The element of second row and third column in the inverse of `[[1, 2, -3], [0, 1, 2], [0, 0, 1]]` is

Options

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

Answer

To find the element of the second row and third column in the inverse of the matrix \( A = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \), 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 \( A \):

\[ A = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \]

We can expand the determinant along the first column:

\[ \text{det}(A) = 1 \cdot \text{det}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} - 0 + 0 \]

Calculating the determinant of the 2x2 matrix:

\[ \text{det}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = (1)(1) - (2)(0) = 1 \]

Thus,

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

### Step 2: Find the Adjoint of Matrix \( A \)

The adjoint of a matrix is the transpose of its cofactor matrix. We need to find the cofactor matrix first.

The cofactor \( C_{ij} \) is calculated by taking the determinant of the submatrix formed by deleting the \( i \)-th row and \( j \)-th column, multiplied by \( (-1)^{i+j} \).

Calculating the cofactors:

- \( C_{11} = \text{det}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = 1 \)
- \( C_{12} = -\text{det}\begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix} = 0 \)
- \( C_{13} = \text{det}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = 0 \)

- \( C_{21} = -\text{det}\begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix} = -2 \)
- \( C_{22} = \text{det}\begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} = 1 \)
- \( C_{23} = -\text{det}\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} = 0 \)

- \( C_{31} = \text{det}\begin{bmatrix} 2 & -3 \\ 1 & 2 \end{bmatrix} = 4 \)
- \( C_{32} = -\text{det}\begin{bmatrix} 1 & -3 \\ 1 & 2 \end{bmatrix} = 1 \)
- \( C_{33} = \text{det}\begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} = 0 \)

So the cofactor matrix \( C \) is:

\[ C = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 4 & -1 & 0 \end{bmatrix} \]

Now, the adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:

\[ \text{adj}(A) = \begin{bmatrix} 1 & -2 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \]

### Step 3: Calculate the Inverse of Matrix \( A \)

Since \( \text{det}(A) = 1 \), we have:

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

Thus,

\[ A^{-1} = \begin{bmatrix} 1 & -2 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \]

### Step 4: Find the Element of the Second Row and Third Column in \( A^{-1} \)

The element in the second row and third column of \( A^{-1} \) is:

\[ A^{-1}_{23} = -1 \]

### Final Answer

The element of the second row and third column in the inverse of the matrix is \( -1 \).

---