Question

If $A=\left|\begin{array}{ccc}x& 1& 1\\ 1& x& 1\\ 1& 1& x\end{array}\right|$ and $B=\left|\begin{array}{cc}x& 1\\ 1& x\end{array}\right|$, then $\frac{dA}{dx}$ is equal to

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

Answer 1

Answer: (B)

Given, $A=\left|\begin{array}{ccc}x& 1& 1\\ 1& x& 1\\ 1& 1& x\end{array}\right|$
$\Rightarrow A=x\left(x^2-1\right)-(x-1)+(1-x)$
$\Rightarrow A=x^3-x-x+1+1-x$
$\Rightarrow A=x^3-3x+2$
On differentiating w.r.t. $x$, we get
$\frac{dA}{dx}=3x^2-3...(i)$
Also, given $B=\left|\begin{array}{cc}x& 1\\ 1& x\end{array}\right|$
$\Rightarrow B=x^2-1$
$\Rightarrow 3B=3x^2-3...(ii)$
From Eqs. (i) and (ii), we get
$\frac{dA}{dx}=3B$