Let k be a positive real number and let A=[ 2k-1 2√ k 2√ k 2√ k 1 -2k -2√ k 2k -1 ] and B=[ 0 2k-1 √ k 1-2k 0 2√ k -√ k -2√ k .0 ] If det (adj A) + det (adj B) = 106, then [k] is equal to......

Question

Let k be a positive real number and let A=[ 2k-1 2√ k 2√ k 2√ k 1 -2k -2√ k 2k -1 ] and B=[ 0 2k-1 √ k 1-2k 0 2√ k -√ k -2√ k .0 ] If det (adj A) + det (adj B) = 106, then [k] is equal to...... (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

| A | = (2k + 1)3,|B |= 0 But det (adj .A) + det (adj B) = 106 ⇒ (2k+1)6 =106 ⇒ k=(9/2) ⇒ [k]=4

Q. 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
$B = ⎣ ⎡ 0 1 - 2 k - k 2 k - 1 0 - 2 k k 2 k .0 ⎦ ⎤$
If $d e t (a d j A) + d e t (a d j B) = 1 0^{6}$ , then $[k]$ is equal to......

$∣ A ∣ = (2 k + 1)^{3}, ∣ B ∣ = 0$
But $d e t (a d j . A) + d e t (a d j B) = 1 0^{6}$
$\Rightarrow (2 k + 1)^{6} = 1 0^{6} \Rightarrow k = \frac{9}{2} \Rightarrow [k] = 4$

Q. Let k be a positive real number and let A=⎣⎡​2k−12k​−2k​​2k​12k​2k​−2k−1​⎦⎤​ and B=⎣⎡​01−2k−k​​2k−10−2k​​k​2k​.0​⎦⎤​ If det(adjA)+det(adjB)=106, then [k] is equal to......

∣A∣=(2k+1)3,∣B∣=0 But det(adj.A)+det(adjB)=106 ⇒(2k+1)6=106⇒k=29​⇒[k]=4

Q. 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
$B = ⎣ ⎡ 0 1 - 2 k - k 2 k - 1 0 - 2 k k 2 k .0 ⎦ ⎤$
If $d e t (a d j A) + d e t (a d j B) = 1 0^{6}$ , then $[k]$ is equal to......

$∣ A ∣ = (2 k + 1)^{3}, ∣ B ∣ = 0$
But $d e t (a d j . A) + d e t (a d j B) = 1 0^{6}$
$\Rightarrow (2 k + 1)^{6} = 1 0^{6} \Rightarrow k = \frac{9}{2} \Rightarrow [k] = 4$