Question

Let $M=\left[ \begin{matrix}
   0 & 1 & a \\
   1 & 2 & 3 \\
   3 & b & 1 \\
\end{matrix} \right]$ and $adj\left( M \right)=\left[ \begin{matrix}
   -1 & 1 & -1 \\
   8 & -6 & 2 \\
   -5 & 3 & -1 \\
\end{matrix} \right]$ where a and b are real numbers.
Which of the following options is/are correct?
[a] $\det \left( adj\left( {{M}^{2}} \right) \right)=81$
[b] $a+b=3$
[c] If \[M\left[ \begin{matrix}
   \alpha \\
   \beta \\
   \gamma \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]\], then $\alpha +\beta +\gamma =3$
[d] $adj\left( {{M}^{-1}} \right)+adj{{\left( M \right)}^{-1}}=-M$

Answer 1

Hint: Use the fact that $Aadj\left( A \right)=\det \left( A \right)I$, where I is the identity matrix of same order as A. Hence determine the values of a, b and det(A). Hence verify which of the options are correct and which are incorrect.

Complete step-by-step answer:

We have
$M=\left[ \begin{matrix}
   0 & 1 & a \\
   1 & 2 & 3 \\
   3 & b & 1 \\
\end{matrix} \right]$ and $adj\left( M \right)=\left[ \begin{matrix}
   -1 & 1 & -1 \\
   8 & -6 & 2 \\
   -5 & 3 & -1 \\
\end{matrix} \right]$
We know that $Aadj\left( A \right)=\det \left( A \right)I$, where I is the identity matrix of same order as A.
Hence, we have
$\left[ \begin{matrix}
   0 & 1 & a \\
   1 & 2 & 3 \\
   3 & b & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   -1 & 1 & -1 \\
   8 & -6 & 2 \\
   -5 & 3 & -1 \\
\end{matrix} \right]=\det \left( M \right)\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$
Hence, we have
$\left[ \begin{matrix}
   8-5a & 3\left( a-2 \right) & 2-a \\
   0 & -2 & 0 \\
   8\left( b-1 \right) & 6\left( 1-b \right) & 2\left( -2+y \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   \det \left( M \right) & 0 & 0 \\
   0 & \det \left( M \right) & 0 \\
   0 & 0 & \det \left( M \right) \\
\end{matrix} \right]$
Hence, we have
$\begin{align}
  & 3\left( a-2 \right)=0\Rightarrow a=2 \\
 & \det \left( M \right)=-2 \\
 & 8\left( b-1 \right)=0\Rightarrow b=1 \\
\end{align}$
Checking option [a] We know that $\det \left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-1}}$ where n is the order of A.
Hence, we have
$\det \left( adj\left( {{M}^{2}} \right) \right)=\det {{\left( {{M}^{2}} \right)}^{3-1}}=\det {{\left( {{M}^{2}} \right)}^{2}}$
We know that $\det \left( {{M}^{2}} \right)=\det {{\left( M \right)}^{2}}$
Hence, we have
$\det \left( adj\left( {{M}^{2}} \right) \right)=\det {{\left( M \right)}^{4}}={{\left( -2 \right)}^{4}}=16$
Checking option [b]: We have a =2 and b= 1. Hence, we have a+b = 2+1 = 3
Checking option [c]:
We have $M\left[ \begin{matrix}
   \alpha \\
   \beta \\
   \gamma \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]\ \ \ \ \left( i \right)$
Since $\det \left( M \right)\ne 0$, we have ${{M}^{-1}}$ exists.
Pre-multiplying equation (i) by ${{M}^{-1}}$, we get
$\left[ \begin{matrix}
   \alpha \\
   \beta \\
   \gamma \\
\end{matrix} \right]={{M}^{-1}}\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]$
We know that ${{A}^{-1}}=\dfrac{1}{\det \left( A \right)}adj\left( A \right)$
Hence, we have
${{M}^{-1}}=\dfrac{1}{-2}\left[ \begin{matrix}
   -1 & 1 & -1 \\
   8 & -6 & 2 \\
   -5 & 3 & -1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   \dfrac{1}{2} & \dfrac{-1}{2} & \dfrac{1}{2} \\
   -4 & 3 & -1 \\
   \dfrac{5}{2} & \dfrac{-3}{2} & \dfrac{1}{2} \\
\end{matrix} \right]$
Hence, we have
$\left[ \begin{matrix}
   \alpha \\
   \beta \\
   \gamma \\
\end{matrix} \right]=\left[ \begin{matrix}
   \dfrac{1}{2} & \dfrac{-1}{2} & \dfrac{1}{2} \\
   -4 & 3 & -1 \\
   \dfrac{5}{2} & \dfrac{-3}{2} & \dfrac{1}{2} \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   -1 \\
   1 \\
\end{matrix} \right]$
Hence, we have
$\alpha =1,\beta =-1,\gamma =1$
Hence, we have
$\alpha +\beta +\gamma =1$
Checking option [d]:
We know that $Aadj\left( A \right)=\det \left( A \right)I$
Hence, we have
${{M}^{-1}}adj\left( {{M}^{-1}} \right)=\det \left( {{M}^{-1}} \right)I$
Pre-multiplying both sides by M, we get
$adj\left( {{M}^{-1}} \right)=\dfrac{1}{\det \left( M \right)}M=-\dfrac{M}{2}$
We know that $A{{A}^{-1}}=I$
Hence, we have
$adj\left( M \right)adj{{\left( M \right)}^{-1}}=I$
Pre-multiplying both sides by M, we get
$\begin{align}
  & Madj\left( M \right)adj{{\left( M \right)}^{-1}}=M \\
 & \det \left( M \right)Iadj{{\left( M \right)}^{-1}}=M \\
 & \Rightarrow adj{{\left( M \right)}^{-1}}=\dfrac{M}{\det \left( M \right)}=-\dfrac{M}{2} \\
\end{align}$
Hence, we have
$adj\left( {{M}^{-1}} \right)+adj{{\left( M \right)}^{-1}}=-\dfrac{M}{2}-\dfrac{M}{2}=-M$

So, the correct answers are “Option b and d”.

Note: [1] The properties of adjoint of a matrix are very important in matrix algebra and students are advised to remember them. In particular the following two properties are very important
[a] $Aadj\left( A \right)=adj\left( A \right)A=\det \left( A \right)I$
[b] $\det \left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-1}}$
The calculations in this problem would have been very long and tedious if we had not used these two properties.