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If $p$ and $q$ are two distinct real values of $\lambda$ for which the system of equations
$\begin{aligned}
(\lambda-1) x+(3 \lambda+1) y+2 \lambda z & =0 \\
(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z & =0 \\
2 x+(3 \lambda+1) y+3(\lambda-1) z & =0
\end{aligned}$
has non-zero solution, then $p^2+q^2-p q=$
Options:
$\begin{aligned}
(\lambda-1) x+(3 \lambda+1) y+2 \lambda z & =0 \\
(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z & =0 \\
2 x+(3 \lambda+1) y+3(\lambda-1) z & =0
\end{aligned}$
has non-zero solution, then $p^2+q^2-p q=$
Solution:
1661 Upvotes
Verified Answer
The correct answer is:
9
We have,
$\begin{array}{r}
(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0 \\
(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0 \\
2 x+(3 \lambda+1) y+3(\lambda-1) z=0
\end{array}$
Now, it can be express as $\quad A X=0$
Where, $A=\left[\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3(\lambda-1)\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
For non-zero solution $|A|=0$
$\left[\begin{array}{ccc}
\lambda-1 & 3 \lambda+1 & 2 \lambda \\
\lambda-1 & 4 \lambda-2 & \lambda+3 \\
2 & 3 \lambda+1 & 3(\lambda-1)
\end{array}\right]=0$
On applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$
$\left[\begin{array}{ccc}
\lambda-1 & 3 \lambda+1 & 2 \lambda \\
0 & \lambda-3 & -\lambda+3 \\
-\lambda+3 & 0 & \lambda-3
\end{array}\right]=0$
$\begin{array}{cc}\Rightarrow & (\lambda-1)\left[(\lambda-3)^2\right]+(-\lambda+3)((3 \lambda+1)(-\lambda+3) \\ & -2 \lambda(\lambda-3))=0 \\ \Rightarrow & (\lambda-1)(\lambda-3)^2-(\lambda-3)[(\lambda-3)(-3 \lambda-1-2 \lambda))=0 \\ \Rightarrow & (\lambda-3)^2[\lambda-1+5 \lambda+1]=0 \\ \Rightarrow & p=3, q=0 \\ \therefore & (\lambda-3)^2(6 \lambda)=0 \Rightarrow \lambda=3,0 \\ \therefore & p^2+q^2-p q=9+0-0=9\end{array}$
$\begin{array}{r}
(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0 \\
(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0 \\
2 x+(3 \lambda+1) y+3(\lambda-1) z=0
\end{array}$
Now, it can be express as $\quad A X=0$
Where, $A=\left[\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3(\lambda-1)\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
For non-zero solution $|A|=0$
$\left[\begin{array}{ccc}
\lambda-1 & 3 \lambda+1 & 2 \lambda \\
\lambda-1 & 4 \lambda-2 & \lambda+3 \\
2 & 3 \lambda+1 & 3(\lambda-1)
\end{array}\right]=0$
On applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$
$\left[\begin{array}{ccc}
\lambda-1 & 3 \lambda+1 & 2 \lambda \\
0 & \lambda-3 & -\lambda+3 \\
-\lambda+3 & 0 & \lambda-3
\end{array}\right]=0$
$\begin{array}{cc}\Rightarrow & (\lambda-1)\left[(\lambda-3)^2\right]+(-\lambda+3)((3 \lambda+1)(-\lambda+3) \\ & -2 \lambda(\lambda-3))=0 \\ \Rightarrow & (\lambda-1)(\lambda-3)^2-(\lambda-3)[(\lambda-3)(-3 \lambda-1-2 \lambda))=0 \\ \Rightarrow & (\lambda-3)^2[\lambda-1+5 \lambda+1]=0 \\ \Rightarrow & p=3, q=0 \\ \therefore & (\lambda-3)^2(6 \lambda)=0 \Rightarrow \lambda=3,0 \\ \therefore & p^2+q^2-p q=9+0-0=9\end{array}$
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