Question

Let A be a non-singular matrix of order $3\times 3$ whose entries are complex numbers which satisfy $2{{A}^{2}}=4A+{{A}^{3}}$, then which of the following is/are true.
(a) $\det \left( A \right)=-8$,
(b) $\det \left( adj\left( \dfrac{A}{2} \right) \right)=1$,
(c) $adjA={{A}^{2}}$,
(d) $tr\left( {{\left( A-2I \right)}^{3}} \right)=24$.

Answer 1

Hint: We start solving the problem by multiplying $2{{A}^{2}}=4A+{{A}^{3}}$ with ${{A}^{-1}}$ and using ${{A}^{2}}=2A+\dfrac{{{A}^{3}}}{2}$ to get matrix ${{A}^{3}}$. We then use the facts that $\det \left( xB \right)={{x}^{n}}\det \left( B \right)$, where ‘n’ is the order of the matrix, $\det \left( {{A}^{n}} \right)={{\left( \det \left( A \right) \right)}^{n}}$ and $\det \left( I \right)=1$ to find value of $\det \left( A \right)$. We then use the fact $\det \left( adj\left( A \right) \right)={{\left( \det \left( A \right) \right)}^{n-1}}$ to find $\det \left( adj\left( \dfrac{A}{2} \right) \right)$. We then multiply the obtained ${{A}^{3}}$ with ${{A}^{-1}}$ on both sides and use the fact $adj\left( A \right)=\left| A \right|{{A}^{-1}}$ to find $adj\left( A \right)$. Now, consider ${{\left( A-2I \right)}^{3}}$ and use the necessary relations to find the matrix and then find the trace of it.

Complete step by step answer:
According to the problem, we are given that A is a non-singular matrix of order $3\times 3$ whose entries are complex numbers and satisfies $2{{A}^{2}}=4A+{{A}^{3}}$. We need to find which of the given options are true.
Since A is a non-singular matrix, it will have an inverse.
Now, let us multiply both sides of $2{{A}^{2}}=4A+{{A}^{3}}$ with ${{A}^{-1}}$.
So, we get $2A.A.{{A}^{-1}}=4A.{{A}^{-1}}+A.A.A.{{A}^{-1}}$.
We know that $A.{{A}^{-1}}=I$. So, we get $2A.I=4I+A.A.I$.
$\Rightarrow 2A=4I+{{A}^{2}}$ ---(1).
But we have $2{{A}^{2}}=4A+{{A}^{3}}$, which gives ${{A}^{2}}=2A+\dfrac{{{A}^{3}}}{2}$. We substitute this in equation (1).
$\Rightarrow 2A=4I+2A+\dfrac{{{A}^{3}}}{2}$.
$\Rightarrow \dfrac{{{A}^{3}}}{2}=-4I$.
$\Rightarrow {{A}^{3}}=-8I$ ---(2).
Now, we have $\dfrac{{{A}^{3}}}{8}=-I$. Let us apply determinant on both sides.
$\Rightarrow \det \left( \dfrac{{{A}^{3}}}{8} \right)=\det \left( -I \right)$.
We know that $\det \left( xB \right)={{x}^{n}}\det \left( B \right)$, where ‘n’ is the order of the matrix.
$\Rightarrow \det \left( {{\left( \dfrac{A}{2} \right)}^{3}} \right)={{\left( -1 \right)}^{3}}\det \left( I \right)$.
We know that $\det \left( {{A}^{n}} \right)={{\left( \det \left( A \right) \right)}^{n}}$ and $\det \left( I \right)=1$.
So, we get $\det {{\left( \dfrac{A}{2} \right)}^{3}}=-1$.
$\Rightarrow \det \left( \dfrac{A}{2} \right)=-1$ ---(3).
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}\times \det \left( A \right)=-1$.
$\Rightarrow \dfrac{1}{8}\times \det \left( A \right)=-1$.
$\Rightarrow \det \left( A \right)=-8$ ---(4).
We know that $\det \left( adj\left( A \right) \right)={{\left( \det \left( A \right) \right)}^{n-1}}$, where ‘n’ is the order of the matrix.
So, we get $\det \left( adj\left( \dfrac{A}{2} \right) \right)={{\left( \det \left( \dfrac{A}{2} \right) \right)}^{3-1}}$.
$\Rightarrow \det \left( adj\left( \dfrac{A}{2} \right) \right)={{\left( \det \left( \dfrac{A}{2} \right) \right)}^{2}}$.
From equation (3), we get $\det \left( adj\left( \dfrac{A}{2} \right) \right)={{\left( -1 \right)}^{2}}$.
$\Rightarrow \det \left( adj\left( \dfrac{A}{2} \right) \right)=1$ ---(5).
From equation (2), we have ${{A}^{3}}=-8I$. Let us multiply both sides with ${{A}^{-1}}$.
$\Rightarrow A.A.A.{{A}^{-1}}=-8I.{{A}^{-1}}$.
$\Rightarrow A.A.I=-8{{A}^{-1}}$.
From equation (4), we have ${{A}^{2}}=\left| A \right|{{A}^{-1}}$.
We know that $adj\left( A \right)=\left| A \right|{{A}^{-1}}$. So, we get ${{A}^{2}}=adj\left( A \right)$ ---(6).
Now, let us consider ${{\left( A-2I \right)}^{3}}$.
$\Rightarrow {{\left( A-2I \right)}^{3}}={{A}^{3}}-6{{A}^{2}}+12A-8I$.
$\Rightarrow {{\left( A-2I \right)}^{3}}={{A}^{3}}-3\left( 4A+{{A}^{3}} \right)+12A-8I$.
$\Rightarrow {{\left( A-2I \right)}^{3}}={{A}^{3}}-12A-3{{A}^{3}}+12A-8I$.
$\Rightarrow {{\left( A-2I \right)}^{3}}=-2{{A}^{3}}-8I$.
From equation (2), we get ${{\left( A-2I \right)}^{3}}=-2\left( -8I \right)-8I$.
$\Rightarrow {{\left( A-2I \right)}^{3}}=16I-8I$.
$\Rightarrow {{\left( A-2I \right)}^{3}}=8I$.
$\Rightarrow {{\left( A-2I \right)}^{3}}=\left[ \begin{matrix}
   8 & 0 & 0 \\
   0 & 8 & 0 \\
   0 & 0 & 8 \\
\end{matrix} \right]$.
$\Rightarrow tr\left( {{\left( A-2I \right)}^{3}} \right)=8+8+8=24$ ---(7).
From equations (4), (5), (6) and (7), we can see all the given options are true.

∴ The correct options for the given problem are (a), (b), (c) and (d).

Note: We can see that this problem contains a heavy amount of calculations, so we need to do each step carefully while solving this problem. We can also assume a matrix with complex numbers to solve this problem but that becomes hectic to make the calculations and get the solution. We can also take the complex cube roots of –1 but, we just need to check whether at least one of the solutions satisfy the options.