Question

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. 31+3x2= 4 2x1+kx2= h

Answer 1

Answer

a) k=7, h=9, the unique solution of the system is  $(x_1,x_2)=(1,1)$

b) If k=6 and h=8 the system has infinite solutions.

c)If k=6 and h=9 the system has no solutions.

Step-by-step explanation:

I am assuming that the system is x1+3x2=4; 2x1+kx2=h

The augmented matrix of the system is $\left[\begin{array}{ccc}1&3&4\\2&k&h\end{array}\right]$. If two times the row 1 is subtracted to row 2 we get the following matrix$\left[\begin{array}{ccc}1&3&4\\0&k-6&h-8\end{array}\right]$.

Then

a) If k=7 and h=9, the unique solution of the system is $x_2=\frac{9-8}{7-6}=\frac{1}{1}=1$ and solviong for $x_1$,

$x_1+3x_2=4\\\\x_1=4-3(1)=1$

Then the solution is $(x_1,x_2)=(1,1)$

b) If k=6 and h=8 the system has infinite solutions because the echelon form of the matrix has a free variable.

c)If k=6 and h=9 the system has no solutions because the last equation of the system of the echelon form of the matrix is $0x_2=1$