Question

Let the trace and determinant of a matrix A=[acbd] be 6 and 16 respectively. The eigenvalues of A are​

Answer 1

SOLUTION

GIVEN

Let the trace and determinant of a matrix

$A = \displaystyle\begin{pmatrix} a & c \\ b & d \end{pmatrix}$

be 6 and 16 respectively.

TO DETERMINE

The eigenvalues of A

FORMULA TO BE IMPLEMENTED

If trace of a matrix and determinant of a matrix is given the characteristic polynomial is

p(t) = t² - ( trace A) t + det A

EVALUATION

Here the given matrix is

$A = \displaystyle\begin{pmatrix} a & c \\ b & d \end{pmatrix}$

So the characteristic polynomial is

p(t) = t² - ( trace A) t + det A

⟹ p(t) = t² - 6t + 16

So the roots of the polynomial is given by

t² - 6t + 16 = 0

$\displaystyle \sf{t = \frac{( - 6) \pm \sqrt{ {( - 6)}^{2} - 4.1.16} }{2.1} }$

$\displaystyle \sf{ \implies \: t = \frac{6 \pm \sqrt{36 - 64}}{2} }$

$\displaystyle \sf{ \implies \: t = \frac{6 \pm \sqrt{ - 28}}{2} }$

$\displaystyle \sf{ \implies \: t = \frac{6 \pm 2i \sqrt{7}}{2} }$

$\displaystyle \sf{ \implies \: t = 3 \pm \: i \sqrt{7} }$

Hence the The eigenvalues of A are

$\sf{3 + i \sqrt{7} \: \: \: and \: \: 3 - i \sqrt{7} }$

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