GATE 2014 MA Eigen Values

Question

Answer 1

Answer is 81

Given that det(A-I)=0 and A, I are 3*3 matrix so there are 3 eigen values $\Lambda1,\Lambda2 and \Lambda3$

Trace(A) =13=($\Lambda1+ \Lambda2+\Lambda3$)

Det(A)=32=($\Lambda1*\Lambda2*\Lambda3$)

Now as det(A-$\Lambda$*I)=0

So $\Lambda$1=1

Substituting it in trace and det(A) we get other two eigen values as 8,4

=8^2+4^2+1^2

=81

Comments

Given that $|A-I|=0$

               $\Rightarrow|A|=|I|=1$---->$(1)$

Characteristic Equation $|A-\lambda I|=0$ 

                                       $\Rightarrow|A|=|\lambda I|$ 

                                       $\Rightarrow|\lambda I|=1$ (put the value of $|A|$ from the equation $1$)

                                        $\Rightarrow\lambda=1$------>$(2)$

Matrix $A$ is $3\times3$,So $3$ eigen values are possible.

Let say $\lambda_{1},\lambda_{2},\lambda_{3}=1$

Given that Trace of the Matrix, $=$Sum of the all leading diagonal elements$=$Sum of all Eigen values$=13$

            $\Rightarrow\lambda_{1}+\lambda_{2}+\lambda_{3}=13$

            Put the value of $\lambda_{3},$

                  $\Rightarrow\lambda_{1}+\lambda_{2}+1=13$

                 $\Rightarrow\lambda_{1}+\lambda_{2}=12$----->$(3)$

and $Det(A)=$Product of all Eigen values$=32$

             $\lambda_{1}\times\lambda_{2}\times\lambda_{3}=32$

                      Put the value of $\lambda_{3},$

             $\Rightarrow\lambda_{1}\times\lambda_{2}\times 1=32$

             $\Rightarrow\lambda_{1}\times\lambda_{2}=32$---->$(4)$

Now i used the concept of Quadratic Equation$:$

If $\lambda_{1},\lambda_{2}$ are two roots of the quadratic equation then,Quadratic Equation are

 $x^{2}-(\lambda_{1}+\lambda_{2})x+\lambda_{1}.\lambda_{2}=0$

   Put the value of $\lambda_{1}+\lambda_{2}=12$ and $\lambda_{1}.\lambda_{2}=32$ in the above equation

         $x^{2}-12x+32=0$

       Now, Solve the equation 

               $x^{2}-8x-4x+32=0$

        $\Rightarrow x(x-8)-4(x-8)=0$

        $\Rightarrow (x-8).(x-4)=0$

        $\Rightarrow x=4,8$

So,$\lambda_{1}=4,\lambda_{2}=8$

Now,we find the value of sum of the square of the Eigenvalues of $A=(\lambda_{1})^{2}+(\lambda_{2})^{2}+(\lambda_{3})^{2}$

                                                                                                                      $=4^{2}+8^{2}+1^{2}$

                                                                                                                      $=16+64+1$

                                                                                                                      $=81$