If $x, y, z$ are in arithmetic progression with common difference $d, x \neq 3 d$, and the determinant of the matrix $\left[\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 4 & 5 \sqrt{2} & y \\ 5 & k & z\end{array}\right]$ is zero, then the value of $k^{2}$ is:

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Determinants

x, y, z

are in arithmetic progression with common difference

d, x \neq = 3 d

, and the determinant of the matrix

345 4252 k x y z

is zero, then the value of

k^{2}

is:

Answer: (B)

With

y = x + d

and

z = x + 2 d

, replace the first row by

R_{1} - 2 R_{2} + R_{3}

, which leaves the determinant unchanged. In that combination the first entry is

3 - 8 + 5 = 0

and the third is

x - 2 (x + d) + (x + 2 d) = 0

, while the second is

42 - 102 + k = k - 62

. Expanding along the new row,

345 4252 k x y z = - (k - 62) 45 y z = - (k - 62) (3 d - x)

Because

x \neq = 3 d

the second factor is non-zero, so the determinant vanishes only when

k = 62

, giving

k^{2} = 72

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option