To test following system of linear equation for equivalency

Question

Let F be field of complex numbers I have two system of equations

$x_1 - x_2 =0 $

$2x_1 + x_2 =0$

And

$3x_1 + x_2 -0$

$x_1 + x_2 =0$

The definition says that each if equation in first system is a linear combination of system of equations for second system ,then they are equivalent .How do i start this ?Hints ?

Thanks !

Answer 1

Here is one way. Let the equations in the first system be $R_1$, $R_2$, and those in the second system be $S_1$, $S_2$. Let $a$ and $b$ be the multipliers for $R_1$, so that

$$R_1=aS_1+bS_2$$

Looking at the coordinates in the two systems, first for $x_1$ then for $x_2$, we get

$$3a+b=1$$ $$a+b=-1$$

Solve these simultaneous equations for $a$ and $b$ and you found the linear combination for $R_1$. Do similarly to find the linear combination for $R_2$ (using variables $c$ and $d$).

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Here is a second way. As a half-way point, we name these equations:

$$e_1:\ 1x_1+0x_2=0$$ $$e_2:\ 0x_1+1x_2=0$$

Obviously we have

$$R_1=e_1-e_2$$ $$R_2=2e_1+e_2$$

You can use elementary row operations to get $e_1$ and $e_2$ in terms of $S_1$ and $S_2$. You will get

$$e_1=\frac 12S_1-\frac 12S_2$$ $$e_2=-\frac 12S_1+\frac 32S_2$$

Substitute the equations for the $e$'s into the equations for the $R$'s, simplify, and you are done.

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Here is a third way that uses matrices. We can see that you are looking for $a$, $b$, $c$, and $d$, as in my first method, such that

$$\begin{bmatrix} 1& -1\\ 2& 1\\ \end{bmatrix} =\begin{bmatrix}a&b\\c&d\\ \end{bmatrix} \cdot \begin{bmatrix} 3& 1\\ 1& 1\\ \end{bmatrix}$$

This is solved with

$$\begin{bmatrix}a&b\\c&d\\ \end{bmatrix} =\begin{bmatrix} 1& -1\\ 2& 1\\ \end{bmatrix} \cdot \begin{bmatrix} 3& 1\\ 1& 1\\ \end{bmatrix}^{-1}$$

Answer 2

Show that they have the same reduced row echelon form: \begin{align} \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & -1 \\ 0 & 3 \end{bmatrix} \\&\to \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align} for the first system and \begin{align} \begin{bmatrix} 3 & 1 \\ 1 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/3 \\ 1 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1/3 \\ 0 & 2/3 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1/3 \\ 0 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align} for the second system.

If $A$ and $B$ are the systems' matrices, you know that the reduced row echelon form for the first matrix is $U=FA$ for some invertible matrix $F$; similarly, the RREF of $B$ is $V=GB$ for an invertible matrix $G$. Since $U=V$, you have $$ FA=GB $$ which can be written as $B=(G^{-1}F)A$.

Note that this method is completely general and works for every linear system.