Question

If $P=\left|\begin{array}{cc}x& 1\\ 1& x\end{array}\right|$ and $Q=\left|\begin{array}{ccc}x& 1& 1\\ 1& x& 1\\ 1& 1& x\end{array}\right|,$ then $\frac{dQ}{dx}=$

• A. $3P+1$
• B. $1-3P$
• C. $-3P$
• D. $3P$

Answer 1

The correct option is D $3P$

$P=\left|\begin{array}{cc}x& 1\\ 1& x\end{array}\right|=x^2-1$

$Q=\left|\begin{array}{ccc}x& 1& 1\\ 1& x& 1\\ 1& 1& x\end{array}\right|$

$\Rightarrow \frac{dQ}{dx}=\left|\begin{array}{ccc}1& 0& 0\\ 1& x& 1\\ 1& 1& x\end{array}\right|+\left|\begin{array}{ccc}x& 1& 1\\ 0& 1& 0\\ 1& 1& x\end{array}\right|+\left|\begin{array}{ccc}x& 1& 1\\ 1& x& 1\\ 0& 0& 1\end{array}\right|$

$=(x^2-1)+(x^2-1)+(x^2-1)$
$=3(x^2-1)=3P$