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A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix} -8 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 8 \end{bmatrix}$$

and

$$M\begin{bmatrix} 1 \\ 5 \end{bmatrix} = \begin{bmatrix} -8 \\ 7 \end{bmatrix}.$$

Evaluate $$M\begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$

What is the question essentially asking? Isn't this just a matrix equation?

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  • $\begingroup$ You could formulate it that way. There is a couple of approaches to the problem, all ending up in the same place if done correctly. $\endgroup$
    – gt6989b
    Jun 17 '20 at 14:55
  • $\begingroup$ What would be the way to approach? $\endgroup$
    – user713999
    Jun 17 '20 at 14:59
  • $\begingroup$ the question is that you have to compute the value of $M\begin{bmatrix}1\\1\end{bmatrix}$ $\endgroup$ Jun 17 '20 at 15:00
  • $\begingroup$ @Daniel there are 4 free variable namely, $M_{11}, M_{12}, M_{21}$ and $M_{22}$. $\endgroup$ Jun 17 '20 at 15:20
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We know that

$$M\begin{pmatrix} -8 & 1 \\ 1 & 5 \end{pmatrix} = \begin{pmatrix} 3 &-8\\ 8 & 7 \end{pmatrix}.$$ Multiply that equation from right by $$ \begin{pmatrix} -8 & 1 \\ 1 & 5 \end{pmatrix}^{-1}=\frac{1}{-41}\begin{pmatrix} 5 & -1 \\ -1 & -8 \end{pmatrix} $$ to get $M$.

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Note that the given input vectors form a basis of $\Bbb{R}^2$. So we can express any vector in $\Bbb{R}^2$ in terms of the given vectors such as: $$\begin{bmatrix}a\\b\end{bmatrix}={\small\left(\frac{b-5a}{41}\right)}\begin{bmatrix}-8\\1\end{bmatrix}+{\small \left(\frac{a+8b}{41}\right)}\begin{bmatrix}1\\5\end{bmatrix}. \tag{1}$$ In particular, $$\begin{bmatrix}1\\1\end{bmatrix}={\small\frac{-4}{41}}\begin{bmatrix}-8\\1\end{bmatrix}+{\small\frac{9}{41}}\begin{bmatrix}1\\5\end{bmatrix}.$$ Thus \begin{align*} M\begin{bmatrix}1\\1\end{bmatrix}& ={\small \frac{-4}{41}}\color{red}{M\begin{bmatrix}-8\\1\end{bmatrix}}+{\small \frac{9}{41}}\color{blue}{M\begin{bmatrix}1\\5\end{bmatrix}}\\ &={\small \frac{-4}{41}}\color{red}{\begin{bmatrix}3\\8\end{bmatrix}}+{\small\frac{9}{41}}\color{blue}{\begin{bmatrix}-8\\7\end{bmatrix}}\\ &=\begin{bmatrix}\frac{-84}{41}\\\frac{31}{41}\end{bmatrix} \end{align*}

Generalization:

In fact, we can answer more generally as to what will $M$ do to any vector in $\Bbb{R}^2$. From equation (1) \begin{align*} M\begin{bmatrix}a\\b\end{bmatrix}&={\small\left(\frac{b-5a}{41}\right)}\color{red}{M\begin{bmatrix}-8\\1\end{bmatrix}}+{\small \left(\frac{a+8b}{41}\right)}\color{blue}{M\begin{bmatrix}1\\5\end{bmatrix}}\\ &={\small\left(\frac{b-5a}{41}\right)}\color{red}{\begin{bmatrix}3\\8\end{bmatrix}}+{\small \left(\frac{a+8b}{41}\right)}\color{blue}{\begin{bmatrix}-8\\7\end{bmatrix}}\\ &={\small \frac{1}{41}}\begin{bmatrix}-23a-61b\\-33a+64b\end{bmatrix} \end{align*}

This also helps us find $M$ as $$M\begin{bmatrix}a\\b\end{bmatrix}={\small \frac{1}{41}}\begin{bmatrix}-23a-61b\\-33a+64b\end{bmatrix}=\begin{bmatrix}\frac{-23}{41}&\frac{-61}{41}\\ \frac{-33}{41}&\frac{64}{41}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}.$$

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If you find two numbers $x,y\mathbb{R}$ such that $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$, then, make a left multiply by $M$:

$$xM\begin{bmatrix}-8\\1\end{bmatrix}+yM\begin{bmatrix}1\\5\end{bmatrix}=M\begin{bmatrix}1\\1\end{bmatrix}$$ finally,

$$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x\begin{bmatrix}3\\8\end{bmatrix}+y \begin{bmatrix}-8\\7\end{bmatrix} = M\begin{bmatrix}1\\1\end{bmatrix}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cdots(I)$$

So, all your problem is reduced to $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$

which is equals to

$$\begin{cases} -8x + y & =1\\ x+5y & = 1\end{cases} $$

with solution $x=-\frac{4}{41},~y=\frac{9}{41}$. Just reemplace $x,y$ in equation $(I)$.

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  • $\begingroup$ Could you elaborate on how did you get from $(I)$ to the reduced form you mentioned? $\endgroup$
    – user713999
    Jun 17 '20 at 15:17
  • $\begingroup$ its because from premises $M\begin{bmatrix}-8\\1\end{bmatrix} =\begin{bmatrix}3\\8\end{bmatrix} $ $\endgroup$ Jun 17 '20 at 15:21
  • $\begingroup$ I meant from $x\begin{bmatrix}3\\8\end{bmatrix}+y \begin{bmatrix}-8\\7\end{bmatrix} = M\begin{bmatrix}1\\1\end{bmatrix}$ to $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$ $\endgroup$
    – user713999
    Jun 17 '20 at 15:33
  • $\begingroup$ just multiply by $M$ to the left , $M(x\begin{bmatrix}-8\\1\end{bmatrix}+...y()) = (Mx\begin{bmatrix}-8\\1\end{bmatrix} + My()...)... = (xM\begin{bmatrix}-8\\1\end{bmatrix} + yM()...)$ $\endgroup$ Jun 17 '20 at 15:36

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