If `A=[(0,1,2),(1,2,3),(3,a,1)]` and `A^(-1)=[(1//2,-1//2,1//2),(-4,3,c),(5//2,-3//2,1//2)]`, then the values of a and c are equal to

Options

A. 1, 1
B. 1, -1
C. 1, 2
D. `-1, 1`

Explanation

To find the values of \( a \) and \( c \) 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:
\[ A = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{pmatrix} \]
\[ A^{-1} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & c \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{pmatrix} \]

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

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

### Step 2: Calculate the first row of the product
The first row of the product is calculated as follows:
- 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:
\[ 0 \cdot \frac{1}{2} + 1 \cdot c + 2 \cdot \frac{1}{2} = c + 1 \]

Thus, the first row of the product is:
\[ \begin{pmatrix} 1 & 0 & c + 1 \end{pmatrix} \]

### Step 3: Calculate the second row of the product
The second row of the product is calculated as follows:
- 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:
\[ 1 \cdot (-\frac{1}{2}) + 2 \cdot 3 + 3 \cdot (-\frac{3}{2}) = -\frac{1}{2} + 6 - \frac{9}{2} = 6 - 5 = 1 \]
- Third element:
\[ 1 \cdot \frac{1}{2} + 2 \cdot c + 3 \cdot \frac{1}{2} = \frac{1}{2} + 2c + \frac{3}{2} = 2c + 2 \]

Thus, the second row of the product is:
\[ \begin{pmatrix} 0 & 1 & 2c + 2 \end{pmatrix} \]

### Step 4: Calculate the third row of the product
The third row of the product is calculated as follows:
- 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:
\[ 3 \cdot \frac{1}{2} + a \cdot c + 1 \cdot \frac{1}{2} = \frac{3}{2} + ac + \frac{1}{2} = ac + 2 \]

Thus, the third row of the product is:
\[ \begin{pmatrix} 4 - 4a & 3a - 3 & ac + 2 \end{pmatrix} \]

### Step 5: Set the product equal to the identity matrix
Now we set the product equal to the identity matrix \( I \):
\[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

From the first row:
1. \( c + 1 = 0 \) → \( c = -1 \)

From the second row:
2. \( 2c + 2 = 0 \) → \( 2(-1) + 2 = 0 \) (This is satisfied)

From the third row:
3. \( 4 - 4a = 0 \) → \( 4 = 4a \) → \( a = 1 \)

4. \( 3a - 3 = 0 \) → \( 3(1) - 3 = 0 \) (This is satisfied)

5. \( ac + 2 = 1 \) → \( 1(-1) + 2 = 1 \) (This is satisfied)

### Final Values
Thus, the values of \( a \) and \( c \) are:
\[ a = 1, \quad c = -1 \]