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`

Explanation

To solve the problem, we need to verify which of the given options holds true for the matrix \( A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \) and the identity matrix \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) for all \( n \geq 1 \) using mathematical induction.

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

We start by calculating \( A^2 \):

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

Calculating the product:

- First row, first column: \( 1 \cdot 1 + 0 \cdot 1 = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( 1 \cdot 1 + 1 \cdot 1 = 2 \)
- Second row, second column: \( 1 \cdot 0 + 1 \cdot 1 = 1 \)

Thus,

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

### 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 the product:

- First row, first column: \( 1 \cdot 1 + 0 \cdot 1 = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( 2 \cdot 1 + 1 \cdot 1 = 3 \)
- Second row, second column: \( 2 \cdot 0 + 1 \cdot 1 = 1 \)

Thus,

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

### Step 3: Identify the pattern

From our calculations, we observe that:

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

It appears that the second row's first element is increasing by 1 for each power of \( A \).

### Step 4: Generalize to \( A^n \)

We can generalize this pattern to:

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

### Step 5: Verify the options

Now we need to check which of the given options matches our result:

1. **Option (A)**: \( A^n = 2^{n-1} A + (n-1)I \)
2. **Option (B)**: \( A^n = nA + (n-1)I \)
3. **Option (C)**: \( A^n = 2^{n-1} A - (n-1)I \)
4. **Option (D)**: \( A^n = nA - (n-1)I \)

Calculating \( nA + (n-1)I \):

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

Adding these:

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

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

Calculating \( nA - (n-1)I \):

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

This matches our generalization for \( A^n \).

### Conclusion

Thus, the correct option is:

**(D) \( A^n = nA - (n-1)I \)**