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 find the value of \( \lambda \) such that \( B = \lambda A \), where \( B = A + A^2 + A^3 + A^4 \) and \( A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \).

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

To find \( A^2 \), we multiply matrix \( A \) 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 2: Calculate \( A^3 \)

Next, we calculate \( A^3 \) using \( A^2 \):

\[ A^3 = A^2 \cdot A = (3A) \cdot A = 3A^2 \]

Since we already found \( A^2 = 3A \):

\[ A^3 = 3(3A) = 9A \]

### Step 3: Calculate \( A^4 \)

Now, we calculate \( A^4 \):

\[ A^4 = A^2 \cdot A^2 = (3A) \cdot (3A) = 9A^2 \]

Using \( A^2 = 3A \):

\[ A^4 = 9(3A) = 27A \]

### Step 4: Calculate \( B \)

Now we can find \( B \):

\[ B = A + A^2 + A^3 + A^4 = A + 3A + 9A + 27A \]

Combining these:

\[ B = (1 + 3 + 9 + 27)A = 40A \]

### Step 5: Find \( \lambda \)

Since we have \( B = 40A \), we can compare this with \( B = \lambda A \):

\[ \lambda = 40 \]

Thus, the value of \( \lambda \) is \( \boxed{40} \).