Introductory Linear Algebra/System of linear equations

Systems of linear equations in matrix form

We should define what is Template:Colored em before expressing it in matrix form. Template:Colored definition Template:Colored remark

We often use the terms 'Template:Colored em' and 'Template:Colored em' to describe the number of solutions of a system of linear equations. Template:Colored definition Template:Colored remark Template:Colored example Template:Colored example Template:Colored example Template:Colored exercise After defining a Template:Colored em, we can express it in matrix form in multiple ways, as in the following definition. Template:Colored definition Template:Colored remark Template:Colored example Template:Colored exercise

Gauss-Jordan algorithm

Proof. Outline: There is a reverse process (i.e. performing the ERO with its reverse process together, in arbitrary orders, will have no effect on matrix) for each type of EROs, which is also a ERO itself, as illustrated below:

• the reverse process of type I ERO ${𝐫}_i\leftrightarrow {𝐫}_j$ is also ${𝐫}_i\leftrightarrow {𝐫}_j$
• the reverse process of type II ERO $k{𝐫}_i\to {𝐫}_i$ is type II ERO $\frac{1}{k}{𝐫}_i\to {𝐫}_i$ (if $k=0$, this ERO is undefined, this is why $k$ must be nonzero for type II ERO, so that it is reversible)
• the reverse process of type III ERO $k{𝐫}_j+{𝐫}_i\to {𝐫}_i$ is type III ERO $-k{𝐫}_j+{𝐫}_i\to {𝐫}_i$

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Proof. Outline: It suffices to show that the solution set is unchanged if we perform one ERO. E.g.

• Type I ERO:

$$\text{Compare }\{\begin{array}{c}x+2y=3\\ 4x+5y=6\end{array}\text{ and }\{\begin{array}{c}4x+5y=6\\ x+2y=3\end{array}$$

• Type II ERO:

$$\text{Compare }\{\begin{array}{c}x+2y=3\\ 4x+5y=6\end{array}\text{ and }\{\begin{array}{c}kx+2ky=3k\\ 4x+5y=6\end{array}\iff \{\begin{array}{c}x+2y=3\\ 4x+5y=6\end{array}(k\ne 0)$$

• Type III ERO:

$$\text{Compare }\{\begin{array}{c}x+2y=3\\ 4x+5y=6\end{array}\text{ and }\{\begin{array}{cc}x+2y=3& (1)\\ (4+k)x+(5+2k)y=6+3k& (2)\end{array}$$ $(1)-k(2):$ $$(4+k-k)x+(5+2k-2k)y=6+3k-3k⟺4x+5y=6$$

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Proof. The augmented matrix of the homogeneous system of $m$ linear equations in $n$ unknowns is $A𝐱=\mathrm{𝟎}$ is $(A|\mathrm{𝟎})$, and thus its RREF has the form $(R|\mathrm{𝟎})$ in which $R$ is of size $m\times n$, since there are $m$ linear equations and $n$ unknowns.

If $R$ has a leading one in each of the first $n$ columns, then there are at least $n$ rows in $R$. However, $R$ has only $m<n$ rows, a contradiction. Thus, the homogeneous SLE does not have a unique solution.

Since a SLE can either have no solutions (which is impossible for a homogeneous SLE), a unique solution (which is impossible in this case), or infinitely many solutions, it follows that the homogeneous SLE must have infinitely many solutions, and thus have a nontrivial solution.

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