The elements in the first row and third column of the inverse of the matrix `[(1,2,3),(0,1,2),(0,0,1)]` is

Options

A. `-2`
B. `0`
C. `7`
D. `1`

Explanation

To find the element in the first row and third column of the inverse of the matrix \( A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of the Matrix
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:
\[ \text{det}(A) = 1(1 \cdot 1 - 2 \cdot 0) - 2(0 \cdot 1 - 2 \cdot 0) + 3(0 \cdot 0 - 1 \cdot 0) = 1(1) - 2(0) + 3(0) = 1 \]

### Step 2: Check if the Inverse Exists
Since the determinant is \( 1 \) (which is not zero), the inverse of the matrix exists.

### Step 3: Find the Adjoint of the Matrix
To find the adjoint, we calculate the cofactor matrix and then transpose it. The cofactor \( C_{ij} \) for each element \( a_{ij} \) is given by:
\[ C_{ij} = (-1)^{i+j} \cdot M_{ij} \]
where \( M_{ij} \) is the determinant of the submatrix formed by removing the \( i \)-th row and \( j \)-th column.

Calculating the cofactors:
- For \( a_{11} = 1 \): \( C_{11} = 1 \) (det of \( \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \))
- For \( a_{12} = 2 \): \( C_{12} = -0 \) (det of \( \begin{pmatrix} 0 & 2 \\ 0 & 1 \end{pmatrix} \))
- For \( a_{13} = 3 \): \( C_{13} = 0 \) (det of \( \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \))

- For \( a_{21} = 0 \): \( C_{21} = 0 \) (det of \( \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} \))
- For \( a_{22} = 1 \): \( C_{22} = 1 \) (det of \( \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \))
- For \( a_{23} = 2 \): \( C_{23} = -2 \) (det of \( \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} \))

- For \( a_{31} = 0 \): \( C_{31} = 0 \) (det of \( \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \))
- For \( a_{32} = 0 \): \( C_{32} = 2 \) (det of \( \begin{pmatrix} 1 & 3 \\ 0 & 2 \end{pmatrix} \))
- For \( a_{33} = 1 \): \( C_{33} = 1 \) (det of \( \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \))

Thus, the cofactor matrix is:
\[ \text{Cof}(A) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 2 & 1 \end{pmatrix} \]

Now, we transpose this matrix to get the adjoint:
\[ \text{Adj}(A) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & -2 & 1 \end{pmatrix} \]

### Step 4: Calculate the Inverse of the Matrix
Using the formula for the inverse:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) \]
Since \( \text{det}(A) = 1 \):
\[ A^{-1} = \text{Adj}(A) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & -2 & 1 \end{pmatrix} \]

### Step 5: Find the Element in the First Row and Third Column
The element in the first row and third column of \( A^{-1} \) is:
\[ A^{-1}_{13} = 0 \]

### Final Answer
The element in the first row and third column of the inverse of the matrix is \( 0 \).

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