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Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If
$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$
then value of $\lambda^{2}$ is equal to $.....$
$4$
$1$
$9$
$16$
Solution
$\left|\begin{array}{lll}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{array}\right|=2$
$C_{2} \rightarrow C_{2}-C_{3}$
$\left|\begin{array}{ccc}x-2 \lambda & \lambda & x+a \\ x-1 & \lambda & x+b \\ x+2 \lambda & \lambda & x+C\end{array}\right|=2$
$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$
$\Rightarrow\left|\begin{array}{ccc}x-2 \lambda & 1 & x+a \\ 2 \lambda-1 & 0 & \lambda \\ 4 \lambda & 0 & 2 \lambda\end{array}\right|=2$
$\Rightarrow 1\left(4 \lambda-4 \lambda^{2}+2 \lambda\right)=2 \Rightarrow \lambda^{2}=1$
