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Quiz 6 Form A \hfil \break
Nov 10, 2017

[10]~~ 1.  ${\cal L}(u_2(t)(t + 5)) =  \underline{~~e^{-2s}({1 \over s^2} + {7 \over s})~~}$

${\cal L}(u_2(t)(t + 5)) = e^{-2s}{\cal L}(t+2 + 5)) =
e^{-2s}{\cal L}(t+7(1))) =e^{-2s}({\cal L}(t)+7{\cal L}(1))) =
e^{-2s}({1 \over s^2} + {7 \over s})$

\v

[10]~~ 2.  Find the eigenvalues and corresponding eigenvectors for $ A = \left(\matrix{1 & 2 \cr 2 & 4}\right)$ 

An eigenvalue of $A$ is    $\underline{~~0~~}$. The eigenvectors corresponding to this eigenvalue



\rightline{ are all nonzero multiples of  $\underline{\left(\matrix{~2 \cr -1}\right)}$.}

An 2nd eigenvalue of $A$ is    $\underline{~~5~~}$. The eigenvectors corresponding to this \v

\rightline{ eigenvalue are all nonzero multiples of  $\underline{~~\left(\matrix{1 \cr 2 }\right)~~}$.}


Note it is ``obvious'' that 0 is an eigenvalue of $A$ since the rows/columns of A are linearly dependent (e.g., row 2 is twice row 1) or equivalently notice $det(A) = 0$.

Since $\left(\matrix{1 & 2 \cr 2 & 4}\right)\left(\matrix{ ~2 \cr -1 }\right) = \left(\matrix{0 \cr 0}\right)$, all eigenvectors corresponding to the eigenvalue 0 are 

\rightline{all nonzero multiples of
$\left(\matrix{~2 \cr -1}\right)$}

To find eigenvalues in general, solve $det(A - rI) = 0$

$\left|\matrix{1-r & 2 \cr 2 & 4-r}\right| = (1-r)(4-r) - 4
= r^2 - 5r + 4 - 4 = r(r-5)= 0$

Thus the eigenvalues of $A$ are $r = 0,~ 5$

For $r = 5$: 
$\left(\matrix{1-5 & 2 \cr 2 & 4-5}\right) =
\left(\matrix{-4 & 2 \cr 2 & -1}\right)$
~~and~~
$\left(\matrix{-4 & 2 \cr 2 & -1}\right)\left(\matrix{1 \cr 2 }\right) = \left(\matrix{0 \cr 0}\right)$



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