Question

If \[{{l}_{1}}\], \[{{m}_{1}}\], \[{{n}_{1}}\]; \[{{l}_{2}}\], \[{{m}_{2}}\], \[{{n}_{2}}\]; \[{{l}_{3}}\], \[{{m}_{3}}\], \[{{n}_{3}}\] are real quantities satisfying six relations \[l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=l_{3}^{2}+m_{3}^{2}+n_{3}^{2}=1\], ${{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}={{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}}={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0$ and $\left| \begin{matrix}
   {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\
   {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\
   {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\
\end{matrix} \right|=\pm k$. Find the value of k?

Answer 1

Hint: We start solving the problem by finding the transpose of the given matrix. We use the fact that the determinant of product of two matrices is equal to the product of the determinant of the individual matrices to multiply the determinants of given matrix and its transpose. We use the fact that the determinant of the matrix is equal to the determinant of the transpose of the matrix and make required calculations to get the value of k.

Complete step by step answer:
Given that we have six real quantities \[{{l}_{1}}\], \[{{m}_{1}}\], \[{{n}_{1}}\]; \[{{l}_{2}}\], \[{{m}_{2}}\], \[{{n}_{2}}\]; \[{{l}_{3}}\], \[{{m}_{3}}\], \[{{n}_{3}}\] which are satisfying the relations \[l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=l_{3}^{2}+m_{3}^{2}+n_{3}^{2}=1\], ${{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}={{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}}={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0$ and $\left| \begin{matrix}
   {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\
   {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\
   {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\
\end{matrix} \right|=\pm k$. We need to find the value of k.
Let us the matrix $\left[ \begin{matrix}
   {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\
   {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\
   {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\
\end{matrix} \right]$ as ‘A’. Let us find the transpose of the matrix ‘A’. We know that transpose of a matrix is found by interchanging the rows and of a matrix.
So, we get ${{A}^{T}}=\left[ \begin{matrix}
   {{l}_{1}} & {{l}_{2}} & {{l}_{3}} \\
   {{m}_{1}} & {{m}_{2}} & {{m}_{3}} \\
   {{n}_{1}} & {{n}_{2}} & {{n}_{3}} \\
\end{matrix} \right]$.
We know that for two matrices P and Q, $\det \left( PQ \right)=\det \left( P \right).\det \left( Q \right)$.
Let us find the value of $\det \left( A{{A}^{T}} \right)$.
$\Rightarrow \det \left( A{{A}^{T}} \right)=\det \left( \left[ \begin{matrix}
   {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\
   {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\
   {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\
\end{matrix} \right]\times \left[ \begin{matrix}
   {{l}_{1}} & {{l}_{2}} & {{l}_{3}} \\
   {{m}_{1}} & {{m}_{2}} & {{m}_{3}} \\
   {{n}_{1}} & {{n}_{2}} & {{n}_{3}} \\
\end{matrix} \right] \right)$.
\[\Rightarrow \det \left( A \right).\det \left( {{A}^{T}} \right)=\det \left[ \begin{matrix}
   \left( {{l}_{1}}\times {{l}_{1}} \right)+\left( {{m}_{1}}\times {{m}_{1}} \right)+\left( {{n}_{1}}\times {{n}_{1}} \right) & \left( {{l}_{1}}\times {{l}_{2}} \right)+\left( {{m}_{1}}\times {{m}_{2}} \right)+\left( {{n}_{1}}\times {{n}_{2}} \right) & \left( {{l}_{1}}\times {{l}_{3}} \right)+\left( {{m}_{1}}\times {{m}_{3}} \right)+\left( {{n}_{1}}\times {{n}_{3}} \right) \\
   \left( {{l}_{2}}\times {{l}_{1}} \right)+\left( {{m}_{2}}\times {{m}_{1}} \right)+\left( {{n}_{2}}\times {{n}_{1}} \right) & \left( {{l}_{2}}\times {{l}_{2}} \right)+\left( {{m}_{2}}\times {{m}_{2}} \right)+\left( {{n}_{2}}\times {{n}_{2}} \right) & \left( {{l}_{2}}\times {{l}_{3}} \right)+\left( {{m}_{2}}\times {{m}_{3}} \right)+\left( {{n}_{2}}\times {{n}_{3}} \right) \\
