# Problem of the Week #26 - September 24th, 2012

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem!

-----

Problem: Let $A$ be the matrix given by

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

Prove that the characteristic polynomial $p(\lambda)$ of $A$ is given by

$p(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A),$

where $\text{tr}(A)$ denotes the trace of the matrix $A$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by BAdhi, dwsmith, Reckoner, Siron and Sudharaka. You can find Siron's solution below.

$$A = \left( \begin{array}{cc} a&b \\ c&d \end{array} \right)$$The characteristic polynomial $p(\lambda)$ can be found by calculating
$\det(A- \lambda I_2) = \det \left( \begin{array}{cc} a-\lambda&b \\ c&d-\lambda \end{array} \right) = (a-\lambda)(d-\lambda) - bc$
$=ad - d\lambda - a\lambda + \lambda^2 - bc$
$=\lambda^2 + (ad-bc) - \lambda (a+d)$

We have
$\det(A) = ad-bc$
and
$\mbox{tr}(A) = a+d$

therefore
$p(\lambda) = \lambda^2 - \lambda \mbox{tr}(A) + \det(A)$

Status
Not open for further replies.