Question and Answers Forum

All Questions      Topic List

Matrices and Determinants Questions

Previous in All Question      Next in All Question      

Previous in Matrices and Determinants      Next in Matrices and Determinants      

Question Number 183965 by nadovic last updated on 01/Jan/23

A linear transformation E, of the x−y plane is defined as E:(x, y) → (2x+y, 2x+3y) Find the equation of the line that remains invariant under the transformation.

$$\mathrm{A}\:\mathrm{linear}\:\mathrm{transformation}\:\boldsymbol{{E}},\:\mathrm{of}\:\mathrm{the} \\ $$$${x}−{y}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{E}}:\left({x},\:{y}\right)\:\rightarrow\:\left(\mathrm{2}{x}+{y},\:\mathrm{2}{x}+\mathrm{3}{y}\right) \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{that} \\ $$$$\mathrm{remains}\:\mathrm{invariant}\:\mathrm{under}\:\mathrm{the} \\ $$$$\mathrm{transformation}. \\ $$

Answered by mr W last updated on 01/Jan/23

say the line is: ax+by+c=0 under transformation: a(2x+y)+b(2x+3y)+c=0 ⇒(2a+2b)x+(a+3b)y+c=0 if c≠0: ((2a+2b)/c)=(a/c) b((a+3b)/c)=(b/c) ⇒a=−2b ⇒−2bx+by+c=0 or 2x−y+k=0 ✓ if c=0: ((2a+2b)/a)=((a+3b)/b) 2((b/a))^2 −((b/a))−1=0 (2(b/a)+1)((b/a)−1)=0 ⇒(b/a)=−(1/2) or (b/a)=1 ⇒2x−y=0 (included in 2x−y+k=0) or x+y=0 ✓ summary: the searched line is 2x−y+k=0 or x+y=0

$${say}\:{the}\:{line}\:{is}: \\ $$$${ax}+{by}+{c}=\mathrm{0} \\ $$$${under}\:{transformation}: \\ $$$${a}\left(\mathrm{2}{x}+{y}\right)+{b}\left(\mathrm{2}{x}+\mathrm{3}{y}\right)+{c}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{a}+\mathrm{2}{b}\right){x}+\left({a}+\mathrm{3}{b}\right){y}+{c}=\mathrm{0} \\ $$$$\underline{\boldsymbol{{if}}\:\boldsymbol{{c}}\neq\mathrm{0}:} \\ $$$$\frac{\mathrm{2}{a}+\mathrm{2}{b}}{{c}}=\frac{{a}}{{c}} \\ $$$${b}\frac{{a}+\mathrm{3}{b}}{{c}}=\frac{{b}}{{c}} \\ $$$$\Rightarrow{a}=−\mathrm{2}{b} \\ $$$$\Rightarrow−\mathrm{2}{bx}+{by}+{c}=\mathrm{0} \\ $$$${or}\:\mathrm{2}{x}−{y}+{k}=\mathrm{0}\:\checkmark \\ $$$$\underline{\boldsymbol{{if}}\:\boldsymbol{{c}}=\mathrm{0}:} \\ $$$$\frac{\mathrm{2}{a}+\mathrm{2}{b}}{{a}}=\frac{{a}+\mathrm{3}{b}}{{b}} \\ $$$$\mathrm{2}\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} −\left(\frac{{b}}{{a}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}\frac{{b}}{{a}}+\mathrm{1}\right)\left(\frac{{b}}{{a}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{b}}{{a}}=−\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\frac{{b}}{{a}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{x}−{y}=\mathrm{0}\:\left({included}\:{in}\:\mathrm{2}{x}−{y}+{k}=\mathrm{0}\right)\: \\ $$$$\:\:\:\:{or}\:{x}+{y}=\mathrm{0}\:\checkmark \\ $$$$\underline{\boldsymbol{{summary}}:} \\ $$$${the}\:{searched}\:{line}\:{is} \\ $$$$\mathrm{2}{x}−{y}+{k}=\mathrm{0}\:{or} \\ $$$${x}+{y}=\mathrm{0} \\ $$

Commented by nadovic last updated on 01/Jan/23

Thanks Sir, but the book gave two answers: 2x−y+c=0 and y=−x+c

$${Thanks}\:{Sir},\:{but}\:\:{the}\:{book}\:{gave}\:{two} \\ $$$${answers}:\:\mathrm{2}{x}−{y}+{c}=\mathrm{0}\:{and}\:{y}=−{x}+{c} \\ $$

Commented by mr W last updated on 01/Jan/23

y=−x+c is not correct. but y=−x is ok. see above.

$${y}=−{x}+{c}\:{is}\:{not}\:{correct}.\:{but}\:{y}=−{x}\:{is} \\ $$$${ok}.\:{see}\:{above}. \\ $$

Commented by nadovic last updated on 01/Jan/23

Alright.. Thank you very much.

$$\mathrm{Alright}..\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com