Respuesta :
Answer:
Step-by-step explanation:
Given
[tex]6x_1-9x_2=8[/tex]
[tex]9x_1+kx_2=-1[/tex]
The given system is [tex]AX=B[/tex] can be represented by
[tex]\begin{bmatrix}6 &-9 \\ 9 & k\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}8\\ -1\end{bmatrix}[/tex]
The given system is consistent when determinant of A is not equal to zero
[tex]|A|[/tex]
[tex]|A|=6k-(-81)=6k+81[/tex]
[tex]k\neq \frac{-27}{2}[/tex]
i.e. system is consistent for all value of k except [tex]k=\frac{-27}{2}[/tex]
[tex]R-\frac{-27}{2}[/tex]
Answer:
[tex]k\neq -\frac{27}{2}[/tex]
Step-by-step explanation:
We have been given a system of equations as [tex]6x_1-9x_2 = 8\\\\9x_1+kx_2 =-1[/tex]. We are asked to find the value of k such that the given system is consistent.
Let us consider two equations as:
[tex]a_1x+b_1y = c_1\\\\a_2x+b_2y =c_2[/tex]
For a system to be consistent the ratio of coefficient of x terms and y terms should not be equal that is:
[tex]\frac{a_1}{a_2}\neq \frac{b_1}{b_2}[/tex]
Upon substituting our given values, we will get:
[tex]\frac{6}{9}\neq \frac{-9}{k}[/tex]
[tex]6k\neq -81[/tex]
[tex]\frac{6k}{6}\neq -\frac{81}{6}[/tex]
[tex]k\neq -\frac{27}{2}[/tex]
Since the given system will be consistent for all value except [tex]-\frac{27}{2}[/tex], therefore, we can choose any values for k such as [tex]-2[/tex] or 2.
