If A = `[ (1,-1,1),(2,1,-3),(1,1,1)] " and " A^(-1) = (1)/(10) [(4,2,2),(-5,0,alpha),(1,-2,1)] ` then `alpha` is equal to

Options

A. `-2`
B. 5
C. 2
D. `-1`

Answer

To find the value of \( \alpha \) in the given matrices, we will follow the steps of calculating the determinant of matrix \( A \) and then using the formula for the inverse of a matrix.

### Step 1: Define the matrix \( A \)

The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \)

The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \( A \):
- \( a = 1, b = -1, c = 1 \)
- \( d = 2, e = 1, f = -3 \)
- \( g = 1, h = 1, i = 1 \)

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

### Step 3: Find the adjoint of matrix \( A \)

To find the adjoint, we need to calculate the cofactor matrix and then take its transpose.

#### Step 3.1: Calculate the minors and cofactors

1. **Minor \( M_{11} \)**:
\[ M_{11} = \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-3)(1) = 1 + 3 = 4 \]
**Cofactor \( C_{11} = M_{11} = 4 \)**

2. **Minor \( M_{12} \)**:
\[ M_{12} = \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-3)(1) = 2 + 3 = 5 \]
**Cofactor \( C_{12} = -M_{12} = -5 \)**

3. **Minor \( M_{13} \)**:
\[ M_{13} = \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1 \]
**Cofactor \( C_{13} = M_{13} = 1 \)**

4. **Minor \( M_{21} \)**:
\[ M_{21} = \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (1)(1) = -1 - 1 = -2 \]
**Cofactor \( C_{21} = -M_{21} = 2 \)**

5. **Minor \( M_{22} \)**:
\[ M_{22} = \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 1 - 1 = 0 \]
**Cofactor \( C_{22} = M_{22} = 0 \)**

6. **Minor \( M_{23} \)**:
\[ M_{23} = \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 1 + 1 = 2 \]
**Cofactor \( C_{23} = -M_{23} = -2 \)**

7. **Minor \( M_{31} \)**:
\[ M_{31} = \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (1)(1) = 3 - 1 = 2 \]
**Cofactor \( C_{31} = M_{31} = 2 \)**

8. **Minor \( M_{32} \)**:
\[ M_{32} = \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} = (1)(-3) - (1)(2) = -3 - 2 = -5 \]
**Cofactor \( C_{32} = -M_{32} = 5 \)**

9. **Minor \( M_{33} \)**:
\[ M_{33} = \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-1)(2) = 1 + 2 = 3 \]
**Cofactor \( C_{33} = M_{33} = 3 \)**

#### Step 3.2: Construct the cofactor matrix

The cofactor matrix is:
\[ \text{Cofactor Matrix} = \begin{pmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{pmatrix} \]

#### Step 3.3: Transpose the cofactor matrix to get the adjoint

The adjoint of \( A \) is:
\[ \text{adj}(A) = \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{pmatrix} \]

### Step 4: Calculate the inverse of matrix \( A \)

Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{10} \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{pmatrix} \]

### Step 5: Compare with the given \( A^{-1} \)

The given \( A^{-1} \) is:
\[ A^{-1} = \frac{1}{10} \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 1 \end{pmatrix} \]

By comparing the second row of both matrices:
\[ -5 = -5 \quad \text{and} \quad 0 = 0 \quad \text{and} \quad \alpha = 5 \]

### Conclusion

Thus, the value of \( \alpha \) is:
\[ \alpha = 5 \]