Find the coordinates of a and b

[chrisdk]

We are given two vectors \vec{a} i \vec{b}. We know that

2\vec{a}-3\vec{b}=\begin{bmatrix}0\\-17\end{bmatrix}

\vec{a}+4\vec{b}=\begin{bmatrix}11\\19\end{bmatrix}

I have tried to solve this through a set of equations with two unknown, but don't really know how to do that:

\begin{cases}2a-3b=[0,-17]\\a+4b=[11,19]\end{cases}

Thank you in advance for any clues:)

 

[Lord Crc]

As you can see from the original equations, the x and y components of the vectors are independent.

 

[Rasalhague]

Try to express what you know as a matrix equation: a matrix containing the components of the unknown vectors, multiplied by a matrix of some numbers equals another matrix of some numbers.

 

[HallsofIvy]

We are given two vectors \vec{a} i \vec{b}. We know that

2\vec{a}-3\vec{b}=\begin{bmatrix}0\\-17\end{bmatrix}

\vec{a}+4\vec{b}=\begin{bmatrix}11\\19\end{bmatrix}

I have tried to solve this through a set of equations with two unknown, but don't really know how to do that:

\begin{cases}2a-3b=[0,-17]\\a+4b=[11,19]\end{cases}

Thank you in advance for any clues:)

You can solve just as you would any pair of equations, just being careful to keep both components on the right. For example, if I had "2a- 3b= " and "a+ 4b= " whether they were number, vectors, matrices, or whatever, I would think "If I multiply the second equation by 2 and subtract, I will eliminate the 'a'!"

Okay, 2a- 3b= [0,-17] and 2a+ 8b= [22, 38]. Subtracting the first from the second, 11b= [22-0, 38+17]= [22, 55] and now I see that b= [2, 5].

 

[Rasalhague]

Oh, yeah, that's simpler than what I had in mind:

\begin{bmatrix}a_1 & b_1\\a_2 & b_2\end{bmatrix}\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix}=\begin{bmatrix}0 & 11\\-17 & 19\end{bmatrix}

\begin{bmatrix}a_1 & b_1\\a_2 & b_2\end{bmatrix}\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix}\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix}^{-1}=\begin{bmatrix}0 & 11\\-17 & 19\end{bmatrix}\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix}^{-1}

\begin{bmatrix}a_1 & b_1\\a_2 & b_2\end{bmatrix}=\begin{bmatrix}0 & 11\\-17 & 19\end{bmatrix}\begin{bmatrix}\frac{4}{11} & -\frac{1}{11}\\\frac{3}{11} & \frac{2}{11}\end{bmatrix} = ...

 

[Redbelly98]

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