Let's solve the problem step by step.
Given matrices:
\[ A = \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
### (i) Find AB
To find \( AB \), we multiply matrix \( A \) by matrix \( B \):
\[
AB = \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}
\]
Calculating the elements:
1. First row, first column: \( 0 \cdot 1 + 2 \cdot 3 = 0 + 6 = 6 \)
2. First row, second column: \( 0 \cdot -1 + 2 \cdot 2 = 0 + 4 = 4 \)
3. Second row, first column: \( 5 \cdot 1 + (-2) \cdot 3 = 5 - 6 = -1 \)
4. Second row, second column: \( 5 \cdot -1 + (-2) \cdot 2 = -5 - 4 = -9 \)
Thus,
\[
AB = \begin{pmatrix} 6 & 4 \\ -1 & -9 \end{pmatrix}
\]
### (ii) Find BA
To find \( BA \), we multiply matrix \( B \) by matrix \( A \):
\[
BA = \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix}
\]
Calculating the elements:
1. First row, first column: \( 1 \cdot 0 + (-1) \cdot 5 = 0 - 5 = -5 \)
2. First row, second column: \( 1 \cdot 2 + (-1) \cdot -2 = 2 + 2 = 4 \)
3. Second row, first column: \( 3 \cdot 0 + 2 \cdot 5 = 0 + 10 = 10 \)
4. Second row, second column: \( 3 \cdot 2 + 2 \cdot -2 = 6 - 4 = 2 \)
Thus,
\[
BA = \begin{pmatrix} -5 & 4 \\ 10 & 2 \end{pmatrix}
\]
### (iii) Find AI
To find \( AI \), we multiply matrix \( A \) by the identity matrix \( I \):
\[
AI = \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
Calculating the elements:
1. First row, first column: \( 0 \cdot 1 + 2 \cdot 0 = 0 + 0 = 0 \)
2. First row, second column: \( 0 \cdot 0 + 2 \cdot 1 = 0 + 2 = 2 \)
3. Second row, first column: \( 5 \cdot 1 + (-2) \cdot 0 = 5 + 0 = 5 \)
4. Second row, second column: \( 5 \cdot 0 + (-2) \cdot 1 = 0 - 2 = -2 \)
Thus,
\[
AI = \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix}
\]
### (iv) Find \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:
\[
A^2 = \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix} \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix}
\]
Calculating the elements:
1. First row, first column: \( 0 \cdot 0 + 2 \cdot 5 = 0 + 10 = 10 \)
2. First row, second column: \( 0 \cdot 2 + 2 \cdot -2 = 0 - 4 = -4 \)
3. Second row, first column: \( 5 \cdot 0 + (-2) \cdot 5 = 0 - 10 = -10 \)
4. Second row, second column: \( 5 \cdot 2 + (-2) \cdot -2 = 10 + 4 = 14 \)
Thus,
\[
A^2 = \begin{pmatrix} 10 & -4 \\ -10 & 14 \end{pmatrix}
\]
### (v) Find \( B^2A \)
To find \( B^2 \), we first calculate \( B \times B \):
\[
B^2 = \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}
\]
Calculating the elements:
1. First row, first column: \( 1 \cdot 1 + (-1) \cdot 3 = 1 - 3 = -2 \)
2. First row, second column: \( 1 \cdot -1 + (-1) \cdot 2 = -1 - 2 = -3 \)
3. Second row, first column: \( 3 \cdot 1 + 2 \cdot 3 = 3 + 6 = 9 \)
4. Second row, second column: \( 3 \cdot -1 + 2 \cdot 2 = -3 + 4 = 1 \)
Thus,
\[
B^2 = \begin{pmatrix} -2 & -3 \\ 9 & 1 \end{pmatrix}
\]
Now, we multiply \( B^2 \) by \( A \):
\[
B^2A = \begin{pmatrix} -2 & -3 \\ 9 & 1 \end{pmatrix} \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix}
\]
Calculating the elements:
1. First row, first column: \( -2 \cdot 0 + -3 \cdot 5 = 0 - 15 = -15 \)
2. First row, second column: \( -2 \cdot 2 + -3 \cdot -2 = -4 + 6 = 2 \)
3. Second row, first column: \( 9 \cdot 0 + 1 \cdot 5 = 0 + 5 = 5 \)
4. Second row, second column: \( 9 \cdot 2 + 1 \cdot -2 = 18 - 2 = 16 \)
Thus,
\[
B^2A = \begin{pmatrix} -15 & 2 \\ 5 & 16 \end{pmatrix}
\]
### Summary of Results:
1. \( AB = \begin{pmatrix} 6 & 4 \\ -1 & -9 \end{pmatrix} \)
2. \( BA = \begin{pmatrix} -5 & 4 \\ 10 & 2 \end{pmatrix} \)
3. \( AI = \begin{pmatrix} 0 & 2 \\ 5 & -2 \end{pmatrix} \)
4. \( A^2 = \begin{pmatrix} 10 & -4 \\ -10 & 14 \end{pmatrix} \)
5. \( B^2A = \begin{pmatrix} -15 & 2 \\ 5 & 16 \end{pmatrix} \)
