Let us consider the system
{ x 1 + 3 x 2 = 2 3 x 1 + h x 2 = k \begin{cases}
x_1 + 3x_2 = 2\\
3x_1+ hx_2 = k
\end{cases} { x 1 + 3 x 2 = 2 3 x 1 + h x 2 = k (*)
Let us find the determinant Δ = ∣ 1 3 3 h ∣ = h − 9 \Delta=\left|\begin{array}{cc}1 & 3 \\ 3 & h\end{array}\right|=h-9 Δ = ∣ ∣ 1 3 3 h ∣ ∣ = h − 9 .
(a) If Δ = 0 \Delta=0 Δ = 0 , then h = 9 h=9 h = 9 , and the system (*) is equivalent to the system { x 1 + 3 x 2 = 2 x 1 + 3 x 2 = k 3 \begin{cases}
x_1 + 3x_2 = 2\\
x_1+ 3x_2 = \frac{k}{3}
\end{cases} { x 1 + 3 x 2 = 2 x 1 + 3 x 2 = 3 k .
If k 3 ≠ 2 \frac{k}{3}\ne 2 3 k = 2 , that is k ≠ 6 k\ne 6 k = 6 , then the system (*) has no solution.
(b) If Δ ≠ 0 \Delta\ne 0 Δ = 0 , that is h ≠ 9 h\ne 9 h = 9 , then for any k k k the system (8) has a unique solution.
(c) If Δ = 0 \Delta=0 Δ = 0 , that is h = 9 h=9 h = 9 , and k = 6 k=6 k = 6 , then the system (*) is equivalent to the system
{ x 1 + 3 x 2 = 2 x 1 + 3 x 2 = 2 \begin{cases}
x_1 + 3x_2 = 2\\
x_1+ 3x_2 =2
\end{cases} { x 1 + 3 x 2 = 2 x 1 + 3 x 2 = 2 , and therefore, the system (*) has infinitely many solutions.
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