If `A=[(2,1),(9,3)]` and `A^(2)-5A+7I=O`, then `A^(-1)=`

Options

A. `[(3,-1),(9,2)]`
B. `(1)/(7)[(3,-1),(-9,2)]`
C. `(1)/(14)[(3,-1),(9,2)]`
D. `[(3,1),(9,2)]`

Answer

To find the inverse of the matrix \( A \) given the equation \( A^2 - 5A + 7I = O \), we will follow these steps:

### Step 1: Write down the given equation
We start with the equation:
\[ A^2 - 5A + 7I = O \]
where \( O \) is the zero matrix.

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ A^2 - 5A = -7I \]

### Step 3: Post-multiply by \( A^{-1} \)
We will post-multiply both sides by \( A^{-1} \):
\[ A^2 A^{-1} - 5A A^{-1} = -7I A^{-1} \]
Using the property \( AA^{-1} = I \), we simplify:
\[ A I - 5I = -7A^{-1} \]
This simplifies to:
\[ A - 5I = -7A^{-1} \]

### Step 4: Isolate \( A^{-1} \)
Now, we isolate \( A^{-1} \):
\[ 7A^{-1} = 5I - A \]
Dividing both sides by 7 gives:
\[ A^{-1} = \frac{1}{7}(5I - A) \]

### Step 5: Substitute the identity matrix and matrix \( A \)
Substituting \( A = \begin{pmatrix} 2 & 1 \\ 9 & 3 \end{pmatrix} \) and the identity matrix \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \):
\[ 5I = 5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]
Now, we calculate \( 5I - A \):
\[ 5I - A = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 9 & 3 \end{pmatrix} = \begin{pmatrix} 5 - 2 & 0 - 1 \\ 0 - 9 & 5 - 3 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -9 & 2 \end{pmatrix} \]

### Step 6: Calculate \( A^{-1} \)
Now substituting back into the equation for \( A^{-1} \):
\[ A^{-1} = \frac{1}{7} \begin{pmatrix} 3 & -1 \\ -9 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{7} & -\frac{1}{7} \\ -\frac{9}{7} & \frac{2}{7} \end{pmatrix} \]

### Final Answer
Thus, the inverse of matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} \frac{3}{7} & -\frac{1}{7} \\ -\frac{9}{7} & \frac{2}{7} \end{pmatrix} \]