   \left( {{l}_{3}}\times {{l}_{1}} \right)+\left( {{m}_{3}}\times {{m}_{1}} \right)+\left( {{n}_{3}}\times {{n}_{1}} \right) & \left( {{l}_{3}}\times {{l}_{2}} \right)+\left( {{m}_{3}}\times {{m}_{2}} \right)+\left( {{n}_{3}}\times {{n}_{2}} \right) & \left( {{l}_{3}}\times {{l}_{3}} \right)+\left( {{m}_{3}}\times {{m}_{3}} \right)+\left( {{n}_{3}}\times {{n}_{3}} \right) \\
\end{matrix} \right]\].
We know that $\det \left( P \right)=\det \left( {{P}^{T}} \right)$. We use this in the above determinant.
\[\Rightarrow \det \left( A \right).\det \left( A \right)=\det \left[ \begin{matrix}
   \left( l_{2}^{2} \right)+\left( m_{2}^{2} \right)+\left( n_{2}^{2} \right) & \left( {{l}_{1}}{{l}_{2}} \right)+\left( {{m}_{1}}{{m}_{2}} \right)+\left( {{n}_{1}}{{n}_{2}} \right) & \left( {{l}_{1}}{{l}_{3}} \right)+\left( {{m}_{1}}{{m}_{3}} \right)+\left( {{n}_{1}}{{n}_{3}} \right) \\
   \left( {{l}_{2}}{{l}_{1}} \right)+\left( {{m}_{2}}{{m}_{1}} \right)+\left( {{n}_{2}}{{n}_{1}} \right) & \left( l_{2}^{2} \right)+\left( m_{2}^{2} \right)+\left( n_{2}^{2} \right) & \left( {{l}_{2}}{{l}_{3}} \right)+\left( {{m}_{2}}{{m}_{3}} \right)+\left( {{n}_{2}}{{n}_{3}} \right) \\
   \left( {{l}_{3}}{{l}_{1}} \right)+\left( {{m}_{3}}{{m}_{1}} \right)+\left( {{n}_{3}}{{n}_{1}} \right) & \left( {{l}_{3}}{{l}_{2}} \right)+\left( {{m}_{3}}{{m}_{2}} \right)+\left( {{n}_{3}}{{n}_{2}} \right) & \left( l_{3}^{2} \right)+\left( m_{3}^{2} \right)+\left( n_{3}^{2} \right) \\
\end{matrix} \right]\].
\[\Rightarrow {{\left( \det \left( A \right) \right)}^{2}}=\det \left[ \begin{matrix}
   l_{2}^{2}+m_{2}^{2}+n_{2}^{2} & {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} & {{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}} \\
   {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} & l_{2}^{2}+m_{2}^{2}+n_{2}^{2} & {{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}} \\
   {{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}} & {{l}_{2}}{{l}_{2}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}} & l_{3}^{2}+m_{3}^{2}+n_{3}^{2} \\
\end{matrix} \right]\].
According to the problem, we have \[l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=l_{3}^{2}+m_{3}^{2}+n_{3}^{2}=1\] and ${{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}={{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}}={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0$. We substitute these values in the matrix. We have also got the value of $\det \left( A \right)=\pm k$.
\[\Rightarrow {{\left( \pm k \right)}^{2}}=\left| \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right|\].
We know that the determinant of the matrix $\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]$ is $\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\times \left| \begin{matrix}
   e & f \\
   h & i \\
\end{matrix} \right|+b\times \left| \begin{matrix}
   d & f \\
   g & i \\
\end{matrix} \right|+c\times \left| \begin{matrix}
   d & e \\
   g & h \\
\end{matrix} \right|$. We apply this to find the determinant \[\left| \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right|\].

\[\Rightarrow {{k}^{2}}=1\times \left| \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right|+0\times \left| \begin{matrix}
   0 & 0 \\
   0 & 1 \\
\end{matrix} \right|+0\times \left| \begin{matrix}
   0 & 1 \\
   0 & 0 \\
\end{matrix} \right|\].
We know that the determinant of the matrix $\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]$ is $\left| \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
\[\Rightarrow {{k}^{2}}=1\times \left( \left( 1\times 1 \right)-\left( 0\times 0 \right) \right)+0+0\].
\[\Rightarrow {{k}^{2}}=1-0\].
\[\Rightarrow {{k}^{2}}=1\].
\[\Rightarrow k=1\].
We have found the value of k as 1.

∴ The value of the k is 1.

Note: Here, we have taken only a positive root for the value of k as the value of the determinant given in the problem is $\pm k$. Whenever we get a problem involving the multiplication of the elements of a given matrix, we need to remember that there can be a possibility of multiplication of two determinants. Similarly, we can expect problems which may involve subtraction and change of rows while performing the determinants to get the required answers.