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Let's solve the problem step by step.

The equation given is:
\[
\begin{pmatrix}
-2 & 4 \\
6 & -8
\end{pmatrix}
\begin{pmatrix}
w & x \\
y & z
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\]

This can be expanded into a system of equations by performing the matrix multiplication on the left-hand side:

1. For the first element (top-left) of the resulting matrix:
   \[
   -2w + 4y = 1
   \]

2. For the second element (top-right) of the resulting matrix:
   \[
   -2x + 4z = 0
   \]

3. For the third element (bottom-left) of the resulting matrix:
   \[
   6w - 8y = 0
   \]

4. For the fourth element (bottom-right) of the resulting matrix:
   \[
   6x - 8z = 1
   \]

Let's solve these equations:

### Solving Equations 3 and 1:
From Equation 3:
\[
6w - 8y = 0 \implies 3w = 4y \implies w = \frac{4}{3}y
\]
Substituting \(w = \frac{4}{3}y\) into Equation 1:
\[
-2\left(\frac{4}{3}y\right) + 4y = 1 \implies -\frac{8}{3}y + 4y = 1
\]
Multiply everything by 3 to eliminate the fraction:
\[
-8y + 12y = 3 \implies 4y = 3 \implies y = \frac{3}{4}
\]
Substitute \(y = \frac{3}{4}\) back to find \(w\):
\[
w = \frac{4}{3} \times \frac{3}{4} = 1
\]

### Solving Equations 4 and 2:
From Equation 4:
\[
6x - 8z = 1
\]
From Equation 2:
\[
-2x + 4z = 0 \implies x = 2z
\]
Substitute \(x = 2z\) into Equation 4:
\[
6(2z) - 8z = 1 \implies 12z - 8z = 1 \implies 4z = 1 \implies z = \frac{1}{4}
\]

The correct answer is not among the listed options, but the calculation gives \( z = \frac{1}{4} \).

Thus, the correct answer is **e. None of the options**.

Would you like more details or have any questions?

Here are some related questions:
1. How would the solution change if the right-hand side matrix were different?
2. Can you solve similar matrix equations involving larger matrices?
3. What does the determinant of the coefficient matrix tell us about the system of equations?
4. How can matrix inversion be used to solve systems of linear equations?
5. How would you interpret the solution in the context of linear transformations?

**Tip:** When solving systems of linear equations derived from matrices, always check if the matrix is invertible. This helps in understanding whether a unique solution exists.

If \(\begin{pmatrix}-2 & 4\\6 & -8\end{pmatrix}\begin{pmatrix}w & x\\y & z\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\), then z = ...

Explore how to solve for z in a matrix equation using matrix multiplication and linear equations. This step-by-step solution is ideal for students in Grades 9-12, covering key mathematical concepts including matrix multiplication and solving systems of linear equations.

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