I'm working through some vector problems in a book of problems I found online (here, see problem C15 on page 28). The problem I'm working on is

Find $\alpha$ and $\beta$ that solve the vector equation.

$\alpha \begin{bmatrix} 2 \\ 1\\ \end{bmatrix} + \beta \begin{bmatrix} 1\\3\\\end{bmatrix} = \begin{bmatrix} 5\\0\\ \end{bmatrix}$

I did $$\alpha \begin{bmatrix} 2 \\ 1\\ \end{bmatrix} + \beta \begin{bmatrix} 1\\3\\\end{bmatrix} = \begin{bmatrix} 5\\0\\ \end{bmatrix}$$ $$\begin{bmatrix} 2 \alpha \\ \alpha\\ \end{bmatrix} + \begin{bmatrix} \beta\\3\beta\\\end{bmatrix} = \begin{bmatrix} 5\\0\\ \end{bmatrix}$$ $$\begin{bmatrix} 2\alpha+\beta\\3\beta+\alpha\\\end{bmatrix} = \begin{bmatrix} 5\\0\\ \end{bmatrix}$$ $$2\alpha+\beta=5$$ $$3\beta+\alpha=0$$ $$-3\beta+5=\alpha$$ $$2(-3\beta+5)+\beta=5$$ $$\beta=-1$$ $$-3(-1)+5=\alpha$$ $$8=a$$

I checked my answer, and I got $\beta$ right but it shows $\alpha$ as 3. The work it shows gives the same system of equations, but it solves them as follows:

which we can solve by row-reducing the augmented matrix of the system,

$$\begin{bmatrix} 2&1&5\\1&3&0\\ \end{bmatrix} \overset{\text{RREF}}{\to} \begin{bmatrix} 1&0&3\\0&1&-1\\ \end{bmatrix}$$ [Please note there should be a box around the two ones in the second matrix; I didn't know how to do that.]

Thus, the only solution is $\alpha=3, \beta=-1$.

I am somewhat (okay, really) new to this, so I don't really understand what is wrong with my method or how how their method works. Any explanation would be helpful.



Where does $-3\beta+5=\alpha$ come from? Your system is $$2\alpha+\beta=5,\quad 3\beta+\alpha=0.$$ From the second one, we get $\alpha=-3\beta$.then from the first one we obtain $$2(-3\beta)+\beta=5\implies \beta=-1$$ and finally $\alpha=-3\beta=-3(-1)=3$.

P.S. In the solution on your book, the system is transformed into equivalent ones by replacing one equation with sum of that equation with a multiple of the other one. See Reduced Row-Echelon Form in your pdf. $$\begin{bmatrix} 2&1&5\\1&3&0\\ \end{bmatrix} \overset{}{\to} \begin{bmatrix} 1&-2&5\\1&3&0\\ \end{bmatrix} \overset{}{\to} \begin{bmatrix} 1&-2&5\\0&5&-5\\ \end{bmatrix}\\ \overset{}{\to} \begin{bmatrix} 1&-2&5\\0&1&-1\\ \end{bmatrix} \overset{}{\to} \begin{bmatrix} 1&0&3\\0&1&-1\\ \end{bmatrix}.$$

  • $\begingroup$ @heather See my P.S. $\endgroup$ – Robert Z Sep 10 '16 at 15:56

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