For matrix `A=[(2,5),(-11,7)]` (adj A)' is equal to :

Options

A. `[(-2,-5),(11,-7)]`
B. `[(7,5),(11,2)]`
C. `[(7,11),(-5,2)]`
D. `[(7,-5),(11,2)]`

Answer

To find \((\text{adj } A)'\) for the matrix \(A = \begin{pmatrix} 2 & 5 \\ -11 & 7 \end{pmatrix}\), we will follow these steps:

### Step 1: Find the Adjoint of Matrix \(A\)

The adjoint of a \(2 \times 2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by:

\[ \text{adj } A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \(A = \begin{pmatrix} 2 & 5 \\ -11 & 7 \end{pmatrix}\):

- \(a = 2\)
- \(b = 5\)
- \(c = -11\)
- \(d = 7\)

Thus, we can compute the adjoint:

\[ \text{adj } A = \begin{pmatrix} 7 & -5 \\ 11 & 2 \end{pmatrix} \]

### Step 2: Find the Transpose of the Adjoint

Next, we need to find the transpose of the adjoint matrix. The transpose of a matrix is obtained by swapping its rows with columns.

So, the adjoint matrix is:

\[ \text{adj } A = \begin{pmatrix} 7 & -5 \\ 11 & 2 \end{pmatrix} \]

Taking the transpose:

\[ (\text{adj } A)' = \begin{pmatrix} 7 & 11 \\ -5 & 2 \end{pmatrix} \]

### Conclusion

Thus, the final result for \((\text{adj } A)'\) is:

\[ \begin{pmatrix} 7 & 11 \\ -5 & 2 \end{pmatrix} \]