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If $P, Q$ and $R$ are $3 \times 3$ matrices such that
$\begin{aligned}
& \left(\begin{array}{ccc}
3 x^2+x+3 & 2 x^2-x+4 & 7 x^2+8 x+5 \\
5 x^2+3 x+2 & 4 x^2-2 x-1 & 7 x^2+5 x+8 \\
3 x^2+2 x+5 & 4 x^2-x-2 & 3 x^2+8 x+7
\end{array}\right) \\
& =P x^2+Q x+R, \text { then } \operatorname{det} R=
\end{aligned}$
Options:
$\begin{aligned}
& \left(\begin{array}{ccc}
3 x^2+x+3 & 2 x^2-x+4 & 7 x^2+8 x+5 \\
5 x^2+3 x+2 & 4 x^2-2 x-1 & 7 x^2+5 x+8 \\
3 x^2+2 x+5 & 4 x^2-x-2 & 3 x^2+8 x+7
\end{array}\right) \\
& =P x^2+Q x+R, \text { then } \operatorname{det} R=
\end{aligned}$
Solution:
1672 Upvotes
Verified Answer
The correct answer is:
$136$
Put $x=0$ in the given matrix
$\left(\begin{array}{ccc}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right)=R$
$\Rightarrow R=3 \times 9+4 \times 26+5=136$
$\left(\begin{array}{ccc}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right)=R$
$\Rightarrow R=3 \times 9+4 \times 26+5=136$
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