Find the K

Algebra Level 5

Let k 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 ] A\quad =\quad \left[ \begin{matrix} 2k-1 & 2\sqrt { k } & 2\sqrt { k } \\ 2\sqrt { k } & 1 & -2k \\ -2\sqrt { k } & 2k & -1 \end{matrix} \right] and B = [ 0 2 k 1 k 1 2 k 0 2 k k 2 k 0 ] . B\quad =\quad \left[ \begin{matrix} 0 & 2k-1 & \sqrt { k } \\ 1-2k & 0 & 2\sqrt { k } \\ -\sqrt { k } & -2\sqrt { k } & 0 \end{matrix} \right] . If det ( a d j A ) + det ( a d j B ) = 10 6 , \text{det}(adj\quad A)\quad +\quad \text{det}(adj\quad B)\quad =\quad { 10 }^{ 6 }, what is the value of [ k ] ? [k]?

Note: [ k ] [k] is the greatest integer less than or equal to k . k.


The answer is 4.

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2 solutions

Suyash Mittal
Apr 24, 2014

A = ( 2 k + 1 ) 3 a n d B = 0 \left| A \right| \quad =\quad { (2k+1) }^{ 3 }\quad {and} \quad \left| B \right| \quad =\quad 0 But d e t ( a d j A ) + d e t ( a d j B ) = 10 6 det(adj\quad A)\quad + \quad det(adj\quad B)\quad =\quad { 10 }^{ 6 } ( 2 k + 1 ) 6 = 10 6 { (2k+1) }^{ 6 }\quad =\quad { 10 }^{ 6 } k = 9 2 k\quad =\quad \cfrac { 9 }{ 2 }

This question is asked in JEE 2007..!!!!

Deepanshu Gupta - 6 years, 8 months ago
汶良 林
Jun 4, 2015

d e t ( a d j A ) = A 2 det(adjA) = |A|^2

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