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

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

First, we calculate \(A^2\):

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \]

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 2: Express \(A^2\) in terms of \(A\)

We can see that:

\[ A^2 = 4A \]

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

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

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 4 & 4 \\ 12 & 12 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \(4 \cdot 1 + 4 \cdot 3 = 4 + 12 = 16\)
- First row, second column: \(4 \cdot 1 + 4 \cdot 3 = 4 + 12 = 16\)
- Second row, first column: \(12 \cdot 1 + 12 \cdot 3 = 12 + 36 = 48\)
- Second row, second column: \(12 \cdot 1 + 12 \cdot 3 = 12 + 36 = 48\)

Thus,

\[ A^3 = \begin{pmatrix} 16 & 16 \\ 48 & 48 \end{pmatrix} \]

### Step 4: Express \(A^3\) in terms of \(A\)

We can see that:

\[ A^3 = 16A \]

### Step 5: Calculate \(A^4\)

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

\[ A^4 = A^3 \cdot A = \begin{pmatrix} 16 & 16 \\ 48 & 48 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \(16 \cdot 1 + 16 \cdot 3 = 16 + 48 = 64\)
- First row, second column: \(16 \cdot 1 + 16 \cdot 3 = 16 + 48 = 64\)
- Second row, first column: \(48 \cdot 1 + 48 \cdot 3 = 48 + 144 = 192\)
- Second row, second column: \(48 \cdot 1 + 48 \cdot 3 = 48 + 144 = 192\)

Thus,

\[ A^4 = \begin{pmatrix} 64 & 64 \\ 192 & 192 \end{pmatrix} \]

### Step 6: Express \(A^4\) in terms of \(A\)

We can see that:

\[ A^4 = 64A \]

### Step 7: 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 8: Find \(\lambda\)

Since \(B = \lambda A\), we have:

\[ 85A = \lambda A \]

Comparing both sides, we find:

\[ \lambda = 85 \]

### Final Answer

Thus, the value of \(\lambda\) is:

\[ \boxed{85} \]