Given the elementary matrices $E_1, E_2, E_3,$ and $E_4$ from part (a), we can express the matrix $B$ in terms of the matrix $A$ using these matrices.

Concept

When you apply a sequence of row operations to a matrix $A$, the resulting matrix $B$ can be expressed as the product of the corresponding elementary matrices multiplied by $A$.

Expression for $B$

If we apply the row operations in the given sequence:

1. The first operation $1.25R_1 \rightarrow R_1$ corresponds to multiplying $A$ by $E_1$.
2. The second operation $R_3 + R_2 \rightarrow R_3$ corresponds to multiplying the result by $E_2$.
3. The third operation $R_2 \rightarrow R_2 - R_3$ corresponds to multiplying the result by $E_3$.
4. The final operation $0.8R_3 \rightarrow R_3$ corresponds to multiplying the result by $E_4$.

Thus, the matrix $B$ is obtained by applying all these operations sequentially:

$B = E_4 \cdot E_3 \cdot E_2 \cdot E_1 \cdot A$

Interpretation

• Matrix Multiplication Order: The order in which the matrices are multiplied is crucial. The first elementary matrix applied (corresponding to the first row operation) is multiplied directly by $A$. Subsequent matrices are multiplied to the left of the result.
• Resulting Matrix: This expression gives $B$ as the final matrix after all row operations have been performed on $A$.

Would you like to see how to compute $B$ using specific matrices $A$ and the elementary matrices we found? Here are some related questions you might explore:

1. How do you determine the product of multiple matrices like $E_4 \cdot E_3 \cdot E_2 \cdot E_1$?
2. What would happen if the order of multiplication of elementary matrices was changed?
3. Can you express $A$ in terms of $B$ using the inverse of the elementary matrices?
4. How can this method be extended to solve linear systems of equations?
5. What is the relationship between elementary matrices and the LU decomposition?

Tip: When working with matrix multiplication, always pay attention to the order, as matrix multiplication is not commutative (i.e., $AB \neq BA$ in general).