Matrices and Determinants Set… | JEE Advanced 2021 P1

JEE Advanced 2021 Paper 1 Mathematics: Paragraph 3

Let

A

be a

3\times3

real matrix whose characteristic polynomial is

\lambda^{3}-6\lambda^{2}+11\lambda-5

. Denote its eigenvalues (counted with multiplicity) by

\alpha,\beta,\gamma

. They satisfy

\alpha+\beta+\gamma=6

\alpha\beta+\beta\gamma+\gamma\alpha=11

and

\alpha\beta\gamma=5

. Note that the eigenvalues need not all be real.

Question 1

Numerical answer

The value of

\operatorname{tr}(A^{4})

is ____.

Show answer and solution

Answer: 74

For a matrix,

\operatorname{tr}(A^k)=p_k=\alpha^k+\beta^k+\gamma^k

, and these power sums follow Newton's identities in terms of

e_1=6

e_2=11

e_3=5

. We get

p_1=6

p_2=e_1p_1-2e_2=36-22=14

p_3=e_1p_2-e_2p_1+3e_3=84-66+15=33

and

p_4=e_1p_3-e_2p_2+e_3p_1=198-154+30=74

. So

\operatorname{tr}(A^4)=74

, even though two eigenvalues are complex conjugates, because the symmetric functions are real. A direct numerical computation with the roots confirms

74

Question 2

Numerical answer

The value of

\operatorname{tr}(A^{-2})

is ____. Give the answer correct to two decimal places.

Show answer and solution

Answer: 2.44

Since

\det A=5\ne0

A

is invertible and its eigenvalues are

1/\alpha,1/\beta,1/\gamma

, so

\operatorname{tr}(A^{-2})=\sum \dfrac1{\alpha^2}

. Now

\sum\dfrac1\alpha=\dfrac{e_2}{e_3}=\dfrac{11}{5}

and

\sum\dfrac1{\alpha\beta}=\dfrac{e_1}{e_3}=\dfrac65

. Using

\sum x^2=(\sum x)^2-2\sum xy

for the reciprocals,

\operatorname{tr}(A^{-2})=\left(\dfrac{11}{5}\right)^2-2\cdot\dfrac65=\dfrac{121}{25}-\dfrac{60}{25}=\dfrac{61}{25}=2.44

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