If `A=[[1,1],[1,1]]` ,prove that `A^n=[[2^(n-1),2^(n-1)],[2^(n-1),2^(n-1)]]`, for all positive integers n.

Options

A. `2^nA`
B. `2^(n-1)A`
C. nA
D. none of these

Answer

To prove that \( A^n = \begin{bmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{bmatrix} \) for the matrix \( A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \), we will calculate the first few powers of \( A \) and observe a pattern.

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

\[ A^1 = A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]

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

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

Calculating the elements:

- First row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- First row, second column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- Second row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- Second row, second column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)

Thus,

\[ A^2 = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \]

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

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

Calculating the elements:

- First row, first column: \( 2 \cdot 1 + 2 \cdot 1 = 4 \)
- First row, second column: \( 2 \cdot 1 + 2 \cdot 1 = 4 \)
- Second row, first column: \( 2 \cdot 1 + 2 \cdot 1 = 4 \)
- Second row, second column: \( 2 \cdot 1 + 2 \cdot 1 = 4 \)

Thus,

\[ A^3 = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} \]

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

\[ A^4 = A^3 \cdot A = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]

Calculating the elements:

- First row, first column: \( 4 \cdot 1 + 4 \cdot 1 = 8 \)
- First row, second column: \( 4 \cdot 1 + 4 \cdot 1 = 8 \)
- Second row, first column: \( 4 \cdot 1 + 4 \cdot 1 = 8 \)
- Second row, second column: \( 4 \cdot 1 + 4 \cdot 1 = 8 \)

Thus,

\[ A^4 = \begin{bmatrix} 8 & 8 \\ 8 & 8 \end{bmatrix} \]

### Step 5: Observe the Pattern

From the calculations:

- \( A^1 = \begin{bmatrix} 2^{1-1} & 2^{1-1} \\ 2^{1-1} & 2^{1-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
- \( A^2 = \begin{bmatrix} 2^{2-1} & 2^{2-1} \\ 2^{2-1} & 2^{2-1} \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \)
- \( A^3 = \begin{bmatrix} 2^{3-1} & 2^{3-1} \\ 2^{3-1} & 2^{3-1} \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} \)
- \( A^4 = \begin{bmatrix} 2^{4-1} & 2^{4-1} \\ 2^{4-1} & 2^{4-1} \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 8 \end{bmatrix} \)

### Conclusion

We can see that the pattern holds true, and we can generalize this to say that for any positive integer \( n \):

\[ A^n = \begin{bmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{bmatrix} \]

Thus, we have proved that \( A^n = \begin{bmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{bmatrix} \).