If `A=[[0, 1,2],[1,2,3],[3,a,1]]and A^(-1)[[1//2,-1//2,1//2],[-4,3,b],[5//2,-3//2,1//2]]` then

Options

A. `a = 1, b = -1`
B. `a = 2, b = -1/2`
C. `a = -1, b=1`
D. `a=1/2, b=1/2`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) in the given matrices \( A \) and \( A^{-1} \). We will use the property that the product of a matrix and its inverse is the identity matrix \( I \).

Given:
\[ A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \]
\[ A^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & b \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix} \]

### Step 1: Compute the product \( A \cdot A^{-1} \)
We need to multiply \( A \) and \( A^{-1} \) and set the result equal to the identity matrix \( I \).

\[ A \cdot A^{-1} = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & b \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix} \]

### Step 2: Calculate each element of the resulting matrix
1. **First row**:
- First element: \( 0 \cdot \frac{1}{2} + 1 \cdot (-4) + 2 \cdot \frac{5}{2} = -4 + 5 = 1 \)
- Second element: \( 0 \cdot (-\frac{1}{2}) + 1 \cdot 3 + 2 \cdot (-\frac{3}{2}) = 3 - 3 = 0 \)
- Third element: \( 0 \cdot \frac{1}{2} + 1 \cdot b + 2 \cdot \frac{1}{2} = b + 1 \)

2. **Second row**:
- First element: \( 1 \cdot \frac{1}{2} + 2 \cdot (-4) + 3 \cdot \frac{5}{2} = \frac{1}{2} - 8 + \frac{15}{2} = \frac{1 + 15 - 16}{2} = 0 \)
- Second element: \( 1 \cdot (-\frac{1}{2}) + 2 \cdot 3 + 3 \cdot (-\frac{3}{2}) = -\frac{1}{2} + 6 - \frac{9}{2} = 0 \)
- Third element: \( 1 \cdot \frac{1}{2} + 2 \cdot b + 3 \cdot \frac{1}{2} = \frac{1}{2} + 2b + \frac{3}{2} = 2 + 2b \)

3. **Third row**:
- First element: \( 3 \cdot \frac{1}{2} + a \cdot (-4) + 1 \cdot \frac{5}{2} = \frac{3}{2} - 4a + \frac{5}{2} = 4 - 4a \)
- Second element: \( 3 \cdot (-\frac{1}{2}) + a \cdot 3 + 1 \cdot (-\frac{3}{2}) = -\frac{3}{2} + 3a - \frac{3}{2} = 3a - 3 \)
- Third element: \( 3 \cdot \frac{1}{2} + a \cdot b + 1 \cdot \frac{1}{2} = \frac{3}{2} + ab + \frac{1}{2} = 2 + ab \)

### Step 3: Set the resulting matrix equal to the identity matrix
The resulting matrix from \( A \cdot A^{-1} \) should equal:
\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

From the calculations, we have:
1. \( b + 1 = 0 \) (from the first row, third column)
2. \( 2 + 2b = 0 \) (from the second row, third column)
3. \( 4 - 4a = 0 \) (from the third row, first column)

### Step 4: Solve the equations
1. From \( b + 1 = 0 \):
\[ b = -1 \]

2. From \( 2 + 2b = 0 \):
\[ 2 - 2 = 0 \quad \text{(This is consistent with } b = -1\text{)} \]

3. From \( 4 - 4a = 0 \):
\[ 4a = 4 \implies a = 1 \]

### Final Values
Thus, we have:
\[ a = 1, \quad b = -1 \]

### Conclusion
The values of \( a \) and \( b \) are:
- \( a = 1 \)
- \( b = -1 \)