Question

# The number of real values of t such that the system of homogeneous equationtx + (t + 1) y + (t â€“ 1) z = 0(t + 1) x + ty + (t + 2) z = 0(t â€“ 1) x + (t + 2) y + tz = 0Has non â€“ trivial solutions, is${\text{A}}{\text{. 3}} \\ {\text{B}}{\text{. 2}} \\ {\text{C}}{\text{. 1}} \\ {\text{D}}{\text{. 4}} \\$

Hint â€“ To find the number of non-trivial solutions of given equations we write the set of equations in matrix form. Then find its determinant and equate it to 0.

For a non-trivial solution the determinant of the respective matrix = 0
$\Rightarrow \left( {\begin{array}{*{20}{c}} {\text{t}}&{{\text{t + 1}}}&{{\text{t - 1}}} \\ {{\text{t + 1}}}&{\text{t}}&{{\text{t + 2}}} \\ {{\text{t - 1}}}&{{\text{t + 2}}}&{\text{t}} \end{array}} \right)$ = 0

Now, we reduce the matrix using row operations
R2 -> R2 â€“ R1
R3 -> R3 â€“ R1
Which gives us,
$\left( {\begin{array}{*{20}{c}} {\text{t}}&{{\text{t + 1}}}&{{\text{t - 1}}} \\ {\text{1}}&{ - 1}&3 \\ { - 1}&1&1 \end{array}} \right) = 0$

For ${\text{A = }}\left( {\begin{array}{*{20}{c}} {\text{a}}&{\text{b}}&{\text{c}} \\ {\text{d}}&{\text{e}}&{\text{f}} \\ {\text{g}}&{\text{h}}&{\text{i}} \end{array}} \right) \\ \\$, Det A = a (ei - fh) - b (di - fg) + c (dh - eg)
âŸ¹t (-1 x 1 â€“ 3 x 1) â€“ (t + 1) (1 x 1 â€“ (3 x -1)) + (t -1) (1 x 1 â€“ (-1 x -1)) = 0
âŸ¹t (-1 -3) â€“ (t + 1)(1 +3) + (t â€“ 1)(1 â€“ 1) = 0
âŸ¹-4t -4t -4 = 0
âŸ¹-8t â€“ 4 = 0
âŸ¹t =$- \dfrac{1}{2}$.

â€˜tâ€™ has only one value for which the system has non- homogeneous equations and has non- trivial solutions. Hence Option C is the correct answer.

Note: The key point to solve such problems is to know that for a non-trivial solution the determinant of the matrix is zero.
A (n x n) homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions.