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# HELP!!

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For a certain value of , the system:

x+y+3z=10

-4x+2y+5z=7

kx+z=3

has no solutions. What is this value of k?

Aug 27, 2024

### 1+0 Answers

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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:

$\boxed{-3}$

Aug 27, 2024