Q46 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
If the system $\left[\begin{array}{ll}2 & 8 \\ 3 & 7\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=k\left[\begin{array}{l}a \\ b\end{array}\right]$ has non-trivial, solution then the positive value of $k$ and a solution of the system for that value of $k$ are
- A$9,\left[\begin{array}{c}3 \\ -8\end{array}\right]$
- B$10,\left[\begin{array}{c}-8 \\ 3\end{array}\right]$
- C$6,\left[\begin{array}{c}-1 \\ -1\end{array}\right]$
- Correct answerD$10,\left[\begin{array}{l}17 \\ 1\end{array}\right]$
Solution
We have,
$$
\begin{gathered}
{\left[\begin{array}{ll}
2 & 8 \\
3 & 7
\end{array}\right]\left[\begin{array}{l}
a \\
b
\end{array}\right]=k\left[\begin{array}{l}
a \\
b
\end{array}\right] \Rightarrow\left[\begin{array}{c}
2 a+8 b \\
3 a+7 b
\end{array}\right]=\left[\begin{array}{l}
k a \\
k b
\end{array}\right]} \\
\Rightarrow\left[\begin{array}{l}
a(2-k)+8 b \\
3 a+(7-k) b
\end{array}\right]=\left[\begin{array}{l}
0 \\
0
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
2-k & 8 \\
3 & 7-k
\end{array}\right]\left[\begin{array}{l}
a \\
b
\end{array}\right]=\left[\begin{array}{l}
0 \\
0
\end{array}\right]
\end{gathered}
$$
It has non-trivial solution.
$$
\begin{aligned}
& \Rightarrow \quad\left|\begin{array}{cc}
2-k & 8 \\
3 & 7-k
\end{array}\right|=0 \\
& \Rightarrow \quad(2-k)(7-k)-24=0 \\
& \Rightarrow \quad k^2-9 k+14-24=0 \\
& \Rightarrow \quad k^2-9 k-10=0 \\
& \Rightarrow \quad(k-10)(k+1)=0 \\
& \Rightarrow k=10, k=-1 \text { (rejected) } k>0 \\
& \text { and } \\
& a=b=1 \\
& \therefore \text { Solution are } 10,10 \text {. } \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'D'.