Let `A=[(1,0),(1,1)]` . Then which of the following is not true ?

Options

A. `lim_(n to oo)1/n^2 A^(-n) =[(0,0),(-1,0)]`
B. `lim_(n to oo)1/nA^(-n)=[(0,0),(-1,0)]`
C. `A^(-n)=[(1,0),(-n,1)]AA n ne N`
D. None of these

Answer

To solve the problem, we need to analyze the matrix \( A \) and its properties, particularly focusing on the inverse and powers of the matrix.

Given:
\[ A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \]

### Step 1: Find the Determinant of \( A \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
\[ \text{det}(A) = (1)(1) - (0)(1) = 1 \]

### Step 2: Find the Adjoint of \( A \)

The adjoint of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
For our matrix \( A \):
\[ \text{adj}(A) = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \]

### Step 3: Find the Inverse of \( A \)

The inverse of a matrix \( A \) is given by:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]
Since \( \text{det}(A) = 1 \):
\[ A^{-1} = \text{adj}(A) = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \]

### Step 4: Find \( A^{-n} \)

To find \( A^{-n} \), we can observe the pattern. We already have:
\[ A^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \]
Calculating \( A^{-2} \):
\[ A^{-2} = A^{-1} \cdot A^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \]
Calculating this 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,
\[ A^{-2} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} \]

Continuing this pattern, we can deduce:
\[ A^{-n} = \begin{pmatrix} 1 & 0 \\ -n & 1 \end{pmatrix} \]

### Step 5: Analyze the Given Options

1. **Option A**: \( \lim_{n \to \infty} \frac{1}{n^2} A^{-n} \)
\[ = \lim_{n \to \infty} \frac{1}{n^2} \begin{pmatrix} 1 & 0 \\ -n & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]
This is true.

2. **Option B**: \( \lim_{n \to \infty} \frac{1}{n} A^{-n} \)
\[ = \lim_{n \to \infty} \frac{1}{n} \begin{pmatrix} 1 & 0 \\ -n & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} \]
This is true.

3. **Option C**: The value of \( A^{-n} \) is given as \( \begin{pmatrix} 1 & 0 \\ -n & 1 \end{pmatrix} \), which we have derived.
This is true.

### Conclusion

The only option that is not true is **Option A**, as it leads to a misunderstanding of the limit behavior.

### Final Answer
The option that is not true is **Option A**.