If A =`[{:( 2,5,3),(3,1,2),(1,2,1) :}]` then `A^(-1) ` is equal to

Options

A. ` [{:( (-3)/(4),(1)/(4) ,(7)/(4) ),( (-1)/( 4) ,(-1)/( 4) ,(5)/(4)) ,( ( 5)/( 4) ,(1)/(4) , (-13)/( 4)):}]`
B. ` [{:( (3)/(4),(1)/(4) ,(7)/(4) ),( (1)/( 4) ,(-1)/( 4) ,(5)/(4)) ,( ( 5)/( 4) ,(1)/(4) , (-13)/( 4)):}]`
C. ` [{:( (-3)/(4),(1)/(4) ,(7)/(4) ),( (-1)/( 4) ,(-1)/( 4) ,(5)/(4)) ,( ( -5)/( 4) ,(1)/(4) , (13)/( 4)):}]`
D. none of these

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 2 & 5 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of A
The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
- \( a = 2, b = 5, c = 3 \)
- \( d = 3, e = 1, f = 2 \)
- \( g = 1, h = 2, i = 1 \)

Calculating the determinant:
\[ \text{det}(A) = 2(1 \cdot 1 - 2 \cdot 2) - 5(3 \cdot 1 - 2 \cdot 1) + 3(3 \cdot 2 - 1 \cdot 1) \]
\[ = 2(1 - 4) - 5(3 - 2) + 3(6 - 1) \]
\[ = 2(-3) - 5(1) + 3(5) \]
\[ = -6 - 5 + 15 = 4 \]

### Step 2: Calculate the Cofactor Matrix
The cofactor matrix \( C \) is calculated by finding the determinant of the 2x2 submatrices formed by removing the row and column of each element, and applying the checkerboard pattern of signs.

Calculating the cofactors:
1. For \( C_{11} = \text{det} \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} = 1 \cdot 1 - 2 \cdot 2 = 1 - 4 = -3 \)
2. For \( C_{12} = -\text{det} \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} = -(3 \cdot 1 - 2 \cdot 1) = -(3 - 2) = -1 \)
3. For \( C_{13} = \text{det} \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} = 3 \cdot 2 - 1 \cdot 1 = 6 - 1 = 5 \)

4. For \( C_{21} = -\text{det} \begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix} = -(5 \cdot 1 - 3 \cdot 2) = -(5 - 6) = 1 \)
5. For \( C_{22} = \text{det} \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} = 2 \cdot 1 - 3 \cdot 1 = 2 - 3 = -1 \)
6. For \( C_{23} = -\text{det} \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix} = -(2 \cdot 2 - 5 \cdot 1) = -(4 - 5) = 1 \)

7. For \( C_{31} = \text{det} \begin{pmatrix} 5 & 3 \\ 1 & 2 \end{pmatrix} = 5 \cdot 2 - 3 \cdot 1 = 10 - 3 = 7 \)
8. For \( C_{32} = -\text{det} \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix} = -(2 \cdot 2 - 3 \cdot 3) = -(4 - 9) = 5 \)
9. For \( C_{33} = \text{det} \begin{pmatrix} 2 & 5 \\ 3 & 1 \end{pmatrix} = 2 \cdot 1 - 5 \cdot 3 = 2 - 15 = -13 \)

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} -3 & -1 & 5 \\ 1 & -1 & 1 \\ 7 & 5 & -13 \end{pmatrix} \]

### Step 3: Calculate the Adjoint of A
The adjoint of \( A \) is the transpose of the cofactor matrix \( C \):
\[ \text{adj}(A) = C^T = \begin{pmatrix} -3 & 1 & 7 \\ -1 & -1 & 5 \\ 5 & 1 & -13 \end{pmatrix} \]

### Step 4: Calculate the Inverse of A
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the determinant and adjoint we found:
\[ A^{-1} = \frac{1}{4} \begin{pmatrix} -3 & 1 & 7 \\ -1 & -1 & 5 \\ 5 & 1 & -13 \end{pmatrix} \]
\[ = \begin{pmatrix} -\frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ -\frac{1}{4} & -\frac{1}{4} & \frac{5}{4} \\ \frac{5}{4} & \frac{1}{4} & -\frac{13}{4} \end{pmatrix} \]

### Final Answer
Thus, the inverse of matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ -\frac{1}{4} & -\frac{1}{4} & \frac{5}{4} \\ \frac{5}{4} & \frac{1}{4} & -\frac{13}{4} \end{pmatrix} \]