Question

Let $ A $ be a square matrix of order $ 2 $ or $ 3 $ and $ I $ will be the identity matrix of the same order. Then the matrix $ A-\lambda I $ is called the characteristic matrix of the matrix $ A $ , where $ \lambda $ is the some complex number. The determinant of the characteristic matrix is called characteristic determinant of the matrix $ A $ which will, of course, be a polynomial of degree $ 3 $ in $ \lambda $ . The equation $ \left| A-\lambda I \right|=0 $ is called the characteristic equation of the matrix $ A $ and its roots (the values of $ \lambda $ ) are called characteristic roots or eigenvalues. It is also known that every square matrix has its characteristic equation.
The eigenvalues of the matrix $ A=\left[ \begin{matrix}
   2 & 1 & 1 \\
   2 & 3 & 4 \\
   -1 & -1 & -2 \\
\end{matrix} \right] $ are
A. $ 2,1,1 $
B. $ 2,3,-2 $
C. $ -1,1,3 $
D. None of these

Answer 1

Hint: In the problem, we have all the required data and formulas that we will use for this problem. We have given a square matrix of order $ 3 $ . In the problem they have also mentioned the procedure to calculate the characteristic equation, so we will calculate the characteristic equation and solve the equation for eigenvalues.

Complete step by step answer:
Given that,
 $ A=\left[ \begin{matrix}
   2 & 1 & 1 \\
   2 & 3 & 4 \\
   -1 & -1 & -2 \\
\end{matrix} \right] $
Let $ \lambda $ be the some complex number and $ I $ be the identity matrix of order $ 3 $ .
Now the value of $ \lambda I $ can be written as
 $ \begin{align}
  & \lambda I=\lambda \left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right] \\
 & \Rightarrow \lambda I=\left[ \begin{matrix}
   \lambda & 0 & 0 \\
   0 & \lambda & 0 \\
   0 & 0 & \lambda \\
\end{matrix} \right] \\
\end{align} $
Now the characteristic matrix can be written as
 $ \begin{align}
  & A-\lambda I=\left[ \begin{matrix}
   2 & 1 & 1 \\
   2 & 3 & 4 \\
   -1 & -1 & -2 \\
\end{matrix} \right]-\left[ \begin{matrix}
   \lambda & 0 & 0 \\
   0 & \lambda & 0 \\
   0 & 0 & \lambda \\
\end{matrix} \right] \\
 & \Rightarrow A-\lambda I=\left[ \begin{matrix}
   2-\lambda & 1 & 1 \\
   2 & 3-\lambda & 4 \\
   -1 & -1 & -2-\lambda \\
\end{matrix} \right] \\
\end{align} $
Now the characteristic equation of the matrix $ A $ can be written as
 $ \begin{align}
  & \left| A-\lambda I \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   2-\lambda & 1 & 1 \\
   2 & 3-\lambda & 4 \\
   -1 & -1 & -2-\lambda \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left( 2-\lambda \right)\left[ \left( 3-\lambda \right)\left( -2-\lambda \right)-\left( 4 \right)\left( -1 \right) \right]-1\left[ \left( 2 \right)\left( -2-\lambda \right)-\left( 4 \right)\left( -1 \right) \right]+1\left[ \left( 2 \right)\left( -1 \right)-\left( 3-\lambda \right)\left( -1 \right) \right]=0 \\
 & \Rightarrow \left( 2-\lambda \right)\left[ -6-3\lambda +2\lambda +{{\lambda }^{2}}+4 \right]-\left[ -4-2\lambda +4 \right]+\left[ -2+3-\lambda \right]=0 \\
 & \Rightarrow \left( 2-\lambda \right)\left( {{\lambda }^{2}}-\lambda -2 \right)-\left( -2\lambda \right)+\left( -\lambda +1 \right)=0 \\
 & \Rightarrow 2{{\lambda }^{2}}-2\lambda -4-{{\lambda }^{3}}+{{\lambda }^{2}}+2\lambda +2\lambda -\lambda +1=0 \\
 & \Rightarrow -{{\lambda }^{3}}+3{{\lambda }^{2}}+\lambda -3=0 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=0 \\
\end{align} $
We got the characteristic equation for the given matrix. To calculate the eigen values we need to find the roots of the above equation. For a cubic equation we can use trial and error method.
Let us take $ \lambda =-1 $ and check whether it satisfies the equation or not.
 $ \begin{align}
  & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3={{\left( -1 \right)}^{3}}-3{{\left( -1 \right)}^{2}}-\left( -1 \right)+3 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=-1-3\left( 1 \right)+1+3 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=-1-3+1+3 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=0 \\
\end{align} $
 $ \therefore \lambda =-1 $ satisfies the characteristic equation.
Let us take $ \lambda =1 $ and check whether it satisfies the equation or not.
 $ \begin{align}
  & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3={{\left( 1 \right)}^{3}}-3{{\left( 1 \right)}^{2}}-\left( 1 \right)+3 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=1-3\left( 1 \right)-1+3 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=1-3-1+3 \\
 & \Rightarrow {{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3=0 \\
\end{align} $
 $ \therefore \lambda =1 $ satisfies the characteristic equation.
Up to now we have the values of $ \lambda $ are $ 1,-1 $ . Now assume the third root as $ k $ , then we can write
 $ \begin{align}
  & \left( \lambda -1 \right)\left( \lambda +1 \right)\left( \lambda -k \right)={{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3 \\
 & \Rightarrow {{\lambda }^{2}}-1\left( \lambda -k \right)={{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3 \\
 & \Rightarrow {{\lambda }^{3}}-{{\lambda }^{2}}k-\lambda +k={{\lambda }^{3}}-3{{\lambda }^{2}}-\lambda +3 \\
\end{align} $
Equating on both sides we got $ k=3 $ .
Finally, the eigen values are $ 1,-1,3 $ .
 $ \therefore $ Option – C is correct answer.

Note:
 In this problem, they have mentioned the concept of characteristic equation and eigenvalues. But don’t think that in all problems we get these data, so it is better to have an idea about the characteristic equation and eigenvalues.