Question

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

Answer 1

Hint – 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.

Complete step-by-step answer:
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 – R1
       R3 -> R3 – 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) \\
    \\
$, Det A = a (ei - fh) - b (di - fg) + c (dh - eg)
⟹t (-1 x 1 – 3 x 1) – (t + 1) (1 x 1 – (3 x -1)) + (t -1) (1 x 1 – (-1 x -1)) = 0
⟹t (-1 -3) – (t + 1)(1 +3) + (t – 1)(1 – 1) = 0
⟹-4t -4t -4 = 0
⟹-8t – 4 = 0
⟹t =$ - \dfrac{1}{2}$.

‘t’ 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.