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.
- Row 1 stays the same.
- Row 2 can be modified by adding 4 times Row 1 to it:
\[ R_2 = R_2 + 4R_1 \ \Rightarrow -4 + 4(1) = 0, \ 2 + 4(1) = 6, \ 5 + 4(3) = 17, \ 7 + 4(10) = 47 \ \Rightarrow R_2 = [0, 6, 17 | 47] \]
- 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:
\[ \boxed{-3} \]