If `A=[(1,-1,1),(2,1,-3),(1,1,1)]` and `10 B=[(4,2,2),(-5,0,alpha),(1,-2,3)]` If B is the inverse of A then `alpha` is

Options

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

Answer

To find the value of \( \alpha \) such that matrix \( B \) is the inverse of matrix \( A \), we will follow these steps:

Given:
\[ A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix} \]
\[ 10B = \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} \]
This implies:
\[ B = \frac{1}{10} \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} \]

### Step 1: Calculate \( AB \)
To find \( \alpha \), we need to calculate the product \( AB \) and set it equal to the identity matrix \( I \).

\[ AB = A \cdot B = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix} \cdot \frac{1}{10} \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} \]

### Step 2: Compute the elements of \( AB \)

**Element (1,1):**
\[ 1 \cdot 4 + (-1) \cdot (-5) + 1 \cdot 1 = 4 + 5 + 1 = 10 \]

**Element (1,2):**
\[ 1 \cdot 2 + (-1) \cdot 0 + 1 \cdot (-2) = 2 + 0 - 2 = 0 \]

**Element (1,3):**
\[ 1 \cdot 2 + (-1) \cdot \alpha + 1 \cdot 3 = 2 - \alpha + 3 = 5 - \alpha \]

**Element (2,1):**
\[ 2 \cdot 4 + 1 \cdot (-5) + (-3) \cdot 1 = 8 - 5 - 3 = 0 \]

**Element (2,2):**
\[ 2 \cdot 2 + 1 \cdot 0 + (-3) \cdot (-2) = 4 + 0 + 6 = 10 \]

**Element (2,3):**
\[ 2 \cdot 2 + 1 \cdot \alpha + (-3) \cdot 3 = 4 + \alpha - 9 = \alpha - 5 \]

**Element (3,1):**
\[ 1 \cdot 4 + 1 \cdot (-5) + 1 \cdot 1 = 4 - 5 + 1 = 0 \]

**Element (3,2):**
\[ 1 \cdot 2 + 1 \cdot 0 + 1 \cdot (-2) = 2 + 0 - 2 = 0 \]

**Element (3,3):**
\[ 1 \cdot 2 + 1 \cdot \alpha + 1 \cdot 3 = 2 + \alpha + 3 = \alpha + 5 \]

### Step 3: Form the product matrix
Putting all these together, we have:
\[ AB = \frac{1}{10} \begin{pmatrix} 10 & 0 & 5 - \alpha \\ 0 & 10 & \alpha - 5 \\ 0 & 0 & \alpha + 5 \end{pmatrix} \]

### Step 4: Set \( AB \) equal to the identity matrix
For \( AB \) to equal the identity matrix \( I \), we need:
\[ AB = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This gives us the following equations:
1. \( \frac{10}{10} = 1 \) (true)
2. \( \frac{10}{10} = 1 \) (true)
3. \( \frac{5 - \alpha}{10} = 0 \) ⇒ \( 5 - \alpha = 0 \) ⇒ \( \alpha = 5 \)
4. \( \frac{\alpha - 5}{10} = 0 \) ⇒ \( \alpha - 5 = 0 \) ⇒ \( \alpha = 5 \)
5. \( \frac{\alpha + 5}{10} = 1 \) ⇒ \( \alpha + 5 = 10 \) ⇒ \( \alpha = 5 \)

### Conclusion
Thus, the value of \( \alpha \) is \( 5 \).