HELP!!

To determine the value of $k$ for which the system of equations has no solutions, we can represent the system in matrix form as follows:

\[
\begin{bmatrix}
1 & 1 & 3 \\
-4 & 2 & 5 \\
k & 0 & 1
\end{bmatrix}

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

=

$$\begin{bmatrix} 10 \\ 7 \\ 3 \end{bmatrix}$$

\]

To analyze the conditions under which the system has no solution, we can use the concept of the rank of the coefficient matrix compared to the augmented matrix. Specifically, a system will have no solutions if the rank of the coefficient matrix is less than the rank of the augmented matrix.

First, we set up the augmented matrix:

$$\left[\begin{array}{ccc|c} 1 & 1 & 3 & 10 \\ -4 & 2 & 5 & 7 \\ k & 0 & 1 & 3 \end{array}\right]$$

Next, we will perform row operations to bring this matrix to row echelon form.

1. **Row 1** stays the same.

2. **Row 2** can be modified by adding 4 times Row 1 to it:

$$R_2 = R_2 + 4R_1 \\ \Rightarrow -4 + 4(1) = 0, \\ 2 + 4(1) = 6, \\ 5 + 4(3) = 17, \\ 7 + 4(10) = 47 \\ \Rightarrow R_2 = [0, 6, 17 | 47]$$

3. Now, we will modify **Row 3**. To eliminate the first term in Row 3, we can do the following. Let's say we perform:

$$R_3 = R_3 - kR_1$$

This gives:

$$R_3 = [k - k, 0 - k, 1 - 3k | 3 - 10k] \\ \Rightarrow R_3 = [0, -k, 1 - 3k | 3 - 10k]$$

The new augmented matrix looks like this:

$$\left[\begin{array}{ccc|c} 1 & 1 & 3 & 10 \\ 0 & 6 & 17 & 47 \\ 0 & -k & 1 - 3k & 3 - 10k \end{array}\right]$$

Next, we need to see when the rank of the coefficient matrix differs from the rank of the augmented matrix. For that, the third row has to be a nontrivial linear combination of the first two rows such that it produces a contradiction.

From the second row of the system, we can conclude that, as long as $k \neq 0$, Row 3 provides a new pivot. Thus, we must examine what happens when $k = -3$:

If $k = -3$

$$R_3 = [0, 3, 10 | 27] \quad \text{(after substituting \( k = -3 \))}$$

Now, Row 3 should be checked against the other equations. If we solve with the last row resulting in a contradiction, we find inconsistent results leading to no solutions.

Thus, the value of $k$ which results in this system having no solutions is:

$$\boxed{-3}$$