Let k be a positive real number and let . If det(Adj(A)) + det(Adj(B)) = 2 then [k] is equal to

Engineering

Mathematics

Special Type of Square Matrices

Introduction to Determinants

Properties of Determinant

Question

Let k be a positive real number and let $A = [\begin{array}{ccc} 2 k - 1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & - 2 k \\ - 2 \sqrt{k} & 2 k & - 1 \end{array}], B = [\begin{array}{ccc} 0 & 2 k - 1 & \sqrt{k} \\ 1 - 2 \sqrt{k} & 0 & - 2 k \\ 2 \sqrt{k} & - \sqrt{k} & 0 \end{array}]$ . If det(Adj(A)) + det(Adj(B)) = 2 then [k] is equal to

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Given : $A = [\begin{array}{ccc} 2 k - 1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & - 2 k \\ - 2 \sqrt{k} & 2 k & - 1 \end{array}], B = [\begin{array}{ccc} 0 & 2 k - 1 & \sqrt{k} \\ 1 - 2 \sqrt{k} & 0 & - 2 k \\ 2 \sqrt{k} & - \sqrt{k} & 0 \end{array}]$

$| A | = (2 k - 1) [- 1 - (- 4 k^{2})] - 2 \sqrt{k} [- 2 \sqrt{k} - 4 k \sqrt{k}] + 2 \sqrt{k} [4 k \sqrt{k} - (- 2 \sqrt{k})]$

$| A | = (2 k - 1) (4 k^{2} - 1) + 2 \sqrt{k} (2 \sqrt{k} + 4 k \sqrt{k}) + 2 \sqrt{k} (4 k \sqrt{k} + 2 \sqrt{k})$

$| A | = (2 k - 1) (4 k^{2} - 1) + 4 \sqrt{k} (2 \sqrt{k} + 4 k \sqrt{k})$

|A| = (2k – 1)(2k – 1)(2k + 1) + 8k(1 + 2k)

|A| = (1 + 2k)(4k² + 4k + 1)

|A| + (2k + 1)³ [∵ (a + b)³ = (a + b)(a² + b² + 2ab)]

Since B is a skew symunctric matrix
So, |B| = 0
Now, det⁡(Adj⁡(A)) + det⁡(Adj⁡(B)) = 2

⇒ |A|ⁿ^–¹ + |B|ⁿ^–¹ = 2

Here, m is order of matrix

⇒ |A|^3–1 + |B|^3–1 = 2

⇒ |A|² + 0 = 2

⇒ ((2K + 1)³)² = 2

⇒ (2K + 1)⁶ = 2

⇒ 2K + 1= 2^1/6

⇒ 2K = 2^1/6 – 1

⇒ 2K = 1.122 – 1

⇒ 2K = 0 (approx)

⇒ K = 0

∴ Option C is correct