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Question Number 57790 by pete last updated on 12/Apr/19

Given N= [((5      3)),((6       4)) ]and P= [((4     −3)),((−6    5)) ],  find NP and deduce the inverse of P.

$$\mathrm{Given}\:\mathrm{N}=\begin{bmatrix}{\mathrm{5}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{6}\:\:\:\:\:\:\:\mathrm{4}}\end{bmatrix}\mathrm{and}\:\mathrm{P}=\begin{bmatrix}{\mathrm{4}\:\:\:\:\:−\mathrm{3}}\\{−\mathrm{6}\:\:\:\:\mathrm{5}}\end{bmatrix}, \\ $$$$\mathrm{find}\:\mathrm{NP}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{P}. \\ $$

Answered by math1967 last updated on 12/Apr/19

 [(5,3),(6,4) ] [(4,(−3)),((−6),5) ]= [((20−18),(−15+15)),((24−24),(−18+20)) ]   [(2,0),(0,2) ]=2I  ∴NP=2I⇒NPP^(−1) =2IP^(−1)   ⇒NI=2P^(−1) ∴2P^(−1) =N  ∴P^(−1) = [(2,((−3)/2)),((−3),(5/2)) ]

$$\begin{bmatrix}{\mathrm{5}}&{\mathrm{3}}\\{\mathrm{6}}&{\mathrm{4}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}&{−\mathrm{3}}\\{−\mathrm{6}}&{\mathrm{5}}\end{bmatrix}=\begin{bmatrix}{\mathrm{20}−\mathrm{18}}&{−\mathrm{15}+\mathrm{15}}\\{\mathrm{24}−\mathrm{24}}&{−\mathrm{18}+\mathrm{20}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}}\end{bmatrix}=\mathrm{2}{I} \\ $$$$\therefore{NP}=\mathrm{2}{I}\Rightarrow{NPP}^{−\mathrm{1}} =\mathrm{2}{IP}^{−\mathrm{1}} \\ $$$$\Rightarrow{NI}=\mathrm{2}{P}^{−\mathrm{1}} \therefore\mathrm{2}{P}^{−\mathrm{1}} ={N} \\ $$$$\therefore{P}^{−\mathrm{1}} =\begin{bmatrix}{\mathrm{2}}&{\frac{−\mathrm{3}}{\mathrm{2}}}\\{−\mathrm{3}}&{\frac{\mathrm{5}}{\mathrm{2}}}\end{bmatrix} \\ $$

Commented by pete last updated on 12/Apr/19

Thanks so much

$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much} \\ $$

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