The inverse of matrix `[[2, -3], [5, 7]]` is

Options

A. `(-1)/(29)[[7, 3], [-5, 2]]`
B. `(1)/(29)[[7, 3], [-5, 2]]`
C. `-[[7, 3], [-5, 2]]`
D. `[[7, 3], [-5, 2]]`

Answer

To find the inverse of the matrix \( A = \begin{bmatrix} 2 & -3 \\ 5 & 7 \end{bmatrix} \), we will follow the steps outlined below:

### Step 1: Calculate the Determinant of Matrix A
The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated using the formula:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
- \( a = 2 \)
- \( b = -3 \)
- \( c = 5 \)
- \( d = 7 \)

Thus, the determinant is:
\[ \text{det}(A) = (2)(7) - (-3)(5) = 14 + 15 = 29 \]

### Step 2: Find the Adjoint of Matrix A
The adjoint of a 2x2 matrix is obtained by swapping the elements on the main diagonal and changing the signs of the elements on the other diagonal. For matrix \( A \):
\[ \text{Adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ -5 & 2 \end{bmatrix} \]

### Step 3: Calculate the Inverse of Matrix A
The inverse of matrix \( A \) can be calculated using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) \]
Substituting the values we found:
\[ A^{-1} = \frac{1}{29} \cdot \begin{bmatrix} 7 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} \frac{7}{29} & \frac{3}{29} \\ -\frac{5}{29} & \frac{2}{29} \end{bmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{bmatrix} \frac{7}{29} & \frac{3}{29} \\ -\frac{5}{29} & \frac{2}{29} \end{bmatrix} \]

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