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?

twinkletoe Aug 27, 2024

#2**0 **

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}

\]

bader Aug 27, 2024