Question

If $A =\begin{vmatrix} {x}&{1} &{1}\\ {1}&{x}& {1} \\ {1}&{1}&{x}\\ \end{vmatrix} $ and $B =\begin{vmatrix} {x}&{1} \\ {1}&{x} \\ \end{vmatrix} $, then $\frac {dA}{dx}$ is equal to

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

Answer 1

Answer: (B)

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