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This is the problem.



The solution is below.

How did they row reduce the matrix like that below? How did the $x_i$ variables become $1$s and $0$s? I'd like someone to further clarify the solution.


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You want to have :

$$t_1L_1 + t_2L_2 = \vec{0}$$

By substituting in, this translates to the vector equation :

$$t_1(7x_1,x_2-5x_3) + t_2(x_1+6x_2+x_3,x_3)=0$$ $$\Rightarrow$$

$$\begin{cases} 7t_1x_1 + t_2(x_1+6x_2 + x_3)=0 \\ (x_2-5x_3)t_1 + t_2x_3 = 0\end{cases}$$

$$\Rightarrow$$

$$\begin{cases} (7t_1+t_2)x_1 + 6t_2x_2 + t_2x_3=0 \\ t_1x_2 + (t_2-5t_1)x_3 = 0 \end{cases}$$

In order for your system to be equal to $0$ for any pair of coordinates, you need the coefficients to be zero, hence :

$$\begin{cases} 7t_1 + t_2= 0, t_2 = 0 , t_2 = 0 \\ t_1 = 0 , t_2 - 5t_1 = 0 \end{cases}$$

$$\Rightarrow$$

$$t_1=t_2 = 0$$

What the author is doing there is that he is converting the pair of equations to a matrix expression factoring out $t_1,t_2$ and applying row operations (alike Gauss' Elimination) to the matrix of the system that gets formed to elementary show that the $X,Y$ corresponding parts must be equal to $0$, since from the last matrix he derives $1\cdot X + 0 = 0$ and $0 + 1\cdot Y = 0$, which ultimately means that $t_1=t_2=0$ as proved above.

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  • $\begingroup$ "In order for your system to be equal to 0 for any pair of coordinates, you need the coefficients to be zero, hence :" This is an unjustified assumption since the question is asking whether or not you need the coefficients to be zero, or if there is a non-zero solution. Hence you have to justify this statement as it essentially answers the question. $\endgroup$ – Max Li Feb 20 '18 at 22:09
  • $\begingroup$ Try solving the system with respect to one unknown and substituting. In this case, you will have a random variable with values over the reals. The only way this random expression which will depend on a variable to be zero would be for the coefficients to be zero. I didn't elaborate that but you can easily figure it out, as you have a 2×2 system with 3 unknowns! $\endgroup$ – Rebellos Feb 20 '18 at 22:13
  • $\begingroup$ What's the explanation for the matrix reductions they did in the provided solution in my original post? $\endgroup$ – Max Li Feb 20 '18 at 22:16
  • $\begingroup$ The system derived by the way I described or by the linearity of the solutions giving you the zero vector 8s reduced into a Gaussian form of elimination, as I mentioned in the post. $\endgroup$ – Rebellos Feb 20 '18 at 22:19
  • $\begingroup$ When I used your method and reduced it into a Gaussian form, I didn't have any "x"s in the matrix to start with since I was solving for (c1t1+c2t2+...) = 0, so I just put the coefficients c1, c2... into the matrix. How would using your method and then putting it into a matrix get "x" variables in the matrix? And how would we "eliminate" the "x" variables as shown in the solution I posted when there are only two rows and so many variables? $\endgroup$ – Max Li Feb 20 '18 at 22:24
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Maybe the notation was confusing...

t1 * $\mathbf{L1}$ + t2 * $\mathbf{L2}$ = $\vec{0}$

or

$\begin{pmatrix}\mathbf{L1} & \mathbf{L2}\end{pmatrix} \begin{pmatrix} t1 \\ t2 \end{pmatrix}$ = $\vec{0}$

or

$\left(\begin{array}{cc|c}\mathbf{L1} & \mathbf{L2} & \vec{0} \end{array}\right)$

Then to row reduce this matrix ...

Add $-((x_2-5x_3)/7x_1)*row1$ to $row2$:

$\left(\begin{array}{cc|c}\begin{matrix} 7x_1 \\ 0 \end{matrix} & \begin{matrix} x_1 + 6x_2 + x_3 \\ BLOB_1 \end{matrix} & \vec{0} \end{array}\right)$

Add $-((x_1+6x_2+x_3)/BLOB_1)*row2$ to $row1$:

$\left(\begin{array}{cc|c}\begin{matrix} 7x_1 \\ 0 \end{matrix} & \begin{matrix} 0 \\ BLOB_1 \end{matrix} & \vec{0} \end{array}\right)$

Then, normalize $row1$ and $row2$

$\left(\begin{array}{cc|c}\begin{matrix} 1 \\ 0 \end{matrix} & \begin{matrix} 0 \\ 1 \end{matrix} & \vec{0} \end{array}\right)$

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  • $\begingroup$ No, I understand that. What I don't understand is how the $x_i$s disappear after that, and how that matrix was reduced. $\endgroup$ – Max Li Feb 20 '18 at 22:07
  • $\begingroup$ Sorry for the delay... $\endgroup$ – Victor Feb 21 '18 at 12:40

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