Question

Let \[{{A}^{-1}}=\left[ \begin{matrix}
   1 & 2017 & 2 \\
   1 & 2017 & 4 \\
   1 & 2018 & 8 \\
\end{matrix} \right]\] . Then \[\left| 2A \right|-\left| 2{{A}^{-1}} \right|\] is equal to.
\[\begin{align}
  & \left( a \right)3 \\
 & \left( b \right)-3 \\
 & \left( c \right)12 \\
 & \left( d \right)-12 \\
\end{align}\]

Answer 1

Hint: Questions like these are often simple to understand and are easy to solve. We need to understand all the underlying concepts behind the problem to solve these type of problems quickly and efficiently. Solving this problem requires some previous background knowledge of matrices and determinants and their various properties and relation. We must be able to find the determinant of the matrix efficiently. Here, to solve the problem, we must remember one relation that, the determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix. In mathematical form we write it as,
\[\left| A \right|=\dfrac{1}{\left| {{A}^{-1}} \right|}\] . Using this relation, we can very easily find out the answer to the problem.

Complete step-by-step solution:
Now we start off with the solution to the given problem by trying to find out the determinant of the given matrix, \[A\] , the determinant can be found out as,
\[\left| {{A}^{-1}} \right|=\left| \begin{matrix}
   1 & 2017 & 2 \\
   1 & 2017 & 4 \\
   1 & 2018 & 8 \\
\end{matrix} \right|=1\left( \left( 2017\times 8 \right)-\left( 2018\times 4 \right) \right)-1\left( \left( 2017\times 8 \right)-\left( 2018\times 2 \right) \right)+1\left( \left( 2017\times 4 \right)-\left( 2017\times 2 \right) \right)\]
Now we evaluate the value of the determinant as,
\[\begin{align}
  & \Rightarrow \left| {{A}^{-1}} \right|=\left| \begin{matrix}
   1 & 2017 & 2 \\
   1 & 2017 & 4 \\
   1 & 2018 & 8 \\
\end{matrix} \right|=1\left( 16136-8072 \right)-1\left( 16136-4036 \right)+1\left( 8068-4034 \right) \\
 & \Rightarrow \left| {{A}^{-1}} \right|=\left| \begin{matrix}
   1 & 2017 & 2 \\
   1 & 2017 & 4 \\
   1 & 2018 & 8 \\
\end{matrix} \right|=8064-12100+4034 \\
 & \Rightarrow \left| {{A}^{-1}} \right|=\left| \begin{matrix}
   1 & 2017 & 2 \\
   1 & 2017 & 4 \\
   1 & 2018 & 8 \\
\end{matrix} \right|=-2 \\
\end{align}\]
Thus we get the value of the determinant of the inverse of the matrix \[\left| {{A}^{-1}} \right|\] as \[-2\] . Now, from the relation \[\left| A \right|=\dfrac{1}{\left| {{A}^{-1}} \right|}\] , we can easily find out the value of the determinant of \[\left| A \right|\] as , \[\left| A \right|=\dfrac{1}{-2}=-\dfrac{1}{2}\] . Now we put these values in the original equation to find out the result to our problem as,
\[\begin{align}
  & \left| 2A \right|-\left| 2{{A}^{-1}} \right|=2\left( -\dfrac{1}{2} \right)-2\left( -2 \right) \\
 & \Rightarrow \left| 2A \right|-\left| 2{{A}^{-1}} \right|=-1+4 \\
 & \Rightarrow \left| 2A \right|-\left| 2{{A}^{-1}} \right|=3 \\
\end{align}\]
Thus, we get the answer to our problem as, \[3\] , which is basically option ‘a’.

Note: For such types of problems, we need to have some previous knowledge of matrices and determinants and their various relations and properties. We need to remember that, when we are given to find the determinant of a matrix, then the matrix should be a square matrix or else we would not be able to find the determinant. One important relation that we need to remember here is, \[\left| A \right|=\dfrac{1}{\left| {{A}^{-1}} \right|}\] .