If `A=[(alpha,0),(1,1)],B=[(1,0),(5,1)]` whenever `A^(2)=B` then the value of `alpha`is

Options

A. 1
B. `-1`
C. 4
D. no real value of `alpha`

Answer

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

\[ A = \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}. \]

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

To find \( A^2 \), we multiply matrix \( A \) by itself:

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

Using matrix multiplication, we calculate each element:

- The element at the first row, first column:
\[ \alpha \cdot \alpha + 0 \cdot 1 = \alpha^2. \]

- The element at the first row, second column:
\[ \alpha \cdot 0 + 0 \cdot 1 = 0. \]

- The element at the second row, first column:
\[ 1 \cdot \alpha + 1 \cdot 1 = \alpha + 1. \]

- The element at the second row, second column:
\[ 1 \cdot 0 + 1 \cdot 1 = 1. \]

Thus, we have:

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

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

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

\[ \begin{pmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}. \]

### Step 3: Create equations from the equality of matrices

From the equality of the matrices, we can derive the following equations:

1. \( \alpha^2 = 1 \)
2. \( 0 = 0 \) (which is always true)
3. \( \alpha + 1 = 5 \)
4. \( 1 = 1 \) (which is always true)

### Step 4: Solve the equations

**From the first equation:**

\[ \alpha^2 = 1 \implies \alpha = 1 \text{ or } \alpha = -1. \]

**From the third equation:**

\[ \alpha + 1 = 5 \implies \alpha = 4. \]

### Step 5: Analyze the solutions

We have two potential solutions from the first equation (\( \alpha = 1 \) or \( \alpha = -1 \)) and one from the third equation (\( \alpha = 4 \)).

However, we need to check if these values satisfy both equations:

- For \( \alpha = 1 \):
- \( \alpha + 1 = 2 \neq 5 \) (not valid)

- For \( \alpha = -1 \):
- \( \alpha + 1 = 0 \neq 5 \) (not valid)

- For \( \alpha = 4 \):
- \( \alpha^2 = 16 \neq 1 \) (not valid)

### Conclusion

None of the values of \( \alpha \) satisfy both equations simultaneously. Therefore, there is no real value of \( \alpha \) that satisfies \( A^2 = B \).

The final answer is that there is **no real value of \( \alpha \)**.