Consider `A=[(a_(11),a_(12)),(a_(21),a_(22))]` and `B=[(1,1),(2,1)]` such that `AB=BA.` then the value of `(a_(12))/(a_(21))+(a_(11))/(a_(22))` is

Options

A. 2
B. 4
C. `(3)/(2)`
D. `(1)/(sqrt2)`

Answer

To solve the problem, we need to find the value of \(\frac{a_{12}}{a_{21}} + \frac{a_{11}}{a_{22}}\) given that \(AB = BA\) for the matrices \(A\) and \(B\).

### Step-by-Step Solution:

1. **Define the Matrices**:
Let
\[ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \]

2. **Calculate \(AB\)**:
\[ AB = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} a_{11} + 2a_{12} & a_{11} + a_{12} \\ a_{21} + 2a_{22} & a_{21} + a_{22} \end{pmatrix} \]

3. **Calculate \(BA\)**:
\[ BA = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} a_{11} + a_{21} & a_{12} + a_{22} \\ 2a_{11} + a_{21} & 2a_{12} + a_{22} \end{pmatrix} \]

4. **Set \(AB = BA\)**:
From the equality \(AB = BA\), we equate the corresponding elements:
- From the first element:
\[ a_{11} + 2a_{12} = a_{11} + a_{21} \implies 2a_{12} = a_{21} \tag{1} \]
- From the second element:
\[ a_{11} + a_{12} = a_{12} + a_{22} \implies a_{11} = a_{22} \tag{2} \]
- From the third element:
\[ a_{21} + 2a_{22} = 2a_{11} + a_{21} \implies 2a_{22} = 2a_{11} \implies a_{22} = a_{11} \tag{3} \]
- From the fourth element:
\[ a_{21} + a_{22} = 2a_{12} + a_{22} \implies a_{21} = 2a_{12} \tag{4} \]

5. **Substitute and Simplify**:
From equation (1), we have:
\[ a_{21} = 2a_{12} \]
From equation (2), we have:
\[ a_{11} = a_{22} \]

6. **Find the Required Expression**:
We need to compute:
\[ \frac{a_{12}}{a_{21}} + \frac{a_{11}}{a_{22}} \]
Substituting \(a_{21} = 2a_{12}\) and \(a_{11} = a_{22}\):
\[ \frac{a_{12}}{2a_{12}} + \frac{a_{11}}{a_{11}} = \frac{1}{2} + 1 = \frac{1}{2} + \frac{2}{2} = \frac{3}{2} \]

### Final Answer:
The value of \(\frac{a_{12}}{a_{21}} + \frac{a_{11}}{a_{22}}\) is \(\frac{3}{2}\).