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如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决! 0.40 Solution of Equations Containing Several Unknown QuantitiesOn the other hand, let us see how we can solve an equation with respect to its various unknowns, and, to this end, we shall limit ourselves to the case of two unknowns\[axy+bxy^{\prime}+cx^{\prime}y+dx^{\prime}y^{\prime}=0.\]First solving with respect to \(x\),\[x=(a^{\prime}y+b^{\prime}y^{\prime})x+(cy+dy^{\prime})x^{\prime}.\]The resultant of the elimination of \(x\) is\[acy+bdy^{\prime}=0.\]If the given equation is true, this resultant is true.Now it is an equation involving \(y\) only; solving it,\[y=(a^{\prime}+c^{\prime})y+bdy^{\prime}.\]Had we eliminated \(y\) first and then \(x\), we would have obtained the solution\[y=(a^{\prime}x+c^{\prime}x^{\prime})y+(bx+dx^{\prime})y^{\prime}\]and the equation in \(x\)\[abx+cdx^{\prime}=0,\]whence the solution\[x=(a^{\prime}+b^{\prime})x+cdx^{\prime}.\]We see that the solution of an equation involving two unknown quantities is not symmetrical with respect to these unknowns; according to the order in which they were eliminated, we have the solution\[x =(a^{\prime}y+b^{\prime}y^{\prime})x+(cy+dy^{\prime})x^{\prime},\] \[y =(a^{\prime}+c^{\prime})y+bdy^{\prime},\]or the solution\[x =(a^{\prime}+b^{\prime})x+cdx,\] \[y =(a^{\prime}x+c^{\prime}x^{\prime})y+(bx+dx^{\prime})y^{\prime}.\]If we replace the terms \(x,y,\) in the second members by indeterminates \(u,v,\) one of the unknowns will depend on only one indeterminate, while the other will depend on two. We shall have a symmetrical solution by combining the two formulas,\[x =(a^{\prime}+b^{\prime})u+cdu^{\prime},\] \[y =(a^{\prime}+c^{\prime})v+bdv^{\prime},\]but the two indeterminates \(u\) and \(v\) will no longer be independent of each other. For if we bring these solutions into the given equation, it becomes\[abcd+ab^{\prime}c^{\prime}uv+a^{\prime}bd^{\prime}uv^{\prime}+a^{\prime}cd^{ \prime}u^{\prime}v+b^{\prime}c^{\prime}du^{\prime}v^{\prime}=0\]or since, by hypothesis, the resultant \(abcd=0\) is verified,\[ab^{\prime}c^{\prime}uv+a^{\prime}bd^{\prime}uv^{\prime}+a^{\prime}cdu^{\prime }v+b^{\prime}c^{\prime}du^{\prime}v^{\prime}=0.\]This is an “equation of condition” which the indeterminates \(u\) and \(v\) must verify; it can always be verified, since its resultant is identically true,\[ab^{\prime}c^{\prime}\cdot a^{\prime}bd^{\prime}\cdot a^{\prime}cd^{\prime} \cdot b^{\prime}c^{\prime}d=aa^{\prime}\cdot bb^{\prime}\cdot cc^{\prime}\cdot dd ^{\prime}=0,\]but it is not verified by any pair of values attributed to \(u\) and \(v\).Some general symmetrical solutions, _i.e._, symmetrical solutions in which the unknowns are expressed in terms of several independent indeterminates, can however be found. This problem has been treated by Schroder44, by Whitehead45 and by Johnson. 46_Algebra der Logik_, Vol. I, §24._Universal Algebra_, Vol. I, §353-37._“Sur la theorie des égalités logiques”, _Bibl du Cong. intern. de Phil._, Vol. III, p. 185 (Paris, 1901).This investigation has only a purely technical interest; for, from the practical point of view, we either wish to eliminate one or more unknown quantities (or even all), or else we seek to solve the equation with respect to one particular unknown. In the first case, we develop the first member with respect to the unknowns to be eliminated and equate the product of its coefficients to 0. In the second case we develop with respect to the unknown that is to be extracted and apply the formula for the solution of the equation of one unknown quantity. If it is desired to have the solution in terms of some unknown quantities or in terms of the known only, the other unknowns (or all the unknowns) must first be eliminated before performing the solution. 0.41 The Problem of BooleAccording to Boole the most general problem of the algebra of logic is the following47:_Laws of Thought_, Chap. IX, §8.Given any equation (which is assumed to be possible)\[f(x,y,z,\ldots)=0,\]and, on the other hand, the expression of a term \(t\) in terms of the variables contained in the preceding equation\[t=\varphi(x,y,z,\ldots)\]to determine the expression of \(t\) in terms of the constants contained in \(f\) and in \(\varphi\).Suppose \(f\) and \(\varphi\) developed with respect to the variables \(x,y,z\ldots\) and let \(p_{1},p_{2},p_{3},\ldots\) be their constituents:\[f(x,y,z,\ldots) =Ap_{1}+Bp_{2}+Cp_{3}+\ldots,\] \[\phi(x,y,z,\ldots) =ap_{1}+bp_{2}+cp_{3}+\ldots.\]Then reduce the equation which expresses \(t\) so that its second member will be \(0\):\[(t\phi^{\prime}+t^{\prime}\phi=0)=[(a^{\prime}p_{1}+b^{\prime}p_ {2}+c^{\prime}p_{3}+\ldots)t\] \[\qquad\qquad+(ap_{1}+bp_{2}+cp_{3}+\ldots)t^{\prime}=0].\]Combining the two equations into a single equation and developing it with respect to \(t\):\[[(A+a^{\prime})p_{1}+(B+b^{\prime})p_{2} +(C+c^{\prime})p_{3}+\ldots]t\] \[+[(A+a)p_{1}+(B+b)p_{2}+(C+c)p_{3}+\ldots]t^{\prime}=0.\]This is the equation which gives the desired expression of \(t\). Eliminating \(t\), we obtain the resultant\[Ap_{1}+Bp_{2}+Cp_{3}+\ldots=0,\]as we might expect. If, on the other hand, we wish to eliminate \(x,y,z,\ldots\) (_i.e._, the constituents \(p_{1},p_{2},p_{3}\ldots\)), we put the equation in the form\[(A+a^{\prime}t+at^{\prime})p_{1}+(B+b^{\prime}t+bt^{\prime})p_{2}+(C+c^{\prime }t+ct^{\prime})p_{3}+\ldots=0,\]and the resultant will be\[(A+a^{\prime}t+at^{\prime})(B+b^{\prime}t+bt^{\prime})(C+c^{\prime}t+ct^{ \prime})\ldots=0,\]an equation that contains only the unknown quantity \(t\) and the constants of the problem (the coefficients of \(f\) and of \(\varphi\)). From this may be derived the expression of \(t\) in terms of these constants. Developing the first member of this equation\[(A+a^{\prime})(B+b^{\prime})(C+c^{\prime})\ldots\times t+(A+a)(B+b)(C+c)\ldots \times t^{\prime}=0.\]The solution is\[t=(A+a)(B+b)(C+c)\ldots+u(A^{\prime}a+B^{\prime}b+C^{\prime}c+\ldots).\]The resultant is verified by hypothesis since it is\[ABC\ldots=0,\]which is the resultant of the given equation\[f(x,y,z,\ldots)=0.\]We can see how this equation contributes to restrict the variability of \(t.\) Since \(t\) was defined only by the function \(\varphi,\) it was determined by the double inclusion\[abc\ldots

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