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Question Number 15 by user1 last updated on 25/Jan/15

If A= [((  3),(−5)),((−4),(   2)) ],  show that A^2 −5A−14I=0

$$\mathrm{If}\:{A}=\begin{bmatrix}{\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\mathrm{2}}\end{bmatrix}, \\ $$$$\mathrm{show}\:\mathrm{that}\:{A}^{\mathrm{2}} −\mathrm{5}{A}−\mathrm{14}{I}=\mathrm{0} \\ $$

Answered by user1 last updated on 30/Oct/14

We have:  A^2 = [((   3),(−5)),((−4),(    2)) ] [((   3),(−5)),((−4),(    2)) ]  = [((3∙3+(−5)(−4)),(3∙(−5)+(−5)∙2)),((−4∙3+2∙(−4)),(−4∙(−5)+2∙2)) ]  = [((   29),(−25)),((−20),(   24)) ]  −5A=(−5) [((   3),(−5)),((−4),(    2)) ]= [((−15),(   25)),((  20),( −10)) ]  −14I=(−14) [(1,0),(0,1) ]= [((−14),(    0)),((    0),(−14)) ]  ∴ A^2 −5A−14I=A^2 +(−5)A+(−14I)  = [((   29),(−25)),((−20),(   24)) ]+ [((−15),(   25)),((  20),( −10)) ]+ [((−14),(    0)),((    0),(−14)) ]  = [(0,0),(0,0) ]  Hence,   A^2 −5A−14I=0

$$\mathrm{We}\:\mathrm{have}: \\ $$$${A}^{\mathrm{2}} =\begin{bmatrix}{\:\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\:\mathrm{2}}\end{bmatrix}\begin{bmatrix}{\:\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\:\mathrm{2}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\mathrm{3}\centerdot\mathrm{3}+\left(−\mathrm{5}\right)\left(−\mathrm{4}\right)}&{\mathrm{3}\centerdot\left(−\mathrm{5}\right)+\left(−\mathrm{5}\right)\centerdot\mathrm{2}}\\{−\mathrm{4}\centerdot\mathrm{3}+\mathrm{2}\centerdot\left(−\mathrm{4}\right)}&{−\mathrm{4}\centerdot\left(−\mathrm{5}\right)+\mathrm{2}\centerdot\mathrm{2}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\:\:\:\mathrm{29}}&{−\mathrm{25}}\\{−\mathrm{20}}&{\:\:\:\mathrm{24}}\end{bmatrix} \\ $$$$−\mathrm{5}{A}=\left(−\mathrm{5}\right)\begin{bmatrix}{\:\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\:\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{−\mathrm{15}}&{\:\:\:\mathrm{25}}\\{\:\:\mathrm{20}}&{\:−\mathrm{10}}\end{bmatrix} \\ $$$$−\mathrm{14}{I}=\left(−\mathrm{14}\right)\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{−\mathrm{14}}&{\:\:\:\:\mathrm{0}}\\{\:\:\:\:\mathrm{0}}&{−\mathrm{14}}\end{bmatrix} \\ $$$$\therefore\:{A}^{\mathrm{2}} −\mathrm{5}{A}−\mathrm{14}{I}={A}^{\mathrm{2}} +\left(−\mathrm{5}\right){A}+\left(−\mathrm{14}{I}\right) \\ $$$$=\begin{bmatrix}{\:\:\:\mathrm{29}}&{−\mathrm{25}}\\{−\mathrm{20}}&{\:\:\:\mathrm{24}}\end{bmatrix}+\begin{bmatrix}{−\mathrm{15}}&{\:\:\:\mathrm{25}}\\{\:\:\mathrm{20}}&{\:−\mathrm{10}}\end{bmatrix}+\begin{bmatrix}{−\mathrm{14}}&{\:\:\:\:\mathrm{0}}\\{\:\:\:\:\mathrm{0}}&{−\mathrm{14}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\mathrm{Hence},\:\:\:{A}^{\mathrm{2}} −\mathrm{5}{A}−\mathrm{14}{I}=\mathrm{0} \\ $$

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