Question

For what value(s) of k is the linear system consistent? (Enter your answers as a comma-separated list.)

Answer 1

Answer:

$k\neq -\frac{27}{2}$

Step-by-step explanation:

We have been given a system of equations as $6x_1-9x_2 = 8\\\\9x_1+kx_2 =-1$. We are asked to find the value of k such that the given system is consistent.

Let us consider two equations as:

$a_1x+b_1y = c_1\\\\a_2x+b_2y =c_2$

For a system to be consistent the ratio of coefficient of x terms and y terms should not be equal that is:

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$

Upon substituting our given values, we will get:

$\frac{6}{9}\neq \frac{-9}{k}$

$6k\neq -81$

$\frac{6k}{6}\neq -\frac{81}{6}$

$k\neq -\frac{27}{2}$

Since the given system will be consistent for all value except $-\frac{27}{2}$, therefore, we can choose any values for k such as $-2$ or 2.

Answer 2

Answer:

Step-by-step explanation:

Given

$6x_1-9x_2=8$

$9x_1+kx_2=-1$

The given system is $AX=B$ can be represented by

$\begin{bmatrix}6 &-9 \\ 9 & k\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}8\\ -1\end{bmatrix}$

The given system is consistent when determinant of A is not equal to zero

$|A|$

$|A|=6k-(-81)=6k+81$

$k\neq \frac{-27}{2}$

i.e. system is consistent for all value of k except $k=\frac{-27}{2}$

$R-\frac{-27}{2}$