Q58 / 160 · TS EAMCET 2013 · Mathematics
The number of real values of $t$ such that the system of homogeneous equations
$$
\begin{aligned}
t x+(t+1) y+(t-1) z & =0 \\
(t+1) x+t y+(t+2) z & =0 \\
(t-1) x+(t+2) y+t z & =0
\end{aligned}
$$
has non-trivial solutions is
- A$3$
- B$2$
- C$1$
- Correct answerDNone of these
Solution
Given,
$$
\begin{aligned}
& t x+(t+1) y+(t-1) z=0 \\
& (t+1) x+t y+(t+2) z=0 \\
& (t-1) x+(t+2) y+t z=0
\end{aligned}
$$
Here,
Coefficient matrix, $A=\left[\begin{array}{ccc}t & t+1 & t-1 \\ t+1 & t & t+2 \\ t-1 & t+2 & t\end{array}\right]$ If $|A|=0$, then system of equations has non-trivial solution and it has infinite solutions.
$$
|A|=\left|\begin{array}{ccc}
t & t+1 & t-1 \\
t+1 & t & t+2 \\
t-1 & t+2 & t
\end{array}\right|=0
$$
Apply operation $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$, we get
$$
=\left|\begin{array}{ccc}
t & t+1 & t-1 \\
1 & -1 & 3 \\
-1 & 1 & 1
\end{array}\right|=0
$$
Apply $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1$, we get
$$
\left|\begin{array}{crr}
t & 1 & -1 \\
1 & -2 & 2 \\
-1 & 2 & 2
\end{array}\right|=0
$$
Expanding along $R_1$, we get
$$
\begin{array}{rlrl}
& & t(-4-4)-1(2+2)-1(2-2)=0 \\
& \Rightarrow & & -8 t-4=0 \\
\therefore & & t=\frac{-1}{2}
\end{array}
$$
So, no option is correct.
Hence, the correct answer is option: 'D'.