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The problem asks for the value of \( k \) that makes the given linear system, represented by the augmented matrix:

\[
\begin{bmatrix}
2 & 6 & 8 \\
1 & k & 4
\end{bmatrix}
\]

have infinitely many solutions.

For a system to have infinitely many solutions, the two rows of the augmented matrix must be linearly dependent. In other words, one row must be a scalar multiple of the other.

We begin by comparing the first and second rows. If the rows are multiples of each other, we should have:

\[
\text{First row} = c \times \text{Second row}
\]

This leads to the following proportionalities:

1. \( 2 = c \times 1 \), so \( c = 2 \)
2. \( 6 = c \times k \), so \( 6 = 2k \), which gives \( k = 3 \)
3. Checking the constants in the augmented part: \( 8 = 2 \times 4 \), which is true.

Thus, for \( k = 3 \), the system has infinitely many solutions, as both rows become proportional, indicating that the equations are the same.

Do you want further details on this solution? Here are some related questions:

1. What does it mean for two rows of a matrix to be linearly dependent?
2. How do we know when a system has no solutions or a unique solution?
3. What role does row-reduction play in determining solution types?
4. Can we generalize this method to larger matrices?
5. What happens when the matrix is non-augmented? How does that affect the system?

**Tip:** Linear dependence between rows in a matrix implies that one equation is a multiple of another, leading to either infinitely many solutions or no solution depending on the constants.

For what value of k does the linear system represented by the augmented matrix have infinitely many solutions?

Find the value of k that makes the linear system represented by the augmented matrix have infinitely many solutions. This problem explores linear dependence, matrix theory, and conditions for infinitely many solutions. Suitable for students in Grades 10-12.

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