If the matrix `{:[(a,b),(c,d)]:}` is commutative with matrix `{:[(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 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is commutative with the matrix \( B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \). This means that \( A \cdot B = B \cdot A \).

### Step-by-Step Solution:

1. **Matrix Multiplication \( A \cdot B \)**:
\[ A \cdot B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]
To perform this multiplication, we calculate each element:
- 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,
\[ A \cdot B = \begin{pmatrix} a & a + b \\ c & c + d \end{pmatrix} \]

2. **Matrix Multiplication \( B \cdot A \)**:
\[ B \cdot A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
Again, we calculate each element:
- 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,
\[ B \cdot A = \begin{pmatrix} a + c & b + d \\ c & d \end{pmatrix} \]

3. **Setting the Products Equal**:
Since \( A \cdot B = B \cdot A \), we equate the two resulting 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:
- From the first row, first column: \( a = a + c \)
- From the first row, second column: \( a + b = b + d \)
- From the second row, first column: \( c = c \) (this is always true)
- From the second row, second column: \( c + d = d \)

4. **Solving the Equations**:
- From \( a = a + c \), we can simplify to \( c = 0 \).
- From \( a + b = b + d \), we simplify to \( a = d \).
- From \( c + d = d \), substituting \( c = 0 \) gives \( 0 + d = d \), which is always true.

5. **Final Conditions**:
Thus, we conclude that the conditions for the matrix \( A \) to be commutative with matrix \( B \) are:
- \( c = 0 \)
- \( a = d \)

### Final Answer:
The conditions are:
- \( c = 0 \)
- \( a = d \)