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Question Number 141152 by ZiYangLee last updated on 16/May/21

((3,5),(2,4) ) ((x),(y) )= (((x′)),((y′)) ) and ((a,b),(c,d) ) (((x′)),((y′)) )= ((x),(y) ) are true for any values of x,y,x′,y′, find the values of a,b,c,d.

$$\begin{pmatrix}{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{4}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}=\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\mathrm{are}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{values}\:\mathrm{of}\:{x},{y},{x}',{y}', \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a},{b},{c},{d}. \\ $$

Answered by iloveisrael last updated on 16/May/21

⇒ (((3 5)),((2 4)) ) (((a b)),((c d)) ) (((x′)),((y′)) ) = (((x′)),((y′)) ) ⇒ (((a b)),((c d)) ) = (((3 5)),((2 4)) )^(−1) = (1/2) ((( 4 −5)),((−2 3)) ) = ((( 2 −5/2)),((−1 3/2)) )

$$\Rightarrow\begin{pmatrix}{\mathrm{3}\:\:\:\:\mathrm{5}}\\{\mathrm{2}\:\:\:\:\mathrm{4}}\end{pmatrix}\:\begin{pmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{pmatrix}\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:{d}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\mathrm{5}}\\{\mathrm{2}\:\:\:\:\:\mathrm{4}}\end{pmatrix}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\:\:\:\mathrm{4}\:\:\:\:−\mathrm{5}}\\{−\mathrm{2}\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\:=\:\begin{pmatrix}{\:\:\:\:\mathrm{2}\:\:\:\:\:\:−\mathrm{5}/\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}/\mathrm{2}}\end{pmatrix} \\ $$

Commented by ZiYangLee last updated on 16/May/21

Can i know how to get the first row?

$$\mathrm{Can}\:\mathrm{i}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{first}\:\mathrm{row}? \\ $$

Commented by iloveisrael last updated on 16/May/21

by substitution

$${by}\:{substitution}\: \\ $$

Answered by mr W last updated on 16/May/21

((3,5),(2,4) ) ((x),(y) )= (((x′)),((y′)) ) as given ⇒ ((x),(y) )= ((3,5),(2,4) )^(−1) (((x′)),((y′)) ) ((x),(y) )= ((a,b),(c,d) ) (((x′)),((y′)) ) as given ⇒ ((a,b),(c,d) )= ((3,5),(2,4) )^(−1) = ((2,(−2.5)),((−1),(1.5)) )

$$\begin{pmatrix}{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{4}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:\:{as}\:{given} \\ $$$$\Rightarrow\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{4}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\: \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\:\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:{as}\:{given} \\ $$$$\Rightarrow\:\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{4}}\end{pmatrix}^{−\mathrm{1}} =\begin{pmatrix}{\mathrm{2}}&{−\mathrm{2}.\mathrm{5}}\\{−\mathrm{1}}&{\mathrm{1}.\mathrm{5}}\end{pmatrix} \\ $$

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