# 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

Verified

326.1k+ views

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.

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.

Last updated date: 01st Jun 2023

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