Question Number 143200 by Guddone last updated on 11/Jun/21
Answered by Dwaipayan Shikari last updated on 11/Jun/21
![2x+3y= [((2 3)),((4 0)) ]⇒6x+9y= [(( 6 9)),(( 12 0)) ]→(a) 3x+2y= [((2 −2)),((−1 5)) ]⇒6x+4y= [(( 4 −4)),(( −2 10)) ]→(b) a−b ⇒5y= [((2 13 )),((14 −10)) ]⇒y= [(((2/5) ((13)/5))),((((14)/5) −2)) ] x=(1/2)( [((2 3)),((4 0)) ]− [(((6/5) ((39)/5))),(( ((42)/5) −6)) ])=(1/2)( [(((4/5) ((−24)/5))),((((−22)/5) 6)) ])= [(((2/5) −((12)/5))),((−((11)/5) 3)) ]](https://www.tinkutara.com/question/Q143214.png)
$$\mathrm{2}{x}+\mathrm{3}{y}=\begin{bmatrix}{\mathrm{2}\:\:\mathrm{3}}\\{\mathrm{4}\:\:\mathrm{0}}\end{bmatrix}\Rightarrow\mathrm{6}{x}+\mathrm{9}{y}=\begin{bmatrix}{\:\mathrm{6}\:\:\:\:\mathrm{9}}\\{\:\mathrm{12}\:\:\mathrm{0}}\end{bmatrix}\rightarrow\left({a}\right) \\ $$$$\mathrm{3}{x}+\mathrm{2}{y}=\begin{bmatrix}{\mathrm{2}\:\:−\mathrm{2}}\\{−\mathrm{1}\:\:\mathrm{5}}\end{bmatrix}\Rightarrow\mathrm{6}{x}+\mathrm{4}{y}=\begin{bmatrix}{\:\mathrm{4}\:\:−\mathrm{4}}\\{\:−\mathrm{2}\:\:\mathrm{10}}\end{bmatrix}\rightarrow\left({b}\right) \\ $$$${a}−{b}\:\Rightarrow\mathrm{5}{y}=\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\:\mathrm{13}\:\:}\\{\mathrm{14}\:−\mathrm{10}}\end{bmatrix}\Rightarrow{y}=\begin{bmatrix}{\frac{\mathrm{2}}{\mathrm{5}}\:\:\frac{\mathrm{13}}{\mathrm{5}}}\\{\frac{\mathrm{14}}{\mathrm{5}}\:−\mathrm{2}}\end{bmatrix} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\begin{bmatrix}{\mathrm{2}\:\mathrm{3}}\\{\mathrm{4}\:\mathrm{0}}\end{bmatrix}−\begin{bmatrix}{\frac{\mathrm{6}}{\mathrm{5}}\:\frac{\mathrm{39}}{\mathrm{5}}}\\{\:\frac{\mathrm{42}}{\mathrm{5}}\:\:−\mathrm{6}}\end{bmatrix}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\begin{bmatrix}{\frac{\mathrm{4}}{\mathrm{5}}\:\frac{−\mathrm{24}}{\mathrm{5}}}\\{\frac{−\mathrm{22}}{\mathrm{5}}\:\:\mathrm{6}}\end{bmatrix}\right)=\begin{bmatrix}{\frac{\mathrm{2}}{\mathrm{5}}\:−\frac{\mathrm{12}}{\mathrm{5}}}\\{−\frac{\mathrm{11}}{\mathrm{5}}\:\:\:\mathrm{3}}\end{bmatrix} \\ $$