If `A=[(1,0),(1,1)] and I=[(1,0),(0,1)]` then which one of the following holds for all `nge1` by the principle of mathematica induction? (A) `A^n=2^(n-1) A+(n-1)I` (B) `A^n=nA+(n-1) I` (C) `A^n=2^(n-1) A-(n-1)I` (D) `A^n=nA-(n-1) AI`

Options

A. `A^n=n^(n-1)A+(n-1)I`
B. `A=nA+(n-1)I`
C. `A^n=2^(n-1)A-(n-1)I`
D. `A=nA-(n-1)I`

Explanation

To solve the problem, we need to verify which of the given options holds true for all \( n \geq 1 \) using the principle of mathematical induction. We start with the matrices:

\[ A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

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

To find \( A^2 \), we multiply \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \]

Calculating this gives:

\[ A^2 = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 1 & 1 \cdot 0 + 0 \cdot 1 \\ 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

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

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \]

Calculating this gives:

\[ A^3 = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 1 & 1 \cdot 0 + 0 \cdot 1 \\ 2 \cdot 1 + 1 \cdot 1 & 2 \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \]

### Step 3: Identify the Pattern

From our calculations, we can see a pattern emerging:

- \( A^1 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \)
- \( A^2 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \)
- \( A^3 = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \)

We can generalize this pattern as:

\[ A^n = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \]

### Step 4: Verify Each Option

Now we will check each option to see which one matches our derived formula.

1. **Option (A)**: \( A^n = 2^{n-1} A + (n-1)I \)

Substituting \( A \) and \( I \):

\[ 2^{n-1} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} + (n-1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^{n-1} + n - 1 & 0 \\ 2^{n-1} + n - 1 & 2^{n-1} \end{pmatrix} \]

This does not match \( A^n \).

2. **Option (B)**: \( A^n = nA + (n-1)I \)

Substituting:

\[ n \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} + (n-1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} n + n - 1 & 0 \\ n & n - 1 \end{pmatrix} \]

This does not match \( A^n \).

3. **Option (C)**: \( A^n = 2^{n-1} A - (n-1)I \)

Substituting:

\[ 2^{n-1} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} - (n-1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^{n-1} - (n - 1) & 0 \\ 2^{n-1} & 2^{n-1} - (n - 1) \end{pmatrix} \]

This does not match \( A^n \).

4. **Option (D)**: \( A^n = nA - (n-1)I \)

Substituting:

\[ n \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} - (n-1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} n - (n - 1) & 0 \\ n & n - 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \]

This matches \( A^n \).

### Conclusion

The correct option is **(D)**: \( A^n = nA - (n-1)I \).