Let's solve the matrix products given in the image.

We have two parts, (a) and (b):

Part (a)

Matrix Multiplication: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix}$

The left matrix, $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, is the identity matrix. Multiplying any matrix by the identity matrix results in the matrix itself. Therefore: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix}$

Part (b)

Matrix Multiplication: $\begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Here, we are again multiplying by the identity matrix, but this time on the right side. Since multiplying by the identity matrix does not change the matrix, the result is: $\begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix}$

Final Answers:

• (a): $\begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix}$
• (b): $\begin{bmatrix} 5 & 0 \\ -4 & 8 \end{bmatrix}$

Would you like more detailed steps on how matrix multiplication works or have any questions?

Related Questions:

1. What is the identity matrix, and why does it leave other matrices unchanged when multiplied?
2. How would the result change if the order of multiplication was reversed for non-identity matrices?
3. Can you explain the concept of matrix transposition and its effect on matrix multiplication?
4. What are the properties of matrix multiplication that make it different from regular multiplication?
5. How does matrix multiplication apply in real-world applications such as computer graphics?

Tip:

Remember that the identity matrix acts as a neutral element in matrix multiplication, just like the number 1 in regular multiplication.