{8}{3} & -\frac{4}{3} \\ -\frac{4}{3} & \frac{5}{3} \end{pmatrix} \]

4. **Calculate \( (A + B)^{-1} \)**:
We already calculated \( A + B = \begin{pmatrix} 3 & 2 \\ 2 & 4 \end{pmatrix} \).
\[ (A + B)^{-1} = \frac{1}{\text{det}(A + B)} \text{adj}(A + B) = \frac{1}{8} \begin{pmatrix} 4 & -2 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{3}{8} \end{pmatrix} \]

5. **Compare \( (A + B)^{-1} \) and \( A^{-1} + B^{-1} \)**:
\[ (A + B)^{-1} \neq A^{-1} + B^{-1} \]
Hence, the second statement is also **false**.

### Conclusion:
Both statements are false. Therefore, the correct option is that neither statement 1 nor statement 2 is correct.