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 matrices given. The relationship between a matrix and its inverse is that the product of a matrix and its inverse equals the identity matrix.

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: Multiply \( A \) and \( A^{-1} \)

We need to calculate \( A \cdot A^{-1} \) and set it equal to the identity matrix \( I \):

\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

### Step 2: Calculate the elements of \( A \cdot A^{-1} \)

We will multiply the matrices \( A \) and \( A^{-1} \):

1. **First Row:**
- First Column: \( 0 \cdot \frac{1}{2} + 1 \cdot (-4) + 2 \cdot \frac{5}{2} = 0 - 4 + 5 = 1 \)
- Second Column: \( 0 \cdot (-\frac{1}{2}) + 1 \cdot 3 + 2 \cdot (-\frac{3}{2}) = 0 + 3 - 3 = 0 \)
- Third Column: \( 0 \cdot \frac{1}{2} + 1 \cdot b + 2 \cdot \frac{1}{2} = 0 + b + 1 = b + 1 \)

2. **Second Row:**
- First Column: \( 1 \cdot \frac{1}{2} + 2 \cdot (-4) + 3 \cdot \frac{5}{2} = \frac{1}{2} - 8 + \frac{15}{2} = \frac{1 + 15 - 16}{2} = 0 \)
- Second Column: \( 1 \cdot (-\frac{1}{2}) + 2 \cdot 3 + 3 \cdot (-\frac{3}{2}) = -\frac{1}{2} + 6 - \frac{9}{2} = -\frac{10}{2} + 6 = 3 - 5 = -2 \)
- Third Column: \( 1 \cdot \frac{1}{2} + 2 \cdot b + 3 \cdot \frac{1}{2} = \frac{1}{2} + 2b + \frac{3}{2} = 2b + 2 \)

3. **Third Row:**
- First Column: \( 3 \cdot \frac{1}{2} + a \cdot (-4) + 1 \cdot \frac{5}{2} = \frac{3}{2} - 4a + \frac{5}{2} = 4 - 4a = 0 \)
- Second Column: \( 3 \cdot (-\frac{1}{2}) + a \cdot 3 + 1 \cdot (-\frac{3}{2}) = -\frac{3}{2} + 3a - \frac{3}{2} = 3a - 3 = 0 \)
- Third Column: \( 3 \cdot \frac{1}{2} + a \cdot b + 1 \cdot \frac{1}{2} = \frac{3}{2} + ab + \frac{1}{2} = ab + 2 \)

### Step 3: Set the products equal to the identity matrix

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

### Step 4: Solve the equations

1. From \( b + 1 = 0 \), we get \( b = -1 \).
2. From \( 2b + 2 = 0 \), substituting \( b = -1 \) gives \( 2(-1) + 2 = 0 \) which is consistent.
3. From \( 4 - 4a = 0 \), we get \( 4a = 4 \) leading to \( a = 1 \).
4. From \( 3a - 3 = 0 \), substituting \( a = 1 \) gives \( 3(1) - 3 = 0 \) which is consistent.
5. From \( ab + 2 = 0 \), substituting \( a = 1 \) and \( b = -1 \) gives \( 1(-1) + 2 = 1 \), which is not satisfied.

### Conclusion

Thus, the values are:
- \( a = 1 \)
- \( b = -1 \)