To find the products $AB$ and $BA$, let's proceed with matrix multiplication if the dimensions allow it.

Given matrices:

$A = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 5 \\ -2 & 1 \end{pmatrix}$

Dimensions

Matrix $A$ is a $2 \times 2$ matrix, and matrix $B$ is also a $2 \times 2$ matrix. Since both are $2 \times 2$, both products $AB$ and $BA$ are defined.

Calculating $AB$

$AB = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix} \begin{pmatrix} 2 & 5 \\ -2 & 1 \end{pmatrix}$

Calculating each element of $AB$:

1. $(AB)_{11} = 2 \cdot 2 + 5 \cdot (-2) = 4 - 10 = -6$
2. $(AB)_{12} = 2 \cdot 5 + 5 \cdot 1 = 10 + 5 = 15$
3. $(AB)_{21} = 4 \cdot 2 + 3 \cdot (-2) = 8 - 6 = 2$
4. $(AB)_{22} = 4 \cdot 5 + 3 \cdot 1 = 20 + 3 = 23$

So,

$AB = \begin{pmatrix} -6 & 15 \\ 2 & 23 \end{pmatrix}$

Calculating $BA$

$BA = \begin{pmatrix} 2 & 5 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}$

Calculating each element of $BA$:

1. $(BA)_{11} = 2 \cdot 2 + 5 \cdot 4 = 4 + 20 = 24$
2. $(BA)_{12} = 2 \cdot 5 + 5 \cdot 3 = 10 + 15 = 25$
3. $(BA)_{21} = -2 \cdot 2 + 1 \cdot 4 = -4 + 4 = 0$
4. $(BA)_{22} = -2 \cdot 5 + 1 \cdot 3 = -10 + 3 = -7$

Thus,

$BA = \begin{pmatrix} 24 & 25 \\ 0 & -7 \end{pmatrix}$

Final Results

$AB = \begin{pmatrix} -6 & 15 \\ 2 & 23 \end{pmatrix}, \quad BA = \begin{pmatrix} 24 & 25 \\ 0 & -7 \end{pmatrix}$

Would you like further details on any step, or do you have additional questions?

Further Questions

1. What are the properties of matrix multiplication that allow for these calculations?
2. How does the order of multiplication affect the product in matrix multiplication?
3. What are some applications of matrix multiplication in real-world problems?
4. Why are the results $AB$ and $BA$ different in this case?
5. Can matrix multiplication be used to solve systems of linear equations?

Tip

Remember, matrix multiplication is not commutative: $AB \neq BA$ in most cases.