Answer:
a) The system has a unique solution for [tex]k\neq 6[/tex] and any value of [tex]h[/tex], and we say the system is consisted
b) The system has infinite solutions for [tex]k=6[/tex] and [tex]h=8[/tex]
c) The system has no solution for [tex]k=6[/tex] and [tex]h\neq 8[/tex]
Step-by-step explanation:
Since we need to base the solutions of the system on one of the independent terms ([tex]h[/tex]), the determinant method is not suitable and therefore we use the Gauss elimination method.
The first step is to write our system in the augmented matrix form:
[tex]\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right][/tex]
The we can use the transformation [tex]r_0\rightarrow r_0 -2r_1[/tex], obtaining:
[tex]\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right][/tex].
Now we can start the analysis:
[tex](k-6)x_2=h-8\\x_2=h-8/(k-6)[/tex]
from where we can see that only in the case of [tex]k=6[/tex] the value of [tex]x_2[/tex] can not be determined.
[tex](k-6)x_2=h-8\\0=0[/tex] which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get [tex]x_1[/tex] as a function of [tex]x_2[/tex] o viceversa. Thus, [tex]x_2[/tex] ([tex]x_1[/tex]) is called a parameter since there are no constraints on what values they can take on.
if [tex]k=6[/tex] and [tex]h\neq 8[/tex] the system has no solution. Again by substituting in the equation resulting from the last row:
[tex](k-6)x_2=h-8\\0=h-8[/tex] which is false for all values of [tex]h\neq 8[/tex] and since we have something that is not possible [tex](0\neq h-8,\ \forall \ h\neq 8)[/tex] the system has no solution
