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

Options

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

Answer

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

### Step 1: Calculate the Determinant of Matrix A
The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \]
We can expand the determinant along the first row:
\[ \text{det}(A) = 1 \cdot \text{det}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - 2 \cdot \text{det}\begin{bmatrix} 2 & 0 \\ -1 & 1 \end{bmatrix} + 1 \cdot \text{det}\begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix} \]

Calculating the determinants of the \( 2 \times 2 \) matrices:
1. \( \text{det}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 1 \)
2. \( \text{det}\begin{bmatrix} 2 & 0 \\ -1 & 1 \end{bmatrix} = 2 \cdot 1 - 0 \cdot (-1) = 2 \)
3. \( \text{det}\begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix} = 2 \cdot 0 - 1 \cdot (-1) = 1 \)

Now substituting back:
\[ \text{det}(A) = 1 \cdot 1 - 2 \cdot 2 + 1 \cdot 1 = 1 - 4 + 1 = -2 \]

### Step 2: Calculate the Cofactor Matrix
The cofactor \( C_{ij} \) is calculated as \( (-1)^{i+j} \) times the determinant of the submatrix formed by deleting the \( i \)-th row and \( j \)-th column.

We need to find \( C_{23} \) (the cofactor for the second row and third column):
\[ C_{23} = (-1)^{2+3} \cdot \text{det}\begin{bmatrix} 1 & 2 \\ -1 & 0 \end{bmatrix} \]
Calculating the determinant:
\[ \text{det}\begin{bmatrix} 1 & 2 \\ -1 & 0 \end{bmatrix} = 1 \cdot 0 - 2 \cdot (-1) = 2 \]
Thus,
\[ C_{23} = -2 \]

### Step 3: Form the Cofactor Matrix
The cofactor matrix \( C \) is formed as follows:
\[ C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \]
We have already calculated \( C_{23} = -2 \). We would need to calculate the other cofactors, but for our purpose, we only need \( C_{32} \) to find the element of the inverse matrix.

### Step 4: Calculate the Inverse Matrix
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Where \( \text{adj}(A) \) is the transpose of the cofactor matrix.

The element at the second row and third column of \( A^{-1} \) is given by:
\[ (A^{-1})_{23} = \frac{-C_{32}}{\text{det}(A)} \]
We need to calculate \( C_{32} \).

### Step 5: Calculate \( C_{32} \)
To find \( C_{32} \):
\[ C_{32} = (-1)^{3+2} \cdot \text{det}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \]
Calculating the determinant:
\[ \text{det}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = 1 \cdot 1 - 2 \cdot 2 = 1 - 4 = -3 \]
Thus,
\[ C_{32} = -(-3) = 3 \]

### Step 6: Final Calculation
Now substituting back into our formula for \( (A^{-1})_{23} \):
\[ (A^{-1})_{23} = \frac{-C_{32}}{\text{det}(A)} = \frac{-3}{-2} = \frac{3}{2} \]

### Conclusion
The element of the second row and third column in the inverse of the matrix \( A \) is:
\[ \frac{3}{2} \]