The Eigen values of a matrix are the roots of its characteristic equation.
The largest eigen value of the matrix
$A = \begin{bmatrix}
1 & 4 & 16 \\[0.3em]
4 & 16 & 1 \\[0.3em]
16 & 1 & 4 \\
\end{bmatrix}$ is
$\textbf{(A)}\qquad 16\qquad \textbf{(B)}\qquad 21\qquad \textbf{(C)}\qquad 48\qquad \textbf{(D)}\qquad 64\qquad $
IIT JAM 2016 Problem Number 13
Finding the largest eigen value of a matrix
6 out of 10
Higher Algebra : S K Mapa
First hint
In order to find the largest eigen value of the given matrix we have to find all the eigen values of the given matrix, then we can find the largest among them. Now it is very easy to find the eigen values. Give it a try!!!
Second Hint
Now the process of rank starts with finding the characteristics polynomial of the given matrix, i.e., $\textbf{ det }(A-\lambda I)=0$. We have to find the value of this $\lambda $ which is the eigen value. Now it is very easy to find the determinant of $(A-\lambda I)$. Try to cook this up.
Final Step
So, now let's do some calculation.
$ \textbf{det}(A-\lambda I) = \left| \begin{matrix}
1-\lambda & 4 & 16 \\[0.3em]
4 & 16-\lambda & 1 \\[0.3em]
16 & 1 & 4 -\lambda \\
\end{matrix} \right| =0 $
Now we are half way done just our calculation part remains.
$|A-\lambda I|=(1-\lambda)[64-20 \lambda + \lambda ^2-1]-4[16-4 \lambda -16]+16[4-256+16 \lambda]=0$
$\Rightarrow \quad (1- \lambda )( \lambda ^2-20 \lambda +63)+16 \lambda +256 \lambda -4032=0$
$\Rightarrow \quad \lambda ^2-20 \lambda +63 - \lambda ^3+20 \lambda ^2-63 \lambda +272 \lambda -4032=0$
$\Rightarrow \quad - \lambda ^3+21 \lambda ^2+189 \lambda -3969=0$
$ \Rightarrow \quad \lambda ^3-21 \lambda ^2-189 \lambda +3969=0 $
$\Rightarrow \quad(\lambda-21)(\lambda^2-189)=0$
Therefore $\lambda=21, \quad -13.747727, \quad 13.747727$
So from here we can clearly see
$\lambda=21$ is the largest among all the eigen values.
The Eigen values of a matrix are the roots of its characteristic equation.
The largest eigen value of the matrix
$A = \begin{bmatrix}
1 & 4 & 16 \\[0.3em]
4 & 16 & 1 \\[0.3em]
16 & 1 & 4 \\
\end{bmatrix}$ is
$\textbf{(A)}\qquad 16\qquad \textbf{(B)}\qquad 21\qquad \textbf{(C)}\qquad 48\qquad \textbf{(D)}\qquad 64\qquad $
IIT JAM 2016 Problem Number 13
Finding the largest eigen value of a matrix
6 out of 10
Higher Algebra : S K Mapa
First hint
In order to find the largest eigen value of the given matrix we have to find all the eigen values of the given matrix, then we can find the largest among them. Now it is very easy to find the eigen values. Give it a try!!!
Second Hint
Now the process of rank starts with finding the characteristics polynomial of the given matrix, i.e., $\textbf{ det }(A-\lambda I)=0$. We have to find the value of this $\lambda $ which is the eigen value. Now it is very easy to find the determinant of $(A-\lambda I)$. Try to cook this up.
Final Step
So, now let's do some calculation.
$ \textbf{det}(A-\lambda I) = \left| \begin{matrix}
1-\lambda & 4 & 16 \\[0.3em]
4 & 16-\lambda & 1 \\[0.3em]
16 & 1 & 4 -\lambda \\
\end{matrix} \right| =0 $
Now we are half way done just our calculation part remains.
$|A-\lambda I|=(1-\lambda)[64-20 \lambda + \lambda ^2-1]-4[16-4 \lambda -16]+16[4-256+16 \lambda]=0$
$\Rightarrow \quad (1- \lambda )( \lambda ^2-20 \lambda +63)+16 \lambda +256 \lambda -4032=0$
$\Rightarrow \quad \lambda ^2-20 \lambda +63 - \lambda ^3+20 \lambda ^2-63 \lambda +272 \lambda -4032=0$
$\Rightarrow \quad - \lambda ^3+21 \lambda ^2+189 \lambda -3969=0$
$ \Rightarrow \quad \lambda ^3-21 \lambda ^2-189 \lambda +3969=0 $
$\Rightarrow \quad(\lambda-21)(\lambda^2-189)=0$
Therefore $\lambda=21, \quad -13.747727, \quad 13.747727$
So from here we can clearly see
$\lambda=21$ is the largest among all the eigen values.
