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 ... ... @@ -1704,7 +1704,7 @@ \end{ans} \begin{ans}{One.I.3.23} In this case the solution set is all of $$\Re^n$$, and can be In this case the solution set is all of $$\Re^n$$ and can be expressed in the required form \begin{equation*} \set{c_1\colvec[r]{1 \\ 0 \\ \vdotswithin{1} \\ 0}
 ... ... @@ -3557,8 +3557,8 @@ by writing $$\vec{s}-\vec{p}$$ as $$\vec{h}$$. For set inclusion the other way, take a vector of the form $\vec{p}+\vec{h}$, where $$\vec{p}$$ solves the system and $$\vec{h}$$ solves the associated homogeneous system, and note that it solves the given system: for any equation index $i$, associated homogeneous system, and note that $\vec{p}+\vec{h}$ solves the given system: for any equation index $i$, \begin{equation*} \begin{split} a_{i,1}(p_1+h_1)+\cdots+a_{i,n}(p_n+h_n) ... ... @@ -3613,7 +3613,7 @@ The associated homogeneous system reduces via the same row operations & &y &- &3z &= &0 \end{linsys} &\grstep{-2\rho_1+\rho_2} \;\grstep{\rho_2\leftrightarrow\rho_3} \;\;\grstep{\rho_2\leftrightarrow\rho_3} &\begin{linsys}{3} x &+ &2y &- &z &= &0 \\ & &y &- &3z &= &0 \\ ... ... @@ -3661,7 +3661,7 @@ homogeneous systems have at least one solution. x &+ &y &+ &3z &+ &2w &= &0 \end{linsys} &\grstep[-\rho_1+\rho_3]{-2\rho_1+\rho_2} \;\grstep{\rho_2+\rho_3} \;\;\grstep{\rho_2+\rho_3} &\begin{linsys}{4} x & & &+ &z &+ &w &= &0 \\ & &-y&- &2z &- &w &= &0 \\ ... ... @@ -3675,7 +3675,7 @@ In fact the solution set of this homogeneous system is infinite. \end{equation*} However, because no particular solution of the original system exists, the general solution set is empty\Dash there are no vectors of the form $\vec{p}+\vec{h}$ because there are no $\vec{p}\,$'s. $\vec{p}+\vec{h}$ because there are no $\vec{p}\:$'s. \end{example} \begin{corollary} ... ... @@ -3685,13 +3685,15 @@ have infinitely many elements. \begin{proof} We've seen examples of all three happening so we need only prove that those are the only possibilities. that there are no other possibilities. First, notice a homogeneous system with at least one non-$$\zero$$ solution $\vec{v}$ has infinitely many solutions because the set of multiples $s\vec{v}$ is infinite\Dash if $s\neq 1$ then $s\vec{v}-\vec{v}=(s-1)\vec{v}$ is non-$\zero$ and so $s\vec{v}\neq \vec{v}$. is infinite\Dash if $s\neq 1$ then $s\vec{v}\neq\vec{v}$ because $s\vec{v}-\vec{v}=(s-1)\vec{v}$ is non-$\zero$, since any non-$0$ component of $\vec{v}$ when rescaled by the factor $s-1$ will give a non-$0$ value. Now, apply \nearbylemma{th:GenEqPartHomo} to conclude that a solution set \begin{equation*} ... ... @@ -3748,16 +3750,14 @@ we find that some variable is free. (We formalize ignoring the constants on the right'' by considering the associated homogeneous system.) Especially interesting examples of this are systems with the same number of equations as variables. A notable special case is systems having the same number of equations as unknowns. Such a system will have a solution, and that solution will be unique, if and only it reduces to an echelon form system where every variable leads its row, if and only if it reduces to an echelon form system where every variable leads its row (since there are the same number of variables as rows), which will happen if and only if the associated homogeneous system has a unique solution. % Thus, the question of uniqueness of solution is especially % interesting when the system has the same number of % equations as variables. \begin{definition} A square matrix is \definend{nonsingular}\index{nonsingular!matrix} ... ... @@ -3770,48 +3770,6 @@ that is, if it is the matrix of coefficients of a homogeneous system with infinitely many