`(i) if A=[{:(1,-1),(2,3):}],`then show that `A^(2)-4A+5I=O.`
` (ii) if f(x)=x^(2)+3x-5and A=[{:(2,-1),(4,3):}],`then find f(A).

Explanation

To solve the given problem, we will break it down into two parts as stated in the question.

### Part (i)
We need to show that \( A^2 - 4A + 5I = O \), where \( O \) is the null matrix of order \( 2 \times 2 \).

1. **Given Matrix A**:
\[ A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \]

2. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \]
\[ = \begin{pmatrix} (1 \cdot 1 + -1 \cdot 2) & (1 \cdot -1 + -1 \cdot 3) \\ (2 \cdot 1 + 3 \cdot 2) & (2 \cdot -1 + 3 \cdot 3) \end{pmatrix} \]
\[ = \begin{pmatrix} 1 - 2 & -1 - 3 \\ 2 + 6 & -2 + 9 \end{pmatrix} = \begin{pmatrix} -1 & -4 \\ 8 & 7 \end{pmatrix} \]

3. **Calculate \( 4A \)**:
\[ 4A = 4 \cdot \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & -4 \\ 8 & 12 \end{pmatrix} \]

4. **Calculate \( 5I \)** (where \( I \) is the identity matrix):
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \implies 5I = 5 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

5. **Combine the results**:
Now we need to compute \( A^2 - 4A + 5I \):
\[ A^2 - 4A + 5I = \begin{pmatrix} -1 & -4 \\ 8 & 7 \end{pmatrix} - \begin{pmatrix} 4 & -4 \\ 8 & 12 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]
\[ = \begin{pmatrix} -1 - 4 + 5 & -4 + 4 + 0 \\ 8 - 8 + 0 & 7 - 12 + 5 \end{pmatrix} \]
\[ = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O \]

Thus, we have shown that \( A^2 - 4A + 5I = O \).

### Part (ii)
We need to find \( f(A) \) where \( f(x) = x^2 + 3x - 5 \) and \( A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} \).

1. **Substituting A into f(x)**:
\[ f(A) = A^2 + 3A - 5I \]

2. **Calculate \( A^2 \)**:
\[ A^2 = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} \cdot \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} \]
\[ = \begin{pmatrix} (2 \cdot 2 + -1 \cdot 4) & (2 \cdot -1 + -1 \cdot 3) \\ (4 \cdot 2 + 3 \cdot 4) & (4 \cdot -1 + 3 \cdot 3) \end{pmatrix} \]
\[ = \begin{pmatrix} 4 - 4 & -2 - 3 \\ 8 + 12 & -4 + 9 \end{pmatrix} = \begin{pmatrix} 0 & -5 \\ 20 & 5 \end{pmatrix} \]

3. **Calculate \( 3A \)**:
\[ 3A = 3 \cdot \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 6 & -3 \\ 12 & 9 \end{pmatrix} \]

4. **Calculate \( 5I \)**:
\[ 5I = 5 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

5. **Combine the results**:
Now we compute \( f(A) = A^2 + 3A - 5I \):
\[ f(A) = \begin{pmatrix} 0 & -5 \\ 20 & 5 \end{pmatrix} + \begin{pmatrix} 6 & -3 \\ 12 & 9 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]
\[ = \begin{pmatrix} 0 + 6 - 5 & -5 - 3 - 0 \\ 20 + 12 - 0 & 5 + 9 - 5 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & -8 \\ 32 & 9 \end{pmatrix} \]

Thus, \( f(A) = \begin{pmatrix} 1 & -8 \\ 32 & 9 \end{pmatrix} \).