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

Answer

To solve the problem, we need to compute the matrix powers of \( A \) and then sum them up to find \( B \). Let's go through the steps systematically.

### Step 1: Define the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \times A = \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \times \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 1 \cdot 3 = 1 + 3 = 4 \)
- First row, second column: \( 1 \cdot 1 + 1 \cdot 3 = 1 + 3 = 4 \)
- Second row, first column: \( 3 \cdot 1 + 3 \cdot 3 = 3 + 9 = 12 \)
- Second row, second column: \( 3 \cdot 1 + 3 \cdot 3 = 3 + 9 = 12 \)

Thus,
\[ A^2 = \begin{pmatrix} 4 & 4 \\ 12 & 12 \end{pmatrix} \]

### Step 3: Express \( A^2 \) in terms of \( A \)
Notice that:
\[ A^2 = 4A \]

### Step 4: Calculate \( A^3 \)
Now we calculate \( A^3 \) using \( A^2 \):
\[ A^3 = A^2 \times A = (4A) \times A = 4A^2 \]
Substituting \( A^2 \):
\[ A^3 = 4(4A) = 16A \]

### Step 5: Calculate \( A^4 \)
Next, we calculate \( A^4 \):
\[ A^4 = A^3 \times A = (16A) \times A = 16A^2 \]
Substituting \( A^2 \):
\[ A^4 = 16(4A) = 64A \]

### Step 6: Calculate \( B \)
Now we can calculate \( B \):
\[ B = A + A^2 + A^3 + A^4 \]
Substituting the values we found:
\[ B = A + 4A + 16A + 64A = (1 + 4 + 16 + 64)A = 85A \]

### Step 7: Relate \( B \) to \( \lambda A \)
We know from the problem statement that:
\[ B = \lambda A \]
From our calculation, we have:
\[ B = 85A \]
Thus, by comparing:
\[ \lambda A = 85A \]
This implies:
\[ \lambda = 85 \]

### Final Answer
The value of \( \lambda \) is:
\[ \lambda = 85 \]