Q80 / 160 · TS EAMCET 2017 · Mathematics
If $\Delta=\left|\begin{array}{lll}1 & 5 & 6 \\ 0 & 1 & 7 \\ 0 & 0 & 1\end{array}\right|$ and $\Delta^{\prime}=\left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$, then
- A$\Delta^2-3 \Delta^{\prime}=0$
- Correct answerB$\left(\Delta+\Delta^{\prime}\right)^2-3\left(\Delta+\Delta^{\prime}\right)+2=0$
- C$\left(\Delta+\Delta^{\prime}\right)^2+3\left(\Delta+\Delta^{\prime}\right)+5=0$
- D$\Delta+3 \Delta^{\prime}+1=0$
Solution
Given,
$$
\begin{aligned}
\Delta & =\left|\begin{array}{lll}
1 & 5 & 6 \\
0 & 1 & 7 \\
0 & 0 & 1
\end{array}\right| \\
& =1(1-0)-5(0-0)+6(0-0) \\
& =1-0+0
\end{aligned}
$$
$$
\Rightarrow \quad \Delta=1
$$
If $\Delta=1$ and $\Delta^{\prime}=\left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$
$$
\begin{aligned}
& =1(0-18)-0(300-12)+1(18-0) \\
& =-18-0+18 \\
\Delta^{\prime} & =0
\end{aligned}
$$
Now, $\left(\Delta+\Delta^{\prime}\right)^2-3\left(\Delta+\Delta^{\prime}\right)+2$
$$
\begin{aligned}
& =(1+0)^2-3(1+0)+2 \\
& =1-3+2 \\
& =-2+2=0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.