The inverse of `[(3,5,7),(2,-3,1),(1,1,2)]` is :

Options

A. `[(7,3,-26),(3,1,-11),(-5,-2,0)]`
B. `[(7,3,-26),(3,1,11),(-5,-2,1)]`
C. `[(3,1,11),(7,3,-26),(-5,2,1)]`
D. None of these

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of A
The determinant of a \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
- \( a = 3, b = 5, c = 7 \)
- \( d = 2, e = -3, f = 1 \)
- \( g = 1, h = 1, i = 2 \)

Calculating the determinant:
\[ \text{det}(A) = 3((-3)(2) - (1)(1)) - 5((2)(2) - (1)(1)) + 7((2)(1) - (-3)(1)) \]
\[ = 3(-6 - 1) - 5(4 - 1) + 7(2 + 3) \]
\[ = 3(-7) - 5(3) + 7(5) \]
\[ = -21 - 15 + 35 = -1 \]

### Step 2: Check if the Inverse Exists
Since the determinant \( \text{det}(A) = -1 \) is not equal to zero, the inverse of \( A \) exists.

### Step 3: Find the Cofactor Matrix
The cofactor matrix \( C \) is calculated by finding the cofactors for each element of \( A \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = (-1)^{1+1} \cdot \text{det}\begin{pmatrix} -3 & 1 \\ 1 & 2 \end{pmatrix} = 1((-3)(2) - (1)(1)) = -6 - 1 = -7 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = (-1)^{1+2} \cdot \text{det}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = -1((2)(2) - (1)(1)) = -3 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = (-1)^{1+3} \cdot \text{det}\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix} = 1((2)(1) - (-3)(1)) = 2 + 3 = 5 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = (-1)^{2+1} \cdot \text{det}\begin{pmatrix} 5 & 7 \\ 1 & 2 \end{pmatrix} = -1((5)(2) - (7)(1)) = -3 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = (-1)^{2+2} \cdot \text{det}\begin{pmatrix} 3 & 7 \\ 1 & 2 \end{pmatrix} = 1((3)(2) - (7)(1)) = -1 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = (-1)^{2+3} \cdot \text{det}\begin{pmatrix} 3 & 5 \\ 1 & 1 \end{pmatrix} = -1((3)(1) - (5)(1)) = 2 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = (-1)^{3+1} \cdot \text{det}\begin{pmatrix} 5 & 7 \\ -3 & 1 \end{pmatrix} = 1((5)(1) - (7)(-3)) = 5 + 21 = 26 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = (-1)^{3+2} \cdot \text{det}\begin{pmatrix} 3 & 7 \\ 2 & 1 \end{pmatrix} = -1((3)(1) - (7)(2)) = -11 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = (-1)^{3+3} \cdot \text{det}\begin{pmatrix} 3 & 5 \\ 2 & -3 \end{pmatrix} = 1((3)(-3) - (5)(2)) = -9 - 10 = -19 \]

The cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} -7 & -3 & 5 \\ -3 & -1 & 2 \\ 26 & -11 & -19 \end{pmatrix} \]

### Step 4: Transpose the Cofactor Matrix to Get the Adjoint
The adjoint of \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{pmatrix} -7 & -3 & 26 \\ -3 & -1 & -11 \\ 5 & 2 & -19 \end{pmatrix} \]

### Step 5: Calculate the Inverse of A
Using the formula for the inverse:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} = -\text{adj}(A) \]
Thus,
\[ A^{-1} = -\begin{pmatrix} -7 & -3 & 26 \\ -3 & -1 & -11 \\ 5 & 2 & -19 \end{pmatrix} = \begin{pmatrix} 7 & 3 & -26 \\ 3 & 1 & 11 \\ -5 & -2 & 19 \end{pmatrix} \]

### Final Answer
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 7 & 3 & -26 \\ 3 & 1 & 11 \\ -5 & -2 & 19 \end{pmatrix} \]