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What are the eigenvalues of the matrix $P$ given below

$$P= \begin{pmatrix} a &1 &0 \\ 1& a& 1\\ 0&1 &a \end{pmatrix}$$                               

  1. $a, a -√2, a + √2$
  2. $a, a, a$
  3. $0, a, 2a$
  4. $-a, 2a, 2a$

8 Answers

Best answer
51 votes
51 votes
$\det(A-\lambda I)=0$

$\implies \begin{vmatrix} a-\lambda &1 &0 \\ 1 & a-\lambda &1 \\ 0& 1 & a-\lambda \end{vmatrix} = 0$

$\implies(a-\lambda)*[(a-\lambda)*(a-\lambda)-1*1] -1*[1*(a-\lambda)-0*1]+0*[1*1 - 0*(a-\lambda)] = 0$

$\implies(a-\lambda)^{3} - 2(a-\lambda)) = 0$

$\implies(a-\lambda)((a-\lambda)^2- 2) = 0$

$\implies(a-\lambda)((a-\lambda)^{2}- (\sqrt{2})^{2}) = 0$

$\implies(a-\lambda)(a- \lambda + \sqrt{2})(a-\lambda-\sqrt{2}) = 0$

Eigen values , $\lambda = a , a+\sqrt{2} , a - \sqrt{2}.$

Correct Answer: $A$
edited by
47 votes
47 votes
ANS (A)

1)Sum of eigen values =Trace of matrix(Sum of Diagonal elements)= 3a.

      Here all solution satisfy the condition.

2) Multiplication of eigen values= determinant of matrix=|A| |A|= a(a*a-1)-1(a)= a^3-2a

0nly option A satisfies it... so Ans is (A)
edited by
9 votes
9 votes
PROPERTY OF A MATRIX

1. SUM OF EIGEN VALUES =  TRACE OF MATRIX

2. PRODUCT OF EIGEN VALUES =  DETERMINANT OF MATRIX

therefore

Trace = Sum of elements of principal diagonal elements = ( a + a + a = 3a)

Determinant = ${a^{3}}$ - 2a

Put the values from given options and match both trace and determinant..

Ans: Option (a)
edited by
Answer:

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