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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-




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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