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Question Number 216755 by Mingma last updated on 18/Feb/25
Answered by mr W last updated on 18/Feb/25
![w=au+bv [(1),(2),(r),(s) ]=a [(1),(0),(5),((−1)) ]+b [(1),(1),(3),(1) ] 1=a×1+b×1 2=a×0+b×1 ⇒b=2 ⇒a=−1 r=a×5+b×3=−5+6=1 s=a×(−1)+b×1=1+2=3](Q216758.png)
$$\boldsymbol{{w}}={a}\boldsymbol{{u}}+{b}\boldsymbol{{v}} \\ $$$$\begin{bmatrix}{\mathrm{1}}\\{\mathrm{2}}\\{{r}}\\{{s}}\end{bmatrix}={a}\begin{bmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{5}}\\{−\mathrm{1}}\end{bmatrix}+{b}\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$\mathrm{1}={a}×\mathrm{1}+{b}×\mathrm{1} \\ $$$$\mathrm{2}={a}×\mathrm{0}+{b}×\mathrm{1}\: \\ $$$$\Rightarrow{b}=\mathrm{2}\:\Rightarrow{a}=−\mathrm{1} \\ $$$${r}={a}×\mathrm{5}+{b}×\mathrm{3}=−\mathrm{5}+\mathrm{6}=\mathrm{1} \\ $$$${s}={a}×\left(−\mathrm{1}\right)+{b}×\mathrm{1}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$
Commented by Mingma last updated on 18/Feb/25
Perfect ��