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 the matrices \(A\) and \(B\) commute, i.e., \(AB = BA\).

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

### Step 2: Calculate \(AB\)
Now, we compute \(AB\):
\[ AB = A \cdot B = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \(a_{11} \cdot 1 + a_{12} \cdot 2 = a_{11} + 2a_{12}\)
- First row, second column: \(a_{11} \cdot 1 + a_{12} \cdot 1 = a_{11} + a_{12}\)
- Second row, first column: \(a_{21} \cdot 1 + a_{22} \cdot 2 = a_{21} + 2a_{22}\)
- Second row, second column: \(a_{21} \cdot 1 + a_{22} \cdot 1 = a_{21} + a_{22}\)

Thus, we have:
\[ AB = \begin{pmatrix} a_{11} + 2a_{12} & a_{11} + a_{12} \\ a_{21} + 2a_{22} & a_{21} + a_{22} \end{pmatrix} \]

### Step 3: Calculate \(BA\)
Next, we compute \(BA\):
\[ BA = B \cdot A = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \]
Calculating the elements:
- First row, first column: \(1 \cdot a_{11} + 1 \cdot a_{21} = a_{11} + a_{21}\)
- First row, second column: \(1 \cdot a_{12} + 1 \cdot a_{22} = a_{12} + a_{22}\)
- Second row, first column: \(2 \cdot a_{11} + 1 \cdot a_{21} = 2a_{11} + a_{21}\)
- Second row, second column: \(2 \cdot a_{12} + 1 \cdot a_{22} = 2a_{12} + a_{22}\)

Thus, we have:
\[ BA = \begin{pmatrix} a_{11} + a_{21} & a_{12} + a_{22} \\ 2a_{11} + a_{21} & 2a_{12} + a_{22} \end{pmatrix} \]

### Step 4: Set \(AB = BA\)
Now, we set the two results equal to each other:
\[ \begin{pmatrix} a_{11} + 2a_{12} & a_{11} + a_{12} \\ a_{21} + 2a_{22} & a_{21} + a_{22} \end{pmatrix} = \begin{pmatrix} a_{11} + a_{21} & a_{12} + a_{22} \\ 2a_{11} + a_{21} & 2a_{12} + a_{22} \end{pmatrix} \]

### Step 5: Compare corresponding elements
From the first row, first column:
\[ a_{11} + 2a_{12} = a_{11} + a_{21} \implies 2a_{12} = a_{21} \implies \frac{a_{12}}{a_{21}} = \frac{1}{2} \]

From the second row, second column:
\[ a_{21} + a_{22} = 2a_{12} + a_{22} \implies a_{21} = 2a_{12} \implies a_{22} = a_{11} \]

### Step 6: Substitute values
Now substituting \(a_{22} = a_{11}\) into the expression we want to evaluate:
\[ \frac{a_{12}}{a_{21}} + \frac{a_{11}}{a_{22}} = \frac{1}{2} + \frac{a_{11}}{a_{11}} = \frac{1}{2} + 1 = \frac{3}{2} \]

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