Question

If \(x,y,z\) are in arithmetic progression with common difference \(d\), \(x\ne 3d\), and the determinant of the matrix

\(\begin{bmatrix}3&4\sqrt{2}&x\\4&5\sqrt{2}&y\\5&k&z\end{bmatrix}\)

is zero, then the value of \(k^2\) is :

• A. 72
• B. 12
• C. 36
• D. 6

Hint:

Expand the determinant and substitute \(x=y-d,\ z=y+d\); it factors as \((y-4d)(k-6\sqrt{2})\).

Answer 1

The Correct Option is A

Expand along the first row: \(\Delta=3(5\sqrt{2}z-ky)-4\sqrt{2}(4z-5y)+x(4k-25\sqrt{2})\).
\(=15\sqrt{2}z-3ky-16\sqrt{2}z+20\sqrt{2}y+4kx-25\sqrt{2}x=\sqrt{2}(-z+20y-25x)+k(4x-3y)\).
Put \(x=y-d,\ z=y+d\): \(-z+20y-25x=-6y+24d=-6(y-4d)\) and \(4x-3y=y-4d\).
So \(\Delta=(y-4d)(k-6\sqrt{2})\). Here \(y-4d=x-3d\ne 0\), so \(\Delta=0\Rightarrow k=6\sqrt{2}\).
\(k^2=72\).
Answer: (1) 72