Matrices & Determinants Question - None

Question

Let k be a positive real number and
$$A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \ 2\sqrt{k} & 1 & -2k \ -2\sqrt{k} & 2k & -1 \end{bmatrix}$$ and $$B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \ 1-2k & 0 & 2 \ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}.$$ If $\det(\text{adj}\, A) + \det(\text{adj}\, B) = 10^6$, 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 $k$.]

Accepted Answer

Step 1: Use the formula det(adj M) = [det(M)]^(n-1) for an n×n matrix. Here n = 3, so:

det(adj A) = [det(A)]² and det(adj B) = [det(B)]²

Step 2: Observe that B is skew-symmetric (check: B^T = -B). For a 3×3 skew-symmetric matrix, det(B) = 0 always. Therefore det(adj B) = 0.

Step 3: The equation becomes [det(A)]² + 0 = 10⁶, so [det(A)]² = 10⁶.

Thus det(A) = ±10³ = ±1000

Step 4: Calculate det(A) by expanding or using row operations. After careful expansion:

det(A) = (2k - 1)[(1)(-1) - (-2k)(2k)] - 2√k[2√k(-1) - (-2k)(-2√k)] + 2√k[2√k(2k) - (1)(-2√k)]

= (2k - 1)[4k² - 1] - 2√k[-2√k - 4k√k] + 2√k[4k√k + 2√k]

= (2k - 1)(2k - 1)(2k + 1) - 2√k(-2√k)(1 + 2k) + 2√k(2√k)(2k + 1)

= (2k - 1)² (2k + 1) + 4k(2k + 1) + 4k(2k + 1)

= (2k + 1)[(2k - 1)² + 8k]

= (2k + 1)[4k² - 4k + 1 + 8k]

= (2k + 1)(4k² + 4k + 1) = (2k + 1)(2k + 1)² = (2k + 1)³

Step 5: Solve (2k + 1)³ = 1000 (taking positive root since k > 0):

2k + 1 = 10

2k = 9

k = 4.5

Step 6: Find [k] = [4.5] = 4

∴ Answer: 4

Question:

Step-by-Step Solution

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