If A is a square matrix of order `2xx2` and `B=[(1,2),(3, 4)]`, such that `AB=BA`, then A can be

Options

A. `[(1,4),(6, 7)]`
B. `[(1, 4),(7, 6)]`
C. `[(2,2),(2,4)]`
D. `[(3,4),(4,9)]`

Answer

To solve the problem where we have a square matrix \( A \) of order \( 2 \times 2 \) and a given matrix \( B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \) such that \( AB = BA \), we can follow these steps:

### Step 1: Define Matrix \( A \)
Let matrix \( A \) be represented as:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

### Step 2: Calculate \( AB \)
Now, we will calculate the product \( AB \):
\[ AB = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( a \cdot 1 + b \cdot 3 = a + 3b \)
- First row, second column: \( a \cdot 2 + b \cdot 4 = 2a + 4b \)
- Second row, first column: \( c \cdot 1 + d \cdot 3 = c + 3d \)
- Second row, second column: \( c \cdot 2 + d \cdot 4 = 2c + 4d \)

Thus, we have:
\[ AB = \begin{pmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{pmatrix} \]

### Step 3: Calculate \( BA \)
Next, we calculate the product \( BA \):
\[ BA = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot a + 2 \cdot c = a + 2c \)
- First row, second column: \( 1 \cdot b + 2 \cdot d = b + 2d \)
- Second row, first column: \( 3 \cdot a + 4 \cdot c = 3a + 4c \)
- Second row, second column: \( 3 \cdot b + 4 \cdot d = 3b + 4d \)

Thus, we have:
\[ BA = \begin{pmatrix} a + 2c & b + 2d \\ 3a + 4c & 3b + 4d \end{pmatrix} \]

### Step 4: Set \( AB = BA \)
Now, we set the two products equal to each other:
\[ \begin{pmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{pmatrix} = \begin{pmatrix} a + 2c & b + 2d \\ 3a + 4c & 3b + 4d \end{pmatrix} \]

### Step 5: Formulate the Equations
From the equality of the matrices, we can form the following equations:
1. \( a + 3b = a + 2c \)
2. \( 2a + 4b = b + 2d \)
3. \( c + 3d = 3a + 4c \)
4. \( 2c + 4d = 3b + 4d \)

### Step 6: Simplify the Equations
1. From the first equation: \( 3b = 2c \) ⇒ \( c = \frac{3}{2}b \)
2. From the second equation: \( 2a + 4b = b + 2d \) ⇒ \( 2a + 3b = 2d \) ⇒ \( d = a + \frac{3}{2}b \)
3. From the third equation: \( c + 3d = 3a + 4c \) ⇒ \( 3d = 3a + 3c \) ⇒ \( d = a + c \)
4. From the fourth equation: \( 2c + 4d = 3b + 4d \) ⇒ \( 2c = 3b \) ⇒ \( c = \frac{3}{2}b \)

### Step 7: Substitute Values
Using \( c = \frac{3}{2}b \) in \( d = a + c \):
\[ d = a + \frac{3}{2}b \]

### Step 8: Formulate Matrix \( A \)
Now, we can express \( A \) in terms of \( a \) and \( b \):
\[ A = \begin{pmatrix} a & b \\ \frac{3}{2}b & a + \frac{3}{2}b \end{pmatrix} \]

### Conclusion
Thus, matrix \( A \) can be expressed as:
\[ A = \begin{pmatrix} a & b \\ \frac{3}{2}b & a + \frac{3}{2}b \end{pmatrix} \]
where \( a \) and \( b \) can take any real values.