For the matrix `A=[{:(2,1),(3,0):}]` , find the numbers 'a' and 'b' such that `A^(2)+aA+bI=O`. Hence , find `A^(-1)`.

Answer

To solve the problem, we need to find the values of 'a' and 'b' such that \( A^2 + aA + bI = O \), where \( A = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} \) and \( I \) is the identity matrix. After finding 'a' and 'b', we will also find the inverse of matrix \( A \).

### Step 1: Calculate \( A^2 \)

First, we need to calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} \]

Calculating the product:
- The element at (1,1): \( 2 \cdot 2 + 1 \cdot 3 = 4 + 3 = 7 \)
- The element at (1,2): \( 2 \cdot 1 + 1 \cdot 0 = 2 + 0 = 2 \)
- The element at (2,1): \( 3 \cdot 2 + 0 \cdot 3 = 6 + 0 = 6 \)
- The element at (2,2): \( 3 \cdot 1 + 0 \cdot 0 = 3 + 0 = 3 \)

Thus,
\[ A^2 = \begin{pmatrix} 7 & 2 \\ 6 & 3 \end{pmatrix} \]

### Step 2: Set up the equation \( A^2 + aA + bI = O \)

We know:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
So, we can write the equation as:
\[ \begin{pmatrix} 7 & 2 \\ 6 & 3 \end{pmatrix} + a \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} + b \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Step 3: Combine the matrices

This gives us:
\[ \begin{pmatrix} 7 + 2a + b & 2 + a \\ 6 + 3a & 3 + b \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Step 4: Set up the system of equations

From the above matrix equation, we can set up the following equations:
1. \( 7 + 2a + b = 0 \) (Equation 1)
2. \( 2 + a = 0 \) (Equation 2)
3. \( 6 + 3a = 0 \) (Equation 3)
4. \( 3 + b = 0 \) (Equation 4)

### Step 5: Solve the equations

From Equation 2:
\[ a = -2 \]

Substituting \( a = -2 \) into Equation 1:
\[ 7 + 2(-2) + b = 0 \implies 7 - 4 + b = 0 \implies b = -3 \]

Now, substituting \( a = -2 \) into Equation 3:
\[ 6 + 3(-2) = 0 \implies 6 - 6 = 0 \quad \text{(True)} \]

And substituting \( b = -3 \) into Equation 4:
\[ 3 + (-3) = 0 \quad \text{(True)} \]

Thus, we have:
\[ a = -2, \quad b = -3 \]

### Step 6: Find \( A^{-1} \)

We know from the equation \( A^2 - 2A = 3I \):
\[ A^2 = 2A + 3I \]

Multiplying both sides by \( A^{-1} \):
\[ A \cdot A = 2A \cdot I + 3I \cdot A^{-1} \]
\[ A = 2I + 3A^{-1} \]
Rearranging gives:
\[ 3A^{-1} = A - 2I \]
\[ A^{-1} = \frac{1}{3}(A - 2I) \]

Calculating \( A - 2I \):
\[ A - 2I = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & -2 \end{pmatrix} \]

Thus,
\[ A^{-1} = \frac{1}{3} \begin{pmatrix} 0 & 1 \\ 3 & -2 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & -\frac{2}{3} \end{pmatrix} \]

### Final Answer
The values are:
\[ a = -2, \quad b = -3 \]
And the inverse of \( A \) is:
\[ A^{-1} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & -\frac{2}{3} \end{pmatrix} \]