If `A=[(alpha,0),(1,1)]` and `B=[(9,a),(b,c)]` and `A^(2)=B`, then the value of `a + b + c` is

Options

A. 1 or-1
B. 5 or -1
C. 5 or 1
D. no real values

Answer

To solve the problem, we need to find the value of \( a + b + c \) given the matrices \( A \) and \( B \) and the equation \( A^2 = B \).

### Step 1: Define the matrices
We have:
\[ A = \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 9 & a \\ b & c \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} \alpha & 0 \\ 1 & 1 \end{pmatrix} \]

Calculating the elements of \( A^2 \):
- The element at (1,1):
\[ \alpha \cdot \alpha + 0 \cdot 1 = \alpha^2 \]
- The element at (1,2):
\[ \alpha \cdot 0 + 0 \cdot 1 = 0 \]
- The element at (2,1):
\[ 1 \cdot \alpha + 1 \cdot 1 = \alpha + 1 \]
- The element at (2,2):
\[ 1 \cdot 0 + 1 \cdot 1 = 1 \]

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

### Step 3: Set \( A^2 \) equal to \( B \)
According to the problem, we have:
\[ A^2 = B \]
This gives us the equation:
\[ \begin{pmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{pmatrix} = \begin{pmatrix} 9 & a \\ b & c \end{pmatrix} \]

### Step 4: Set up equations from the matrix equality
From the equality of the matrices, we can derive the following equations:
1. \( \alpha^2 = 9 \)
2. \( 0 = a \)
3. \( \alpha + 1 = b \)
4. \( 1 = c \)

### Step 5: Solve for \( \alpha \), \( a \), \( b \), and \( c \)
From the first equation \( \alpha^2 = 9 \), we find:
\[ \alpha = 3 \quad \text{or} \quad \alpha = -3 \]

Using \( a = 0 \) from the second equation.

Now, substituting \( \alpha = 3 \) into the third equation:
\[ b = 3 + 1 = 4 \]
And from the fourth equation:
\[ c = 1 \]

Thus, we have:
- \( a = 0 \)
- \( b = 4 \)
- \( c = 1 \)

### Step 6: Calculate \( a + b + c \)
Now we can find:
\[ a + b + c = 0 + 4 + 1 = 5 \]

### Final Answer
The value of \( a + b + c \) is \( \boxed{5} \).