JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 12)
If x, y, z are in arithmetic progression with common difference d, x $$\ne$$ 3d, and the determinant of the matrix $$\left[ {\matrix{
3 & {4\sqrt 2 } & x \cr
4 & {5\sqrt 2 } & y \cr
5 & k & z \cr
} } \right]$$ is zero, then the value of k2 is :
72
12
36
6
Explanation
$$\left| {\matrix{
3 & {4\sqrt 2 } & x \cr
4 & {5\sqrt 2 } & y \cr
5 & k & z \cr
} } \right| = 0$$
$${R_1} \to {R_1} + {R_3} - 2{R_2}$$
$$ \Rightarrow $$ $$\left| {\matrix{ 0 & {4\sqrt 2 - k - 10\sqrt 2 } & 0 \cr 4 & {5\sqrt 2 } & y \cr 5 & k & z \cr } } \right| = 0$$ { $$ \because $$ 2y = x + z}
$$ \Rightarrow (k - 6\sqrt 2 )(4z - 5y) = 0$$
$$ \Rightarrow $$ k = $$6\sqrt 2 $$ or 4z = 5y (Not possible $$ \because $$ x, y, z in A.P.)
So, k2 = 72
$$ \therefore $$ Option (A)
$${R_1} \to {R_1} + {R_3} - 2{R_2}$$
$$ \Rightarrow $$ $$\left| {\matrix{ 0 & {4\sqrt 2 - k - 10\sqrt 2 } & 0 \cr 4 & {5\sqrt 2 } & y \cr 5 & k & z \cr } } \right| = 0$$ { $$ \because $$ 2y = x + z}
$$ \Rightarrow (k - 6\sqrt 2 )(4z - 5y) = 0$$
$$ \Rightarrow $$ k = $$6\sqrt 2 $$ or 4z = 5y (Not possible $$ \because $$ x, y, z in A.P.)
So, k2 = 72
$$ \therefore $$ Option (A)
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