Question

Let k be positive real number 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]andB = \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 $\det (adjA) + \det (adjB) = {10^6}$ find $[k]$

Answer 1

Here we are asked to find the value of$[k]$. They have also given the matrices A and B and a condition that$\det (adjA) + \det (adjB) = {10^6}$. We don’t want to find the values of adj A and adj B and substitute them in the given condition to solve for k. We can use the property of matrices to simplify our calculation.

Formula Used:
The property of matrices that we need to know to solve this problem:
A square matrix is called skew-symmetric if the transpose of it is equal to the negative of itself, that is if A is a square matrix and if ${A^T} = - A$then A is a skew-symmetric matrix.
The determinant of a skew-symmetric matrix is zero.
If A is a square matrix with the order$n \times n$ then$\det {(adjA)^n} = \det {(A)^{n - 1}}$.

Complete step-by-step solution:
It is given that 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]
We aim to find the value of$[k]$. It is also given that$\det (adjA) + \det (adjB) = {10^6}$.
By seeing this we will urge to find the values of adj A and adj B and then find its determinant to substitute them in the given condition to solve for the value of k. But this idea will take so much time. So, we will easily solve this problem by using the properties of matrices.
We know that a square matrix A is said to be a skew-symmetric matrix if${A^T} = - A$.
By this property, we can see that the given matrix B is a skew-symmetric matrix. Let us check this.

0&{1 - 2k}&{ - \sqrt k } \\\ {2k - 1}&0&{ - 2\sqrt k } \\\ {\sqrt k }&{2\sqrt k }&0 \end{array}} \right]$$ $$ - 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] = \left[ {\begin{array}{*{20}{c}} 0&{1 - 2k}&{ - \sqrt k } \\\ {2k - 1}&0&{ - 2\sqrt k } \\\ {\sqrt k }&{2\sqrt k }&0 \end{array}} \right]$$ From the above, we can see that $${B^T} = - B$$which implies that matrix B is a skew-symmetric matrix. Now by the property, $$\det {(adjA)^n} = \det {(A)^{n - 1}}$$ the given condition can be rewritten as$$\det {(A)^{n - 1}} + \det {(B)^{n - 1}} = {10^6}$$ here the order of the matrices A and B is $$3 \times 3$$thus we get $$\det {(A)^{3 - 1}} + \det {(B)^{3 - 1}} = {10^6}$$$$ \Rightarrow \det {(A)^2} + \det {(B)^2} = {10^6}$$ Now let us find the determinant value of A and B. $$\det (A) = \left| {\begin{array}{*{20}{c}} {2k - 1}&{2\sqrt k }&{2\sqrt k } \\\ {2\sqrt k }&1&{ - 2k} \\\ { - 2\sqrt k }&{2k}&{ - 1} \end{array}} \right|$$ $$ = (2k - 1)[ - 1 + 4{k^2}] - 2\sqrt k [ - 2\sqrt k - 4k\sqrt k ] + 2\sqrt k [4k\sqrt k + 2\sqrt k ]$$ $$ = (2k - 1)[4{k^2} - 1] + 4k[2k + 1] + 4k[2k + 1]$$ $$ = 8{k^3} + 1 + 12{k^2} + 6k$$ $$ = {(2k + 1)^3}$$ Thus, we have found that$$\det (A) = {(2k + 1)^3}$$. Now we will find the value of$$\det (B)$$. We already know that matrix B is a skew-symmetric matrix. Then by the property that the determinant of a skew-symmetric matrix is zero we get$$\det (B) = 0$$. Now let us substitute the values of $$\det (A)$$and $$\det (B)$$in the condition$$\det {(A)^2} + \det {(B)^2} = {10^6}$$. $$\det {(A)^2} + \det {(B)^2} = {10^6}$$$$ \Rightarrow {[{(2k + 1)^3}]^2} + 0 = {10^6}$$ $$ \Rightarrow {(2k + 1)^6} = {10^6}$$ $$ \Rightarrow 2k + 1 = 10$$ $$ \Rightarrow 2k = 9$$ $$ \Rightarrow k = 4.5$$ Thus, we got the value of k as $$4.5$$but the greatest of the number $$4.5$$is $$4$$thus the value of $$[k]$$is $$4$$. **Note:** Here it is a good idea to use the properties of a matrix to solve this problem which will save our time and doesn’t lead us to the wrong solution. In this question, the value $$[k]$$denotes the greatest integer$$ \leqslant k$$. Also, the term adj represents the adjoint of that matrix.