By using properties of determinants, show that:
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$
$\Delta=\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2} R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}3 y+k & 3 y+k & 3 y+k \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
$=(3 y+k)\left|\begin{array}{ccc}1 & 1 & 1 \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Delta=(3 y+k)\left|\begin{array}{lll}1 & 0 & 0 \\ y & k & 0 \\ y & 0 & k\end{array}\right|$
$=k^{2}(3 x+k)\left|\begin{array}{lll}1 & 0 & 0 \\ y & 1 & 0 \\ y & 0 & 1\end{array}\right|$
Expanding alone $C_{3},$ we have:
$\Delta=k^{2}(3 y+k)\left|\begin{array}{ll}1 & 0 \\ y & 1\end{array}\right|=k^{2}(3 y+k)$
Hence, the given result is proved.
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- [JEE MAIN 2021]
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