The inverse of the matrix `[(1,0,0),(3,3,0),(5,2,-1)]` is

Options

A. `-(1)/(3)[(-3,0,0),(3,1,0),(9,2,-3)]`
B. `-(1)/(3)[(-3,0,0),(3,-1,0),(-9,-2,3)]`
C. `-(1)/(3)[(3,0,0),(3,-1,0),(-9,-2,3)]`
D. `-(1)/(3)[(-3,0,0),(-3,-1,0),(-9,-2,3)]`

Explanation

To find the inverse of the matrix \( A = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A

The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

- \( a = 1, b = 0, c = 0 \)
- \( d = 3, e = 3, f = 0 \)
- \( g = 5, h = 2, i = -1 \)

Calculating the determinant:

\[ \text{det}(A) = 1 \cdot (3 \cdot (-1) - 0 \cdot 2) - 0 \cdot (3 \cdot (-1) - 0 \cdot 5) + 0 \cdot (3 \cdot 2 - 3 \cdot 5) \]

\[ = 1 \cdot (-3) - 0 + 0 = -3 \]

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

The adjoint of a matrix is found by calculating the cofactor matrix and then taking its transpose.

#### Step 2.1: Calculate the Cofactor Matrix

The cofactor \( C_{ij} \) of an element \( a_{ij} \) is given by:

\[ C_{ij} = (-1)^{i+j} \cdot M_{ij} \]

where \( M_{ij} \) is the determinant of the matrix obtained by deleting the \( i \)-th row and \( j \)-th column.

- **Cofactor \( C_{11} \)**:
\[ M_{11} = \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = 3 \cdot (-1) - 0 \cdot 2 = -3 \quad \Rightarrow \quad C_{11} = -3 \]

- **Cofactor \( C_{12} \)**:
\[ M_{12} = \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} = 3 \cdot (-1) - 0 \cdot 5 = -3 \quad \Rightarrow \quad C_{12} = 3 \]

- **Cofactor \( C_{13} \)**:
\[ M_{13} = \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} = 3 \cdot 2 - 3 \cdot 5 = 6 - 15 = -9 \quad \Rightarrow \quad C_{13} = -9 \]

- **Cofactor \( C_{21} \)**:
\[ M_{21} = \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = 0 \cdot (-1) - 0 \cdot 2 = 0 \quad \Rightarrow \quad C_{21} = 0 \]

- **Cofactor \( C_{22} \)**:
\[ M_{22} = \begin{vmatrix} 1 & 0 \\ 5 & -1 \end{vmatrix} = 1 \cdot (-1) - 0 \cdot 5 = -1 \quad \Rightarrow \quad C_{22} = -1 \]

- **Cofactor \( C_{23} \)**:
\[ M_{23} = \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = 1 \cdot 2 - 0 \cdot 5 = 2 \quad \Rightarrow \quad C_{23} = -2 \]

- **Cofactor \( C_{31} \)**:
\[ M_{31} = \begin{vmatrix} 0 & 0 \\ 3 & 0 \end{vmatrix} = 0 \quad \Rightarrow \quad C_{31} = 0 \]

- **Cofactor \( C_{32} \)**:
\[ M_{32} = \begin{vmatrix} 1 & 0 \\ 3 & 0 \end{vmatrix} = 0 \quad \Rightarrow \quad C_{32} = 0 \]

- **Cofactor \( C_{33} \)**:
\[ M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix} = 1 \cdot 3 - 0 \cdot 3 = 3 \quad \Rightarrow \quad C_{33} = 3 \]

Now, the cofactor matrix is:

\[ \text{Cofactor Matrix} = \begin{pmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{pmatrix} \]

#### Step 2.2: Transpose the Cofactor Matrix to Get the Adjoint

\[ \text{Adjoint}(A) = \begin{pmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{pmatrix} \]

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

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

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

Substituting the values:

\[ A^{-1} = \frac{1}{-3} \cdot \begin{pmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{pmatrix} \]

\[ = \begin{pmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{pmatrix} \]

### Final Answer

Thus, the inverse of the matrix \( A \) is:

\[ A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{pmatrix} \]