The inverse of the matrix `[[1, 2, 3], [0, 2, 4], [0, 0, 5]]` is

Options

A. `(1)/(5)[[5, -5, 1], [0, 5, -4], [0, 0, 2]]`
B. `(-1)/(5)[[5, -5, 1], [0, 5, -4], [0, 0, 2]]`
C. `(1)/(10)[[10, -10, 2], [0, 5, -4], [0, 0, 2]]`
D. `(-1)/(10)[[10, -10, 2], [0, 5, -4], [0, 0, 2]]`

Answer

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

### Step 1: Calculate the Determinant of Matrix A
The determinant of a 3x3 matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
\[ \text{det}(A) = 1 \cdot (2 \cdot 5 - 4 \cdot 0) - 2 \cdot (0 \cdot 5 - 4 \cdot 0) + 3 \cdot (0 \cdot 0 - 2 \cdot 0) \]
Calculating this gives:
\[ \text{det}(A) = 1 \cdot (10) - 2 \cdot (0) + 3 \cdot (0) = 10 \]

### Step 2: Check if the Determinant is Non-Zero
Since \( \text{det}(A) = 10 \) which is not equal to 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 will first calculate the cofactor matrix.

#### Step 3.1: Calculate the Cofactor Matrix
The cofactor \( C_{ij} \) of an element \( a_{ij} \) is given by:
\[ C_{ij} = (-1)^{i+j} \cdot \text{det}(M_{ij}) \]
where \( M_{ij} \) is the minor matrix obtained by deleting the \( i \)-th row and \( j \)-th column.

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

- **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{bmatrix} 0 & 4 \\ 0 & 5 \end{bmatrix} = -(0 \cdot 5 - 4 \cdot 0) = 0 \]

- **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det}\begin{bmatrix} 0 & 2 \\ 0 & 2 \end{bmatrix} = (0 \cdot 2 - 2 \cdot 0) = 0 \]

- **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det}\begin{bmatrix} 2 & 3 \\ 0 & 5 \end{bmatrix} = -(2 \cdot 5 - 3 \cdot 0) = -10 \]

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

- **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det}\begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix} = -(1 \cdot 2 - 2 \cdot 0) = -2 \]

- **Cofactor \( C_{31} \)**:
\[ C_{31} = \text{det}\begin{bmatrix} 2 & 3 \\ 2 & 4 \end{bmatrix} = (2 \cdot 4 - 3 \cdot 2) = 2 \]

- **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{bmatrix} 1 & 3 \\ 0 & 4 \end{bmatrix} = -(1 \cdot 4 - 3 \cdot 0) = -4 \]

- **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det}\begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix} = (1 \cdot 2 - 2 \cdot 0) = 2 \]

The cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} 10 & 0 & 0 \\ -10 & 5 & -2 \\ 2 & -4 & 2 \end{bmatrix} \]

#### Step 3.2: Transpose the Cofactor Matrix
The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & -2 & 2 \end{bmatrix} \]

### Step 4: Calculate the Inverse of Matrix A
Using the formula for the inverse:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values we have:
\[ A^{-1} = \frac{1}{10} \cdot \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & -2 & 2 \end{bmatrix} \]
This simplifies to:
\[ A^{-1} = \begin{bmatrix} 1 & -1 & 0.2 \\ 0 & 0.5 & -0.4 \\ 0 & -0.2 & 0.2 \end{bmatrix} \]

### Final Answer
The inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{bmatrix} 1 & -1 & 0.2 \\ 0 & 0.5 & -0.4 \\ 0 & -0.2 & 0.2 \end{bmatrix} \]