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,

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{aligned} & A.adjA=\left| A \right|.I \\\ & \Rightarrow 2A.adj(2A)=\left| 2A \right|.I \\\ \end{aligned}$$ Let us consider matrix A $$=\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{aligned} & \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{aligned}$$ We got that $${{\left| A \right|}^{2}}=k$$. Cancel out $$\left| A \right|$$ on both sides of the expression, $$\begin{aligned} & {{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{aligned}$$ Thus we got det(adj2A) = 64 k. Option D is the right answer. Note:We have said that $$\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).