Find the inverse of each of the following :
`[{:(1,2,3),(0,2,4),(0,0,5):}]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{pmatrix} \), we 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 \( A \):

- \( a = 1, b = 2, c = 3 \)
- \( d = 0, e = 2, f = 4 \)
- \( g = 0, h = 0, i = 5 \)

Calculating the determinant:

\[ \text{det}(A) = 1(2 \cdot 5 - 4 \cdot 0) - 2(0 \cdot 5 - 4 \cdot 0) + 3(0 \cdot 0 - 2 \cdot 0) \]
\[ = 1(10) - 2(0) + 3(0) = 10 \]

### Step 2: Check if the Determinant is Non-Zero

Since \( \text{det}(A) = 10 \neq 0 \), the inverse of the matrix exists.

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

The adjoint of a matrix is the transpose of the cofactor matrix. We need to calculate the cofactors for each element of matrix \( A \).

1. **Cofactor \( A_{11} \)**:
\[ A_{11} = \text{det} \begin{pmatrix} 2 & 4 \\ 0 & 5 \end{pmatrix} = (2 \cdot 5 - 4 \cdot 0) = 10 \]

2. **Cofactor \( A_{12} \)**:
\[ A_{12} = -\text{det} \begin{pmatrix} 0 & 4 \\ 0 & 5 \end{pmatrix} = -(0 \cdot 5 - 4 \cdot 0) = 0 \]

3. **Cofactor \( A_{13} \)**:
\[ A_{13} = \text{det} \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix} = (0 \cdot 0 - 2 \cdot 0) = 0 \]

4. **Cofactor \( A_{21} \)**:
\[ A_{21} = -\text{det} \begin{pmatrix} 2 & 4 \\ 0 & 5 \end{pmatrix} = -(2 \cdot 5 - 4 \cdot 0) = -10 \]

5. **Cofactor \( A_{22} \)**:
\[ A_{22} = \text{det} \begin{pmatrix} 1 & 3 \\ 0 & 5 \end{pmatrix} = (1 \cdot 5 - 3 \cdot 0) = 5 \]

6. **Cofactor \( A_{23} \)**:
\[ A_{23} = -\text{det} \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} = -(1 \cdot 0 - 2 \cdot 0) = 0 \]

7. **Cofactor \( A_{31} \)**:
\[ A_{31} = \text{det} \begin{pmatrix} 2 & 4 \\ 2 & 4 \end{pmatrix} = (2 \cdot 4 - 4 \cdot 2) = 0 \]

8. **Cofactor \( A_{32} \)**:
\[ A_{32} = -\text{det} \begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix} = -(1 \cdot 4 - 3 \cdot 0) = -4 \]

9. **Cofactor \( A_{33} \)**:
\[ A_{33} = \text{det} \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix} = (1 \cdot 2 - 2 \cdot 0) = 2 \]

Now, we can construct the cofactor matrix:

\[ \text{Cofactor Matrix} = \begin{pmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 0 & -4 & 2 \end{pmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{Adjoint}(A) = \begin{pmatrix} 10 & -10 & 0 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix} \]

### Step 4: Calculate the Inverse of Matrix A

Using the formula for the inverse:

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

Substituting the values we found:

\[ A^{-1} = \frac{1}{10} \begin{pmatrix} 10 & -10 & 0 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix} \]

This simplifies to:

\[ A^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{pmatrix} \]

### Final Answer

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

\[ A^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{pmatrix} \]