Question

If the system of equations $kx + y + 2z = 1$, $3x - y - 2z = 2$, $-2x - 2y - 4z = 3$ has infinitely many solutions, then k is equal to ________

Hint:

$\Delta = 0$ is a necessary but not sufficient condition for infinite solutions; the system must also be consistent ($\Delta_x = \Delta_y = \Delta_z = 0$).

Answer 1

Correct Answer: 21

Step 1: For infinitely many solutions, the determinant $\Delta = 0$.
Step 2: $\Delta = \begin{vmatrix} k & 1 & 2
3 & -1 & -2
-2 & -2 & -4 \end{vmatrix} = k(4-4) - 1(-12-4) + 2(-6-2) = 0 + 16 - 16 = 0$.
Step 3: $\Delta$ is 0 for all $k$. However, we must check $\Delta_x, \Delta_y, \Delta_z$.
Step 4: Observe Eq 2 and Eq 3: $3x - (y+2z) = 2$ and $-2x - 2(y+2z) = 3$.
Step 5: Let $u = y+2z$. Then $3x - u = 2$ and $-2x - 2u = 3$. These two lines in $(x, u)$ have a unique solution.
Step 6: For the system to have infinitely many solutions, the first equation $kx + u = 1$ must be dependent. Calculating consistent $\Delta_i$ shows the system is inconsistent (0 solutions) for these constants. \itextbf{*Note: In competitive exams, if the question assumes they exist, k is often found by making rows proportional, but here the constants 1, 2, 3 prevent consistency.*}