Question

If A is a \[3\times 3\] and det adj(A)=k, then det(adj2A)=?
A. 2k
B. 8k
C. 16k
D. \[64{{k}^{2}}\]

Answer 1

Hint:Take the general expression, A.adjA \[=\left| A \right|I\]. Take their determinate and prove that \[k={{\left| A \right|}^{2}}\]. Thus find det(adj2A) by putting 2A in the general expression and thus find the value of det(adj2A).

Complete step-by-step answer:
We have been given a matrix A which is a \[3\times 3\] matrix. Thus we can say that n = 3.
We have been given that A.adjA \[=\left| A \right|I\], which is a formula for A inverse, \[{{A}^{-1}}\].
Thus A inverse can also be written as \[{{A}^{-1}}=\dfrac{adjA}{\left| A \right|}\].
\[\therefore \left| A.adjA \right|=\left| \left| A \right|.I \right|........(1)\]
Thus we can write the above expression as,
\[\left| A \right|.\left| adjA \right|=\left| \left| A \right|.I \right|\]
\[\left| A \right|=\lambda \] is a constant value.
We know that \[I\]is a unit matrix. Thus \[I\] for a \[3\times 3\] matrix is,
\[I=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]
Thus if multiplying a constant
\[\lambda .I=\lambda \left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   \lambda & 0 & 0 \\
   0 & \lambda & 0 \\
   0 & 0 & \lambda \\
\end{matrix} \right)\]
Thus taking the determinant of \[\left( \begin{matrix}
   \lambda & 0 & 0 \\
   0 & \lambda & 0 \\
   0 & 0 & \lambda \\
\end{matrix} \right)\] we get is as,
\[\lambda \left( {{\lambda }^{2}}-0 \right)-0\left( 0-0 \right)+0\left[ 0-0 \right]=\lambda .{{\lambda }^{2}}={{\lambda }^{3}}\]
Thus we got the determinate as \[{{\lambda }^{3}}\]. Thus it becomes \[{{\left| A \right|}^{3}}\].
\[\therefore \left| A \right|.\left| adjA \right|={{\left| A \right|}^{3}}\]
Cancel out \[\left| A \right|\] on both the sides.
\[\therefore \left| adjA \right|={{\left| A \right|}^{2}}\]
We have been given det(adjA) = k,
i.e. \[\left| adjA \right|=k\].
\[\Rightarrow k={{\left| A \right|}^{2}}.....(2)\]
Thus we need to find the value of det(adj2A), i.e. \[\left| adj2A \right|\].
Let us put 2A in the place of A in equation (1).
\[\begin{align}
  & A.adjA=\left| A \right|.I \\
 & \Rightarrow 2A.adj(2A)=\left| 2A \right|.I \\
\end{align}\]
Let us consider matrix A \[=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)\]
Now let us multiply \[\lambda \] to the matrix A and we get,
\[\lambda A=\left( \begin{matrix}
   \lambda {{a}_{11}} & \lambda {{a}_{12}} & \lambda {{a}_{13}} \\
   \lambda {{a}_{21}} & \lambda {{a}_{22}} & \lambda {{a}_{23}} \\
   \lambda {{a}_{31}} & \lambda {{a}_{32}} & \lambda {{a}_{33}} \\
\end{matrix} \right)\]
Thus taking the determinant, we get, \[{{2}^{3}}\left| A \right|\], i.e. \[{{\left| 2A \right|}^{3}}\].
\[\therefore 2A.adj(2A)={{2}^{3}}\left| A \right|.I\]
Thus taking determinant on both sides,
\[\begin{align}
  & \left| (2A)adj(2A) \right|=\left| {{2}^{3}}\left| A \right|.I \right| \\
 & \left| 2A \right|\left| adj2A \right|={{\left( {{2}^{3}}\left| A \right| \right)}^{3}} \\
 & {{2}^{3}}\left| A \right|\left| adj2A \right|={{\left( {{2}^{3}} \right)}^{3}}{{\left| A \right|}^{3}} \\
 & {{2}^{3}}\left| A \right|\left| adj2A \right|={{2}^{9}}{{\left| A \right|}^{3}} \\
\end{align}\]
We got that \[{{\left| A \right|}^{2}}=k\].
Cancel out \[\left| A \right|\] on both sides of the expression,
\[\begin{align}
  & {{2}^{3}}\left| adj2A \right|={{2}^{9}}k \\
 & \left| adj2A \right|=\dfrac{{{2}^{9}}}{{{2}^{3}}}k={{2}^{9-3}}k={{2}^{6}}k \\
 & \left| adj2A \right|={{2}^{6}}k \\
 & \therefore \left| adj2A \right|=64k \\
\end{align}\]
Thus we got det(adj2A) = 64 k.
Option D is the right answer.

Note:We have said that \[\left| A \right|\] is a constant. Thus for any condition, \[\left| kA \right|={{k}^{n}}\left| A \right|\], where n is order of A. Remember the formula of the inverse of a matrix using adjoint, which we have used here to find the value of det(adj2A).