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 the steps outlined below:

### Step 1: Calculate the Determinant of Matrix 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 \( A \):

\[ \text{det}(A) = 3((-3)(2) - (1)(1)) - 5((2)(2) - (1)(1)) + 7((2)(1) - (-3)(1)) \]

Calculating each term:

1. \( -3 \cdot 2 - 1 \cdot 1 = -6 - 1 = -7 \)
2. \( 2 \cdot 2 - 1 \cdot 1 = 4 - 1 = 3 \)
3. \( 2 \cdot 1 - (-3) \cdot 1 = 2 + 3 = 5 \)

Now substituting back into the determinant formula:

\[ \text{det}(A) = 3(-7) - 5(3) + 7(5) \]
\[ = -21 - 15 + 35 = -1 \]

### Step 2: Calculate the Adjoint of Matrix A

The adjoint of a matrix is the transpose of its cofactor matrix. We will calculate the cofactor for each element in \( A \).

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

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = -((2)(2) - (1)(1)) = -(4 - 1) = -3 \]

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

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det}\begin{pmatrix} 5 & 7 \\ 1 & 2 \end{pmatrix} = -((5)(2) - (7)(1)) = -(10 - 7) = -3 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det}\begin{pmatrix} 3 & 7 \\ 1 & 2 \end{pmatrix} = (3)(2) - (7)(1) = 6 - 7 = -1 \]

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

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

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{pmatrix} 3 & 7 \\ 2 & 1 \end{pmatrix} = -((3)(1) - (7)(2)) = -(3 - 14) = 11 \]

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

Now, the cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{pmatrix} -7 & -3 & 5 \\ -3 & -1 & 2 \\ 26 & 11 & -19 \end{pmatrix} \]

Taking the transpose to get the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} -7 & -3 & 26 \\ -3 & -1 & 11 \\ 5 & 2 & -19 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix A

The inverse of matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

Substituting the values we found:

\[ A^{-1} = \frac{1}{-1} \cdot \begin{pmatrix} -7 & -3 & 26 \\ -3 & -1 & 11 \\ 5 & 2 & -19 \end{pmatrix} \]
\[ = \begin{pmatrix} 7 & 3 & -26 \\ 3 & 1 & -11 \\ -5 & -2 & 19 \end{pmatrix} \]

### Final Answer

The inverse of the matrix \( A \) is:

\[ A^{-1} = \begin{pmatrix} 7 & 3 & -26 \\ 3 & 1 & -11 \\ -5 & -2 & 19 \end{pmatrix} \]