Question #147539
Choose h and k such that the system below has (a) no solution, (b) a unique solution, and (c)
many solutions.

x1 + 3x2 = 2

3x1+ hx2 = k.
1
Expert's answer
2020-12-01T03:01:31-0500

Let us consider the system


{x1+3x2=23x1+hx2=k\begin{cases} x_1 + 3x_2 = 2\\ 3x_1+ hx_2 = k \end{cases} (*)


Let us find the determinant Δ=133h=h9\Delta=\left|\begin{array}{cc}1 & 3 \\ 3 & h\end{array}\right|=h-9 .

(a) If Δ=0\Delta=0, then h=9h=9, and the system (*) is equivalent to the system {x1+3x2=2x1+3x2=k3\begin{cases} x_1 + 3x_2 = 2\\ x_1+ 3x_2 = \frac{k}{3} \end{cases} .

If k32\frac{k}{3}\ne 2, that is k6k\ne 6, then the system (*) has no solution.



(b) If Δ0\Delta\ne 0, that is h9h\ne 9, then for any kk the system (8) has a unique solution.



(c) If Δ=0\Delta=0, that is h=9h=9, and k=6k=6, then the system (*) is equivalent to the system


{x1+3x2=2x1+3x2=2\begin{cases} x_1 + 3x_2 = 2\\ x_1+ 3x_2 =2 \end{cases} , and therefore, the system (*) has infinitely many solutions.




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS