Question

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1 + kx2 = h

Answer 1

Answer:

a) k=1, h=1, the unique solution of the system is  $(x_1,x_2)=(\frac{1}{5},\frac{7}{5})$

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

c)If k=6 and h=3 the system has no solutions.

Step-by-step explanation:

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=1 and h=1, the unique solution of the system is $x_2=\frac{1-8}{1-6}=\frac{-7}{-5}=\frac{7}{5}$ and solviong for $x_1$,

$x_1+3x_2=4\\\\x_1=4-3(\frac{7}{5})=\frac{1}{5}$

Then the solution is $(x_1,x_2)=(\frac{1}{5},\frac{7}{5})$

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=3, the system has no solutions because the last equation of the system of the echelon form of the matrix is $0x_2=-5$