If the matrix A = `[{:(a,b),(c, d):}]` is commutative on product with the matrix B = `[{:(1,1),(0,1):}]` , then

Options

A. a = 0, b = c
B. b = 0 , c = d
C. c = 0 , d = a
D. d = 0 , a = b

Answer

To solve the problem, we need to find the conditions under which the matrices A and B commute. The matrices are given as follows:

Matrix A:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

Matrix B:
\[ B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]

We need to show that \( AB = BA \).

### Step 1: Calculate \( AB \)
To find \( AB \), we multiply matrix A by matrix B:

\[ AB = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]

Calculating the product:

- First row, first column: \( a \cdot 1 + b \cdot 0 = a \)
- First row, second column: \( a \cdot 1 + b \cdot 1 = a + b \)
- Second row, first column: \( c \cdot 1 + d \cdot 0 = c \)
- Second row, second column: \( c \cdot 1 + d \cdot 1 = c + d \)

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

### Step 2: Calculate \( BA \)
Now, we calculate \( BA \):

\[ BA = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

Calculating the product:

- First row, first column: \( 1 \cdot a + 1 \cdot c = a + c \)
- First row, second column: \( 1 \cdot b + 1 \cdot d = b + d \)
- Second row, first column: \( 0 \cdot a + 1 \cdot c = c \)
- Second row, second column: \( 0 \cdot b + 1 \cdot d = d \)

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

### Step 3: Set \( AB = BA \)
Now we equate the two matrices:

\[ \begin{pmatrix} a & a + b \\ c & c + d \end{pmatrix} = \begin{pmatrix} a + c & b + d \\ c & d \end{pmatrix} \]

This gives us the following equations:

1. \( a = a + c \)
2. \( a + b = b + d \)
3. \( c = c \) (this is always true)
4. \( c + d = d \)

### Step 4: Solve the equations
From the first equation:
\[ a = a + c \implies c = 0 \]

From the second equation:
\[ a + b = b + d \implies a = d \]

From the fourth equation:
\[ c + d = d \implies c = 0 \text{ (which we already found)} \]

### Conclusion
Thus, we have:
- \( c = 0 \)
- \( a = d \)

So, the final relation is:
\[ c = 0 \quad \text{and} \quad a = d \]