If `A=[{:(,4,-4),(,-3,3):}], B=[{:(,6,5),(,3,0):}] and C=[{:(,2,3),(,-1,-2):}]` show that AB=AC. Write the conclusion, if any, that you can draw from the result obtained above.

Answer

To show that \( AB = AC \), we will perform the matrix multiplication of \( A \) with \( B \) and \( A \) with \( C \) separately and then compare the results.

### Given Matrices:
- \( A = \begin{pmatrix} 4 & -4 \\ -3 & 3 \end{pmatrix} \)
- \( B = \begin{pmatrix} 6 & 5 \\ 3 & 0 \end{pmatrix} \)
- \( C = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} \)

### Step 1: Calculate \( AB \)

To find \( AB \), we use the formula for matrix multiplication:

\[ AB = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \]

Calculating each element of \( AB \):

1. **Element (1,1)**:
\[ 4 \cdot 6 + (-4) \cdot 3 = 24 - 12 = 12 \]

2. **Element (1,2)**:
\[ 4 \cdot 5 + (-4) \cdot 0 = 20 + 0 = 20 \]

3. **Element (2,1)**:
\[ -3 \cdot 6 + 3 \cdot 3 = -18 + 9 = -9 \]

4. **Element (2,2)**:
\[ -3 \cdot 5 + 3 \cdot 0 = -15 + 0 = -15 \]

So, we have:
\[ AB = \begin{pmatrix} 12 & 20 \\ -9 & -15 \end{pmatrix} \]

### Step 2: Calculate \( AC \)

Now, we will calculate \( AC \):

1. **Element (1,1)**:
\[ 4 \cdot 2 + (-4) \cdot (-1) = 8 + 4 = 12 \]

2. **Element (1,2)**:
\[ 4 \cdot 3 + (-4) \cdot (-2) = 12 + 8 = 20 \]

3. **Element (2,1)**:
\[ -3 \cdot 2 + 3 \cdot (-1) = -6 - 3 = -9 \]

4. **Element (2,2)**:
\[ -3 \cdot 3 + 3 \cdot (-2) = -9 - 6 = -15 \]

So, we have:
\[ AC = \begin{pmatrix} 12 & 20 \\ -9 & -15 \end{pmatrix} \]

### Step 3: Compare \( AB \) and \( AC \)

We found:
\[ AB = \begin{pmatrix} 12 & 20 \\ -9 & -15 \end{pmatrix}, \quad AC = \begin{pmatrix} 12 & 20 \\ -9 & -15 \end{pmatrix} \]

Since \( AB = AC \), we conclude that:

\[ AB = AC \]

### Conclusion

The result \( AB = AC \) indicates that the matrices \( B \) and \( C \) have a specific relationship such that when multiplied by the same matrix \( A \), they yield the same product. This can imply that \( B \) and \( C \) could be equivalent in terms of their effect when transformed by \( A \).