solutions. \end{definition} The word singular means departing from general expectation'' and here expresses that we could expect that systems with the same number of equations as unknowns will typically have a unique solution. (That singular' applies to systems having more than one solution is ironic, but it is the standard term.) \begin{example} The systems from \nearbyexample{ex:FirstExHomoSys}, \nearbyexample{ex:HomoZeroOnlySol}, and \nearbyexample{ex:IllusGenEqPartHomo} each have an associated homogeneous system with a unique solution. Thus these matrices are nonsingular. \begin{equation*} \begin{mat}[r] 3 &4 \\ 2 &-1 \end{mat} \qquad \begin{mat}[r] 3 &2 &1 \\ 6 &-4 &0 \\ 0 &1 &1 \end{mat} \qquad \begin{mat}[r] 1 &2 &-1 \\ 2 &4 &0 \\ 0 &1 &-3 \end{mat} \end{equation*} The Chemistry problem from \nearbyexample{ex:SolnChemProb} is a homogeneous system with more than one solution so its matrix is singular. \begin{equation*} \begin{mat}[r] 7 &0 &-7 &0 \\ 8 &1 &-5 &-2 \\ 0 &1 &-3 &0 \\ 0 &3 &-6 &-1 \end{mat} \end{equation*} \end{example} \begin{example} The first of these matrices is nonsingular while the second is singular ... ... @@ -3867,8 +3825,51 @@ has either no solutions or else has infinitely many, as with these. 3x &+ &6y &= &3 \end{linsys} \end{equation*} Thus, singular' can be thought of as connoting troublesome,'' or at least not ideal''. \end{example} The word singular means departing from general expectation'' (people often, naively, expect that systems with the same number of variables as equations will have a unique solution). Thus, it can be thought of as connoting troublesome,'' or at least not ideal.'' (That singular' applies those systems that do not have one solution is ironic, but it is the standard term.) \begin{example} The systems from \nearbyexample{ex:FirstExHomoSys}, \nearbyexample{ex:HomoZeroOnlySol}, and \nearbyexample{ex:IllusGenEqPartHomo} each have an associated homogeneous system with a unique solution. Thus these matrices are nonsingular. \begin{equation*} \begin{mat}[r] 3 &4 \\ 2 &-1 \end{mat} \qquad \begin{mat}[r] 3 &2 &1 \\ 6 &-4 &0 \\ 0 &1 &1 \end{mat} \qquad \begin{mat}[r] 1 &2 &-1 \\ 2 &4 &0 \\ 0 &1 &-3 \end{mat} \end{equation*} The Chemistry problem from \nearbyexample{ex:SolnChemProb} is a homogeneous system with more than one solution so its matrix is singular. \begin{equation*} \begin{mat}[r] 7 &0 &-7 &0 \\ 8 &1 &-5 &-2 \\ 0 &1 &-3 &0 \\ 0 &3 &-6 &-1 \end{mat} \end{equation*} \end{example} The above table has two dimensions. ... ... @@ -4439,13 +4440,12 @@ of Gauss' method itself in the rest of this chapter. (at most) one element.) \end{answer} \item To tell the whole truth, there is another tricky point to the In the proof of \nearbylemma{le:HomoSltnSpanVecs}. What happens if there are no non-$$0=0$$' equations? (There aren't any more tricky points after this one.) \nearbylemma{le:HomoSltnSpanVecs}, what happens if there are no non-$$0=0$$' equations? \begin{answer} In this case the solution set is all of $$\Re^n$$, and can be In this case the solution set is all of $$\Re^n$$ and can be expressed in the required form \begin{equation*} \set{c_1\colvec[r]{1 \\ 0 \\ \vdotswithin{1} \\ 0} ... ... @@ -4463,7 +4463,7 @@ of Gauss' method itself in the rest of this chapter. \partsitem $$3\vec{s}$$ \partsitem $$k\vec{s}+m\vec{t}$$ for $$k,m\in\Re$$ \end{exparts*} What's wrong with: These three show that if a homogeneous What's wrong with this argument: `These three show that if a homogeneous system has one solution then it has many solutions\Dash any multiple of a solution is another solution, and any sum of solutions is a solution also\Dash so there are no ... ...
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