Gabriel Cramer on Cramer's Rule

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When a problem contains several unknowns whose relationships are so complicated that one is obliged to form several equations; then, to discover the values of the unknowns, one makes all of them vanish, except one, which combined only with known quantities, gives, if the problem is determined, a final Equation, whose resolution reveals this first unknown, and then by this means all the others.

Algebra provides rules for this, whose success is infallible, provided one has the patience to follow them. But the calculation becomes extremely long when he number of equations and unknowns are very great and also when the unknowns in the proposed equations rise to much higher degrees. In this second case one falls, through ordinary methods, into another drawback; namely, to be led to more compound equations which are unnecessary and which contain superfluous roots, so that it is not always easy to disentangle those giving the true solution to the problem. I propose in the two following appendices to remedy these drawbacks.

Let there be several unknowns \( z, y, x, v,\) and so on, and as many equations

\[ \begin{align*}
A^1 &= Z^1 z +Y^1 y + X^1 x + V^1 v + \cdots \\
A^2 &= Z^2 z +Y^2 y + X^2 x + V^2 v + \cdots \\
A^3 &= Z^3 z +Y^3 y + X^3 x + V^3 v + \cdots \\
A^4 &= Z^4 z +Y^4 y + X^4 x + V^4 v + \cdots \\
\end{align*} \]

and so on, where the letters \( A^1, A^2, A^3, A^4,\) etc. don't signify, as usual, the powers of \( A,\) but the first member is supposed to be a known value for the first equation, the second for the second, and so on. Similarly \( Z^1, Z^2,\) etc. are the coefficients of \( z, \; Y^1, Y^2,\) etc. are the coefficients of \( y, \; X^1, X^2,\) etc. are the coefficients of \( x, \; V^1, V^2,\) etc. are the coefficients of \( v, \) and so on in the first, second, and other equations.

Assuming this notation, if there is only one equation and one unknown \( z,\) one has \( z = {A^1 / Z^1}. \) If there are two equations and two unknowns \( z \) and \( y,\) one finds:

\[ z = {{A^1 Y^2 - A^2 Y^1} \over {Z^1 Y^2 - Z^2 Y^1}}, \hskip{20pt} y = {{Z^1 A^2 - Z^2 A^1} \over {Z^1 Y^2 - Z^2 Y^1}}. \]

If there are three equations and three unknowns \( z, y, \) and \( x, \) one finds

\[ \begin{align*}
z &= {{A^1 Y^2 X^3 - A^1 Y^3 X^2 - A^2 Y^1 X^3 + A^2 Y^3 X^1 + A^3 Y^1 X^2 - A^3 Y^2 X^1} \over {Z^1 Y^2 X^3 - Z^1 Y^3 X^2 - Z^2 Y^1 X^3 + Z^2 Y^3 X^1 + Z^3 Y^1 X^2 - Z^3 Y^2 X^1}} \\
\end{align*} \]

\[ \begin{align*}
y &= {{Z^1 A^2 X^3 - Z^1 A^3 X^2 - Z^2 A^1 X^3 + Z^2 A^3 X^1 + Z^3 A^1 X^2 - Z^3 A^2 X^1} \over {Z^1 Y^2 X^3 - Z^1 Y^3 X^2 - Z^2 Y^1 X^3 + Z^2 Y^3 X^1 + Z^3 Y^1 X^2 - Z^3 Y^2 X^1}} \\
\end{align*} \]

\[ \begin{align*}
x &= {{Z^1 Y^2 A^3 - Z^1 Y^3 A^2 - Z^2 Y^1 A^3 + Z^2 Y^3 A^1 + Z^3 Y^1 A^2 - Z^3 Y^2 A^1} \over {Z^1 Y^2 X^3 - Z^1 Y^3 X^2 - Z^2 Y^1 X^3 + Z^2 Y^3 X^1 + Z^3 Y^1 X^2 - Z^3 Y^2 X^1}} \\
\end{align*} \]

