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 \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) commutes 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 compute this, we multiply the rows of \( A \) by the columns of \( B \):
\[ A \cdot B = \begin{pmatrix} a \cdot 1 + b \cdot 0 & a \cdot 1 + b \cdot 1 \\ c \cdot 1 + d \cdot 0 & c \cdot 1 + d \cdot 1 \end{pmatrix} = \begin{pmatrix} a & a + b \\ c & c + d \end{pmatrix} \]

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 multiply the rows of \( B \) by the columns of \( A \):
\[ B \cdot A = \begin{pmatrix} 1 \cdot a + 1 \cdot c & 1 \cdot b + 1 \cdot d \\ 0 \cdot a + 1 \cdot c & 0 \cdot b + 1 \cdot d \end{pmatrix} = \begin{pmatrix} a + c & b + d \\ c & d \end{pmatrix} \]

3. **Setting the Two Results Equal**:
Since \( A \cdot B = B \cdot A \), we set the two matrices equal:
\[ \begin{pmatrix} a & a + b \\ c & c + d \end{pmatrix} = \begin{pmatrix} a + c & b + d \\ c & d \end{pmatrix} \]

4. **Equating Corresponding Elements**:
From the equality of the matrices, we can derive the following equations:
- From the first element: \( a = a + c \)
- From the second element: \( a + b = b + d \)
- From the third element: \( c = c \) (This is always true)
- From the fourth element: \( c + d = d \)

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

6. **Final Conditions**:
The conditions we have derived are:
- \( c = 0 \)
- \( a = d \)

Thus, the matrix \( A \) can be expressed as:
\[ A = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \]