Find the matrix X such that AX = I where `A = [(6,17),(1,3)]`

Answer

To find the matrix \( X \) such that \( AX = I \), where

\[ A = \begin{pmatrix} 6 & 17 \\ 1 & 3 \end{pmatrix} \]

we need to calculate the inverse of matrix \( A \), denoted as \( A^{-1} \). The relationship we are using is \( AX = I \) implies \( X = A^{-1} \).

### Step 1: Calculate the Determinant of Matrix \( A \)

The determinant of a 2x2 matrix

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

is given by the formula:

\[ \text{det}(A) = ad - bc \]

For our matrix \( A \):

\[ \text{det}(A) = (6)(3) - (17)(1) = 18 - 17 = 1 \]

### Step 2: Find the Adjoint of Matrix \( A \)

The adjoint of a 2x2 matrix

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

is obtained by interchanging the elements on the main diagonal and changing the signs of the off-diagonal elements:

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

For our matrix \( A \):

\[ \text{adj}(A) = \begin{pmatrix} 3 & -17 \\ -1 & 6 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix \( A \)

The inverse of a matrix \( A \) is given by the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

Since we have already calculated the determinant \( \text{det}(A) = 1 \) and the adjoint \( \text{adj}(A) = \begin{pmatrix} 3 & -17 \\ -1 & 6 \end{pmatrix} \), we can find \( A^{-1} \):

\[ A^{-1} = \frac{1}{1} \cdot \begin{pmatrix} 3 & -17 \\ -1 & 6 \end{pmatrix} = \begin{pmatrix} 3 & -17 \\ -1 & 6 \end{pmatrix} \]

### Step 4: Conclusion

Thus, the matrix \( X \) such that \( AX = I \) is:

\[ X = A^{-1} = \begin{pmatrix} 3 & -17 \\ -1 & 6 \end{pmatrix} \]