If `A=[{:(alpha,0),(1,1):}]and B=[{:(1,0),(2,1):}]` such that `A^(2)=B`, then what is the value of `alpha` ?

Options

A. -1
B. 1
C. 2
D. 4

Explanation

To solve the problem, we need to find the value of \( \alpha \) such that \( A^2 = B \), where

\[ A = \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}. \]

### Step 1: Calculate \( A^2 \)

We need to compute \( A^2 = A \times A \):

\[ A^2 = \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix} \times \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix}. \]

### Step 2: Perform the multiplication

To find the elements of the resulting matrix, we will multiply the rows of the first matrix by the columns of the second matrix.

- **First row, first column**:
\[ \alpha \cdot \alpha + 0 \cdot 1 = \alpha^2 + 0 = \alpha^2. \]

- **First row, second column**:
\[ \alpha \cdot 0 + 0 \cdot 1 = 0 + 0 = 0. \]

- **Second row, first column**:
\[ 1 \cdot \alpha + 1 \cdot 1 = \alpha + 1. \]

- **Second row, second column**:
\[ 1 \cdot 0 + 1 \cdot 1 = 0 + 1 = 1. \]

Thus, we have:

\[ A^2 = \begin{pmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{pmatrix}. \]

### Step 3: Set \( A^2 \) equal to \( B \)

Now we set \( A^2 \) equal to \( B \):

\[ \begin{pmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}. \]

### Step 4: Compare corresponding elements

From the equality of the matrices, we can set up the following equations:

1. \( \alpha^2 = 1 \)
2. \( 0 = 0 \) (which is always true)
3. \( \alpha + 1 = 2 \)
4. \( 1 = 1 \) (which is always true)

### Step 5: Solve the equations

From the first equation \( \alpha^2 = 1 \):

\[ \alpha = 1 \quad \text{or} \quad \alpha = -1. \]

From the third equation \( \alpha + 1 = 2 \):

\[ \alpha = 2 - 1 = 1. \]

### Conclusion

The only value of \( \alpha \) that satisfies both equations is:

\[ \alpha = 1. \]

Thus, the value of \( \alpha \) is \( 1 \).

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