Find the inverse of each of the following :
`[{:(1,0,0),(3,3,0),(5,2,-1):}]`

Answer

To find the inverse of the given 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 \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) can be calculated using the formula:

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

For our matrix \( A \):

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

Calculating this gives:

\[ \text{det}(A) = 1 \cdot (-3) = -3 \]

### Step 2: Calculate the Cofactor Matrix

Next, we need to find the cofactor matrix of \( A \). The cofactor \( C_{ij} \) is calculated by taking the determinant of the \( 2 \times 2 \) matrix that remains after removing the \( i^{th} \) row and \( j^{th} \) column, and then applying a sign based on the position.

1. **Cofactor \( C_{11} \)** (remove 1st row, 1st column):
\[ C_{11} = \text{det}\begin{pmatrix} 3 & 0 \\ 2 & -1 \end{pmatrix} = (3)(-1) - (0)(2) = -3 \]

2. **Cofactor \( C_{12} \)** (remove 1st row, 2nd column):
\[ C_{12} = -\text{det}\begin{pmatrix} 3 & 0 \\ 5 & -1 \end{pmatrix} = -((3)(-1) - (0)(5)) = 3 \]

3. **Cofactor \( C_{13} \)** (remove 1st row, 3rd column):
\[ C_{13} = \text{det}\begin{pmatrix} 3 & 3 \\ 5 & 2 \end{pmatrix} = (3)(2) - (3)(5) = 6 - 15 = -9 \]

4. **Cofactor \( C_{21} \)** (remove 2nd row, 1st column):
\[ C_{21} = -\text{det}\begin{pmatrix} 0 & 0 \\ 2 & -1 \end{pmatrix} = -((0)(-1) - (0)(2)) = 0 \]

5. **Cofactor \( C_{22} \)** (remove 2nd row, 2nd column):
\[ C_{22} = \text{det}\begin{pmatrix} 1 & 0 \\ 5 & -1 \end{pmatrix} = (1)(-1) - (0)(5) = -1 \]

6. **Cofactor \( C_{23} \)** (remove 2nd row, 3rd column):
\[ C_{23} = -\text{det}\begin{pmatrix} 1 & 0 \\ 5 & 2 \end{pmatrix} = -((1)(2) - (0)(5)) = -2 \]

7. **Cofactor \( C_{31} \)** (remove 3rd row, 1st column):
\[ C_{31} = \text{det}\begin{pmatrix} 0 & 0 \\ 3 & 0 \end{pmatrix} = (0)(0) - (0)(3) = 0 \]

8. **Cofactor \( C_{32} \)** (remove 3rd row, 2nd column):
\[ C_{32} = -\text{det}\begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix} = -((1)(0) - (0)(3)) = 0 \]

9. **Cofactor \( C_{33} \)** (remove 3rd row, 3rd column):
\[ C_{33} = \text{det}\begin{pmatrix} 1 & 0 \\ 3 & 3 \end{pmatrix} = (1)(3) - (0)(3) = 3 \]

The cofactor matrix is:

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

### Step 3: Transpose the Cofactor Matrix to Get the Adjoint

The adjoint of matrix \( A \) is the transpose of the cofactor matrix:

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

### Step 4: 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:

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

Calculating this gives:

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

### Final Answer

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

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

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