Examining these formulas provides the general rule. The number of equations and unknowns being \( n, \) one finds the value of each unknown by forming \( n \) fractions whose common denominator has as many terms as there are different arrangements of \( n \) different things. Each term is composed of the letters \( ZYXV, \) etc., always written in the same order, but which one distributes like exponents, the first \( n \) figures arranged in all ways possible. Likewise, when one has three unknowns, the denominator has \( [1 \times 2 \times 3=] \; 6 \) terms composed of the three letters \( ZYX, \) which successively receive the exponents \( 123, 132, 213, 231, 312, 321. \) One gives these terms the signs \( + \) or \( -, \) according to the following rule. When an exponent is followed in the same term immediately or later on by an exponent smaller than itself, I call that a derangement. Then one counts, for each term, the number of derangements: if it is even or zero, the term will have the sign \( +; \) if it is odd, the term will have the sign \( -. \) For example, in the term \( Z^1 Y^2 V^3, \) there is no derangement: the term will then have the sign \( +. \) The term \( Z^3 Y^1 X^2 \) also has the sign \( +, \) because it has two derangement, \( 3 \) before \( 1 \) and \( 3 \) before \( 2.\) But the term \( Z^3 Y^2 X^1, \) which has three derangements, \( 3 \) before \( 2, 3 \) before \( 1, \) and \( 2 \) before \( 1, \) will have the sign \( -.\)

The common denominator being formed like this, one will have the value of \( z \) by giving the denominator the numerator formed by changing all the \( Z \) and \( A \) terms. And the value for \( y \) is the fraction with the same denominator and with numerator the quantity resulting when changing \( Y \) to \( A \) in all terms in the denominator. And one finds in a similar way of the values of the other unknowns.

Generally speaking, the problem is determined. But there may be specific cases where it remains undetermined; and others where it becomes impossible. That is when the common denominator is found equal to zero; that is to say, if there are two equations, when \( Z^1 Y^2 - Z^2 Y^1 = 0; \) if there are three, when \( Z^1 Y^2 X^3 - Z^1 Y^3 X^2 - Z^2 Y^1 X^3 + X^2 Y^3 Z^1 + X^3 Y^1 Z^2 - X^3 Y^2 Z^1 = 0, \) and so on. So if the quantities \( A^1, A^2, A^3, \) etc. are such that the numerators are also equal to zero, the problem is indeterminate, because the fractions \( 0/0, \) which ought to give the values of the unknowns, are indeterminate. But if the quantities \( A^1, A^2, A^3, \) etc. are such that, the common denominator being zero, some of the numerators are not zero, the problem is impossible, or at least the unknown variables that can solve it are all or in part equal to infinity.

For example, if one has these two equations

\[ \begin{align*}
2 &= 3z - 2y \\
5 &= 6z - 4y,
\end{align*} \]

one finds \( z = 2/0 \) and \(y = 3/0.\) So \( z \) and \( y \) are infinite quantities, which are due respectively to the \( 2 \) and the \( 3.\) In clearing these unknowns by the ordinary methods, one would fall into the absurd equation \( 2/3 = 5/6. \) For the first equation gives \( z = 2/3 y + 2/3 \) and the second \( z = 4/6 y + 5/6. \) So \( 2/3 y + 2/3 = 4/6 y + 5/6, \) or \( 2/3 = 5/6, \) which is absurd if \( z \) and \( y \) are finite quantities. But if they are infinite, one can say without absurdity that \( z = 2/3 y + 5/6 \) and at the same time that \( z = 2/3 y + 5/6, \) because the finite quantities \( 2/3 \) and \( 5/6 \) not being comparable to the infinite quantities \( z \) and \( 2/3y, \) the two equations \( z = 2/3y + 2/3 \) and \( z = 2/3y + 5/6 \) are both reduced to \( z = 2/3 y, \) which has nothing contradictory.

Translated by Mike Bertrand

April 16, 2016