Find all $t_1$ and $t_2$ such that $t_1L_1+t_2L_2=\vec{0}^{\,}$

Question

This is the problem.

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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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Answer 1

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.

Answer 2

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 $row2$

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