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Question Number 25623 by Sunilll last updated on 12/Dec/17

solve for A and B if 2A+B  [((6   3)),((6  −2)) ]  and 3A+2B  [((1    0)),((0    5)) ]

$${solve}\:{for}\:{A}\:{and}\:{B}\:{if}\:\mathrm{2}{A}+{B}\:\begin{bmatrix}{\mathrm{6}\:\:\:\mathrm{3}}\\{\mathrm{6}\:\:−\mathrm{2}}\end{bmatrix} \\ $$$${and}\:\mathrm{3}{A}+\mathrm{2}{B}\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{5}}\end{bmatrix} \\ $$

Answered by mrW1 last updated on 12/Dec/17

2A+B=C  ...(i)  3A+2B=D  ...(ii)  2(i)−(ii):  ⇒A=2C−D= [((11),6),((12),(−9)) ]  ⇒B=C−2A= [((−16),(−9)),((−18),(16)) ]

$$\mathrm{2}{A}+{B}={C}\:\:...\left({i}\right) \\ $$$$\mathrm{3}{A}+\mathrm{2}{B}={D}\:\:...\left({ii}\right) \\ $$$$\mathrm{2}\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow{A}=\mathrm{2}{C}−{D}=\begin{bmatrix}{\mathrm{11}}&{\mathrm{6}}\\{\mathrm{12}}&{−\mathrm{9}}\end{bmatrix} \\ $$$$\Rightarrow{B}={C}−\mathrm{2}{A}=\begin{bmatrix}{−\mathrm{16}}&{−\mathrm{9}}\\{−\mathrm{18}}&{\mathrm{16}}\end{bmatrix} \\ $$

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