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 some complex number. The determinant of the characteristic matrix is called characteristic determinant of matrix A which will, of course, be a polynomial of degree 3 in $\lambda$. The equation det (A - $\lambda$I) $=$ 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 $A=\left[\begin{array}{ccc}2& 1& 1\\ 2& 3& 4\\ -1& -1& -2\end{array}\right]$ are

• A. 2, 1, 1
• B. 2, 3, -2
• C. -1, 1, 3
• D. none of these

Answer 1

Answer: (C)

-1, 1, 3

Given, $A=\left[\begin{array}{ccc}2& 1& 1\\ 2& 3& 4\\ -1& -1& -2\end{array}\right]$
The characteristic equation of A is given by

$|A-\lambda I|=0$

$\Rightarrow \left|\begin{array}{ccc}2-\lambda & 1& 1\\ 2& 3-\lambda & 4\\ -1& -1& -2-\lambda \end{array}\right|=0$

$\Rightarrow {\lambda }^3-3{\lambda }^2-\lambda +3=0$

Here, $\lambda =1$ satisfies the equation

$\Rightarrow (\lambda -1)({\lambda }^2-2\lambda -3)=0$

$\Rightarrow (\lambda -1)\left(\lambda \right(\lambda -3)+1(\lambda -3\left)\right)=0$

$\Rightarrow (\lambda -1)(\lambda -3)(\lambda +1)=0$

$\Rightarrow \lambda =1,-1,3$

Hence, the eigen values are $1,-1,3$