If `A^(-1) = ((3,1),(-1,2))`, find the matrix A.

Answer

To find the matrix \( A \) given that \( A^{-1} = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \), we can follow these steps:

### Step 1: Understand the relationship between \( A \) and \( A^{-1} \)
The relationship between a matrix and its inverse is given by the equation:
\[ A \cdot A^{-1} = I \]
where \( I \) is the identity matrix.

### Step 2: Find the determinant of \( A^{-1} \)
The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as:
\[ \text{det} = ad - bc \]
For \( A^{-1} = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \):
- \( a = 3 \)
- \( b = 1 \)
- \( c = -1 \)
- \( d = 2 \)

Calculating the determinant:
\[ \text{det}(A^{-1}) = (3)(2) - (1)(-1) = 6 + 1 = 7 \]

### Step 3: Find the determinant of \( A \)
The determinant of \( A \) is the reciprocal of the determinant of \( A^{-1} \):
\[ \text{det}(A) = \frac{1}{\text{det}(A^{-1})} = \frac{1}{7} \]

### Step 4: Find the adjoint of \( A \)
The adjoint of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Since we know that:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]
We can express the adjoint of \( A \) as:
\[ \text{adj}(A) = A^{-1} \cdot \text{det}(A) \]
Substituting the values we have:
\[ \text{adj}(A) = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \cdot \frac{1}{7} = \begin{pmatrix} \frac{3}{7} & \frac{1}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{pmatrix} \]

### Step 5: Find the matrix \( A \)
Using the adjoint to find \( A \):
\[ A = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A = 7 \cdot \begin{pmatrix} \frac{3}{7} & \frac{1}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \]

### Final Result
Thus, the matrix \( A \) is:
\[ A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \]