GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 26 of 65

Marks: +1, -0

Consider a non-singular

2 × 2

square matrix

A

. If trace

(A)=4

and trace

(A^{2})=5

, the determinant of the matrix

{A}

is (up 1 decimal place).

Solution:

A is

2 × 2

matrix
Let

λ_{1}

and

λ_{2}

be the eigen value of matrix

{A}

.
Since, sum of eigen values

=

trace of a matrix
Therefore,

λ_{1}+λ_{2}=4

And

λ_{1}^{2}+λ_{2}^{2}=5

(λ_{1}+λ_{2})^{2}-(λ_{1}^{2}+λ_{2}^{2})=2 λ_{1} λ_{2}=4^{2}-5=11

λ_{1} λ_{2}=\;{11}/{2}=5.5

\det(A)=λ_{1} λ_{2}=5.5

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