Question

Let K be a positive real number and let A = $\left[ {\begin{array}{*{20}{c}}
  {2k - 1}&{2\sqrt k }&{2\sqrt k } \\
  {2\sqrt k }&1&{ - 2k} \\
  { - 2\sqrt k }&{2k}&{ - 1}
\end{array}} \right]$ and B = $\left[ {\begin{array}{*{20}{c}}
  0&{2k - 1}&{\sqrt k } \\
  {1 - 2k}&0&{2\sqrt k } \\
  { - \sqrt k }&{ - 2\sqrt k }&0
\end{array}} \right]$. If det (adj A) + det (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].

Answer 1

Hint: In this particular question use the concept that det (adj A) = ${\left| A \right|^{n - 1}}$, where n is the order of the matrix, order of matrix in a square matrix is nothing but the number of rows or number of columns, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
A = $\left[ {\begin{array}{*{20}{c}}
  {2k - 1}&{2\sqrt k }&{2\sqrt k } \\
  {2\sqrt k }&1&{ - 2k} \\
  { - 2\sqrt k }&{2k}&{ - 1}
\end{array}} \right]$ and B = $\left[ {\begin{array}{*{20}{c}}
  0&{2k - 1}&{\sqrt k } \\
  {1 - 2k}&0&{2\sqrt k } \\
  { - \sqrt k }&{ - 2\sqrt k }&0
\end{array}} \right]$.
Now as we know that the det (adj A) = ${\left| A \right|^{n - 1}}$, where n is the order of the matrix, order of matrix in a square matrix is nothing but the number of rows or number of columns3.
So in the given matrix the order is 3, so n = 3.
\[ \Rightarrow \det \left( {{\text{adjA}}} \right) = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}\]............ (1)
So first find out determinant of A we have,
\[ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
  {2k - 1}&{2\sqrt k }&{2\sqrt k } \\
  {2\sqrt k }&1&{ - 2k} \\
  { - 2\sqrt k }&{2k}&{ - 1}
\end{array}} \right|\]
Now expand the determinant we have,
\[ \Rightarrow \left| A \right| = \left[ {\left( {2k - 1} \right)\left| {\begin{array}{*{20}{c}}
  1&{ - 2k} \\
  {2k}&{ - 1}
\end{array}} \right| - 2\sqrt k \left| {\begin{array}{*{20}{c}}
  {2\sqrt k }&{ - 2k} \\
  { - 2\sqrt k }&{ - 1}
\end{array}} \right| + 2\sqrt k \left| {\begin{array}{*{20}{c}}
  {2\sqrt k }&1 \\
  { - 2\sqrt k }&{2k}
\end{array}} \right|} \right]\]
\[ \Rightarrow \left| A \right| = \left[ {\left( {2k - 1} \right)\left( { - 1 + 4{k^2}} \right) - 2\sqrt k \left( { - 2\sqrt k - 4k\sqrt k } \right) + 2\sqrt k \left( {4k\sqrt k + 2\sqrt k } \right)} \right]\]
\[ \Rightarrow \left| A \right| = - 2k + 1 + 8{k^3} - 4{k^2} + 4k + 8{k^2} + 8{k^2} + 4k\]
\[ \Rightarrow \left| A \right| = 8{k^3} + 12{k^2} + 6k + 1\]
\[ \Rightarrow \left| A \right| = {\left( {2k + 1} \right)^3}\]
Now from equation (1) we have,
\[ \Rightarrow \det \left( {{\text{adjA}}} \right) = {\left( {{{\left( {2k + 1} \right)}^3}} \right)^2} = {\left( {2k + 1} \right)^6}\].................. (2)
Similarly,
\[ \Rightarrow \det \left( {{\text{adjB}}} \right) = {\left| B \right|^{3 - 1}} = {\left| B \right|^2}\]................ (3)
So first find out determinant of A we have,
\[ \Rightarrow \left| B \right| = \left| {\begin{array}{*{20}{c}}
  0&{2k - 1}&{\sqrt k } \\
  {1 - 2k}&0&{2\sqrt k } \\
  { - \sqrt k }&{ - 2\sqrt k }&0
\end{array}} \right|\]
Now expand the determinant we have,
\[ \Rightarrow \left| B \right| = \left[ {0 - \left( {2k - 1} \right)\left| {\begin{array}{*{20}{c}}
  {1 - 2k}&{2\sqrt k } \\
  { - \sqrt k }&0
\end{array}} \right| + \sqrt k \left| {\begin{array}{*{20}{c}}
  {1 - 2k}&0 \\
  { - \sqrt k }&{ - 2\sqrt k }
\end{array}} \right|} \right]\]
\[ \Rightarrow \left| B \right| = \left[ {0 - \left( {2k - 1} \right)\left( {0 + 2k} \right) + \sqrt k \left( { - 2\left( {1 - 2k} \right)\sqrt k + 0} \right)} \right]\]
\[ \Rightarrow \left| B \right| = \left[ { - 4{k^2} + 2k - 2k + 4{k^2}} \right]\]
\[ \Rightarrow \left| B \right| = 0\]
Now from equation (3) we have,
\[ \Rightarrow \det \left( {{\text{adjB}}} \right) = {0^2} = 0\].................. (4)
Now add equation (2) and (4) we have,
$ \Rightarrow {\text{det}}\left( {{\text{adjA}}} \right) + \det \left( {{\text{adjB}}} \right) = {\left( {2k + 1} \right)^6} + 0 = {\left( {2k + 1} \right)^6}$
But it is given that det (adj A) + det (adj B) = ${10^6}$
$ \Rightarrow {\left( {2k + 1} \right)^6} = {10^6}$
$ \Rightarrow \left( {2k + 1} \right) = 10$
$ \Rightarrow 2k = 9$
$ \Rightarrow k = 4.5$
Now we have to find out the value of [k], where [k] denotes the greatest integer function less than or equal to K.
Therefore,
[K] = [4.5] = 4.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always remember some basic determinant properties and how to expand the determinant which is stated above and always recall that the greatest integer of x (say x = 0.1) i.e. [x] = [0.1] = 0 i.e. less than or equal to x.