Q86 / 180 · KCET 2024 · Mathematics
If $A=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|$ and $B=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|$, then $\frac{d B}{d x}$ is
- Correct answerA$3 A$
- B$-3 B$
- C$3 B+1$
- D$1-3 A$
Solution
$B=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|$
$=\left|\begin{array}{lll}1 & 1 & 1 \\ 0 & x & 1 \\ 0 & 1 & x\end{array}\right|+\left|\begin{array}{lll}x & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & x\end{array}\right|+\left|\begin{array}{lll}x & 1 & 0 \\ 1 & x & 0 \\ 1 & 1 & 1\end{array}\right|$
$=\left(x^2-1\right)+\left(x^2-1\right)+\left(x^2-1\right)$
$=3\left(x^2-1\right)$
$=3 A \quad\left[\because A=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|=x^2-1\right]$
Hence, the correct answer is option: 'A'.