Let the matrix `A=[(1,1),(2,2)]` and `B=A+A^(2)+A^(3)+A^(4)`. If `B=lambdaA, AA lambda in R`, then the vlaue of `lambda` is equal to

Answer

To solve the problem, we need to calculate the powers of the matrix \( A \) and then find the matrix \( B \) as defined in the question. Finally, we will compare \( B \) with \( \lambda A \) to find the value of \( \lambda \).

### Step 1: Define the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 1 \cdot 2 = 1 + 2 = 3 \)
- First row, second column: \( 1 \cdot 1 + 1 \cdot 2 = 1 + 2 = 3 \)
- Second row, first column: \( 2 \cdot 1 + 2 \cdot 2 = 2 + 4 = 6 \)
- Second row, second column: \( 2 \cdot 1 + 2 \cdot 2 = 2 + 4 = 6 \)

Thus,
\[ A^2 = \begin{pmatrix} 3 & 3 \\ 6 & 6 \end{pmatrix} = 3A \]

### Step 3: Calculate \( A^3 \)
To find \( A^3 \), we multiply \( A^2 \) by \( A \):
\[ A^3 = A^2 \cdot A = (3A) \cdot A = 3A^2 \]
Since \( A^2 = 3A \), we have:
\[ A^3 = 3(3A) = 9A \]

### Step 4: Calculate \( A^4 \)
To find \( A^4 \), we multiply \( A^3 \) by \( A \):
\[ A^4 = A^3 \cdot A = (9A) \cdot A = 9A^2 \]
Using \( A^2 = 3A \), we get:
\[ A^4 = 9(3A) = 27A \]

### Step 5: Calculate \( B \)
Now we can calculate \( B \):
\[ B = A + A^2 + A^3 + A^4 \]
Substituting the values we found:
\[ B = A + 3A + 9A + 27A = (1 + 3 + 9 + 27)A = 40A \]

### Step 6: Compare with \( \lambda A \)
We are given that \( B = \lambda A \). From our calculation:
\[ B = 40A \]
Thus, we can equate:
\[ \lambda A = 40A \]
This implies:
\[ \lambda = 40 \]

### Final Answer
The value of \( \lambda \) is \( \boxed{40} \).