Let k be a positive real number and let \[ A=\left[\begin{array}{ccc} 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0 \end{array}\right] . \] If \(\operatorname{det}(\operatorname{adj} \mathrm{A})+\operatorname{det}(\operatorname{adj} \mathrm{B})=10^{6}\), then \([\mathrm{k}]\) is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Matrices

Let k be a positive real number and let

A = 2 k - 1 2 k - 2 k 2 k 1 2 k 2 k - 2 k - 1 and B = 0 1 - 2 k - k 2 k - 1 0 - 2 k k 2 k 0 .

det (adj A) + det (adj B) = 1 0^{6}

, then

[k]

is equal to

[Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].

Answer: 4

For a

3 \times 3

matrix

M

, we have

det (adj M) = (det M)^{n - 1} = (det M)^{2}

.

Matrix

B

is skew-symmetric (

B^{T} = - B

). For a skew-symmetric matrix of odd order,

det B = 0

. Therefore,

det (adj B) = 0

.

Thus:

det (adj A) + det (adj B) = (det A)^{2} + 0 = 1 0^{6}

det A = \pm 1 0^{3}

Computing

det A

(expanding along the first row or using properties):

det A = (2 k - 1) (1 \cdot (- 1) - (- 2 k) (2 k)) + 2 k (2 k \cdot (- 1) - (- 2 k) (- 2 k)) + 2 k (2 k \cdot 2 k - 1 \cdot (- 2 k))

After simplification (which involves careful algebra), we find:

det A = (2 k + 1)^{3}

Setting

(2 k + 1)^{3} = 1 0^{3}

2 k + 1 = 10 ⟹ k = 4.5

Therefore,

[k] = 4

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer