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Cramer's rule

From Wikipedia, the free encyclopedia

Formula for systems of linear equations

Inlinear algebra,Cramer's rule is an explicit formula for the solution of asystem of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of thedeterminants of the (square)coefficient matrix and ofmatrices obtained from it by replacing one column by the column vector of right-sides of the equations. It is named afterGabriel Cramer, who published the rule for an arbitrary number of unknowns in 1750,^[1]^[2] althoughColin Maclaurin also published special cases of the rule in 1748,^[3] and possibly knew of it as early as 1729.^[4]^[5]^[6]

Cramer's rule, implemented in a naive way, is computationally inefficient for systems of more than two or three equations.^[7] In the case ofn equations inn unknowns, it requires computation ofn + 1 determinants, whileGaussian elimination produces the result with the same (up to a constant factor independent of⁠ $n {\displaystyle n}$ ⁠)computational complexity as the computation of a single determinant.^[8]^[9] Moreover,Bareiss algorithm is a simple modification of Gaussian elimination that produces in a single computation a matrix whose nonzero entries are the determinants involved in Cramer's rule.

General case

Consider a system ofn linear equations forn unknowns, represented in matrix multiplication form as follows:

A\mathbf {x} =\mathbf {b}

where then ×n matrixA has a nonzero determinant, and the vector $\mathbf {x} =(x_{1},\ldots ,x_{n})^{\mathsf {T}}$ is the column vector of the variables. Then the theorem states that in this case the system has a unique solution, whose individual values for the unknowns are given by:

x_{i}={\frac {\det(A_{i})}{\det(A)}}\qquad i=1,\ldots ,n

where $A_{i}$ is the matrix formed by replacing thei-th column ofA by the column vectorb.

A more general version of Cramer's rule^[10] considers the matrix equation

AX=B

where then ×n matrixA has a nonzero determinant, andX,B aren ×m matrices. Given sequences $1\leq i_{1}<i_{2}<\cdots <i_{k}\leq n$ and $1\leq j_{1}<j_{2}<\cdots <j_{k}\leq m$ , let $X_{I,J}$ be thek ×k submatrix ofX with rows in $I:=(i_{1},\ldots ,i_{k})$ and columns in $J:=(j_{1},\ldots ,j_{k})$ . Let $A_{B}(I,J)$ be then ×n matrix formed by replacing the $i_{s}$ column ofA by the $j_{s}$ column ofB, for all $s=1,\ldots ,k$ . Then

\det X_{I,J}={\frac {\det(A_{B}(I,J))}{\det(A)}}.

In the case $k=1$ , this reduces to the normal Cramer's rule.

The rule holds for systems of equations with coefficients and unknowns in anyfield, not just in thereal numbers.

Proof

The proof for Cramer's rule uses the followingproperties of the determinants: linearity with respect to any given column and the fact that the determinant is zero whenever two columns are equal, which is implied by the property that the sign of the determinant flips if you switch two columns.

Fix the indexj of a column, and consider that the entries of the other columns have fixed values. This makes the determinant a function of the entries of thejth column. Linearity with respect to this column means that this function has the form

D_{j}(a_{1,j},\ldots ,a_{n,j})=C_{1,j}a_{1,j}+\cdots ,C_{n,j}a_{n,j},

where the $C_{i,j}$ are coefficients that depend on the entries ofA that are not in columnj. So, one has

\det(A)=D_{j}(a_{1,j},\ldots ,a_{n,j})=C_{1,j}a_{1,j}+\cdots ,C_{n,j}a_{n,j}

(Laplace expansion provides a formula for computing the $C_{i,j}$ but their expression is not important here.)

If the function $D_{j}$ is applied to anyother columnk ofA, then the result is the determinant of the matrix obtained fromA by replacing columnj by a copy of columnk, so the resulting determinant is 0 (the case of two equal columns).

Now consider a system ofn linear equations inn unknowns $x_{1},\ldots ,x_{n}$ , whose coefficient matrix isA, with det(A) assumed to be nonzero:

{\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&b_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&b_{2}\\&\vdots &\\a_{n1}x_{1}+a_{n2}x_{2}+\cdots +a_{nn}x_{n}&=&b_{n}.\end{matrix}}

If one combines these equations by takingC_1,j times the first equation, plusC_2,j times the second, and so forth untilC_n,j times the last, then for everyk the resulting coefficient ofx_k becomes

D_{j}(a_{1,k},\ldots ,a_{n,k}).

So, all coefficients become zero, except the coefficient of $x_{j}$ that becomes $\det(A).$ Similarly, the constant coefficient becomes $D_{j}(b_{1},\ldots ,b_{n}),$ and the resulting equation is thus

\det(A)x_{j}=D_{j}(b_{1},\ldots ,b_{n}),

which gives the value of $x_{j}$ as

x_{j}={\frac {1}{\det(A)}}D_{j}(b_{1},\ldots ,b_{n}).

As, by construction, the numerator is the determinant of the matrix obtained fromA by replacing columnj byb, we get the expression of Cramer's rule as a necessary condition for a solution.

It remains to prove that these values for the unknowns form a solution. LetM be then ×n matrix that has the coefficients of $D_{j}$ asjth row, for $j=1,\ldots ,n$ (this is theadjugate matrix forA). Expressed in matrix terms, we have thus to prove that

\mathbf {x} ={\frac {1}{\det(A)}}M\mathbf {b}

is a solution; that is, that

A\left({\frac {1}{\det(A)}}M\right)\mathbf {b} =\mathbf {b} .

For that, it suffices to prove that

A\,\left({\frac {1}{\det(A)}}M\right)=I_{n},

where $I_{n}$ is theidentity matrix.

The above properties of the functions $D_{j}$ show that one hasMA = det(A)I_n, and therefore,

\left({\frac {1}{\det(A)}}M\right)\,A=I_{n}.

This completes the proof, since aleft inverse of a square matrix is also a right-inverse (seeInvertible matrix theorem).

For other proofs, seebelow.

Finding inverse matrix

Main article:Invertible matrix § Methods of matrix inversion

LetA be ann ×n matrix with entries in afieldF. Then

A\,\operatorname {adj} (A)=\operatorname {adj} (A)\,A=\det(A)I

whereadj(A) denotes theadjugate matrix,det(A) is the determinant, andI is theidentity matrix. Ifdet(A) is nonzero, then the inverse matrix ofA is

A^{-1}={\frac {1}{\det(A)}}\operatorname {adj} (A).

This gives a formula for the inverse ofA, provideddet(A) ≠ 0. In fact, this formula works wheneverF is acommutative ring, provided thatdet(A) is aunit. Ifdet(A) is not a unit, thenA is not invertible over the ring (it may be invertible over a larger ring in which some non-unit elements ofF may be invertible).

Applications

Explicit formulas for small systems

Consider the linear system

\left\{{\begin{matrix}a_{1}x+b_{1}y&={\color {red}c_{1}}\\a_{2}x+b_{2}y&={\color {red}c_{2}}\end{matrix}}\right.

which in matrix format is

{\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}{\color {red}c_{1}}\\{\color {red}c_{2}}\end{bmatrix}}.

Assumea₁b₂ −b₁a₂ is nonzero. Then, with the help ofdeterminants,x andy can be found with Cramer's rule as

{\begin{aligned}x&={\frac {\begin{vmatrix}{\color {red}{c_{1}}}&b_{1}\\{\color {red}{c_{2}}}&b_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{vmatrix}}}={{\color {red}c_{1}}b_{2}-b_{1}{\color {red}c_{2}} \over a_{1}b_{2}-b_{1}a_{2}},\quad y={\frac {\begin{vmatrix}a_{1}&{\color {red}{c_{1}}}\\a_{2}&{\color {red}{c_{2}}}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{vmatrix}}}={a_{1}{\color {red}c_{2}}-{\color {red}c_{1}}a_{2} \over a_{1}b_{2}-b_{1}a_{2}}\end{aligned}}.

The rules for3 × 3 matrices are similar. Given

\left\{{\begin{matrix}a_{1}x+b_{1}y+c_{1}z&={\color {red}d_{1}}\\a_{2}x+b_{2}y+c_{2}z&={\color {red}d_{2}}\\a_{3}x+b_{3}y+c_{3}z&={\color {red}d_{3}}\end{matrix}}\right.

which in matrix format is

{\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}{\color {red}d_{1}}\\{\color {red}d_{2}}\\{\color {red}d_{3}}\end{bmatrix}}.

Then the values ofx, y andz can be found as follows:

x={\frac {\begin{vmatrix}{\color {red}d_{1}}&b_{1}&c_{1}\\{\color {red}d_{2}}&b_{2}&c_{2}\\{\color {red}d_{3}}&b_{3}&c_{3}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}},\quad y={\frac {\begin{vmatrix}a_{1}&{\color {red}d_{1}}&c_{1}\\a_{2}&{\color {red}d_{2}}&c_{2}\\a_{3}&{\color {red}d_{3}}&c_{3}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}},{\text{ and }}z={\frac {\begin{vmatrix}a_{1}&b_{1}&{\color {red}d_{1}}\\a_{2}&b_{2}&{\color {red}d_{2}}\\a_{3}&b_{3}&{\color {red}d_{3}}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}}.

Differential geometry

Ricci calculus

Cramer's rule is used in theRicci calculus in various calculations involving theChristoffel symbols of the first and second kind.^[11]

In particular, Cramer's rule can be used to prove that the divergence operator on aRiemannian manifold is invariant with respect to change of coordinates. We give a direct proof, suppressing the role of the Christoffel symbols.Let $(M,g)$ be a Riemannian manifold equipped withlocal coordinates $(x^{1},x^{2},\dots ,x^{n})$ . Let $A=A^{i}{\frac {\partial }{\partial x^{i}}}$ be avector field. We use thesummation convention throughout.

Theorem.

Thedivergence of $A {\displaystyle A}$ ,

\operatorname {div} A={\frac {1}{\sqrt {\det g}}}{\frac {\partial }{\partial x^{i}}}\left(A^{i}{\sqrt {\det g}}\right),

is invariant under change of coordinates.

Proof

Let $(x^{1},x^{2},\ldots ,x^{n})\mapsto ({\bar {x}}^{1},\ldots ,{\bar {x}}^{n})$ be acoordinate transformation withnon-singular Jacobian. Then the classicaltransformation laws imply that $A={\bar {A}}^{k}{\frac {\partial }{\partial {\bar {x}}^{k}}}$ where ${\bar {A}}^{k}={\frac {\partial {\bar {x}}^{k}}{\partial x^{j}}}A^{j}$ . Similarly, if $g=g_{mk}\,dx^{m}\otimes dx^{k}={\bar {g}}_{ij}\,d{\bar {x}}^{i}\otimes d{\bar {x}}^{j}$ , then ${\bar {g}}_{ij}=\,{\frac {\partial x^{m}}{\partial {\bar {x}}^{i}}}{\frac {\partial x^{k}}{\partial {\bar {x}}^{j}}}g_{mk}$ . Writing this transformation law in terms of matrices yields ${\bar {g}}=\left({\frac {\partial x}{\partial {\bar {x}}}}\right)^{\text{T}}g\left({\frac {\partial x}{\partial {\bar {x}}}}\right)$ , which implies $\det {\bar {g}}=\left(\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right)\right)^{2}\det g$ .

Now one computes

{\begin{aligned}\operatorname {div} A&={\frac {1}{\sqrt {\det g}}}{\frac {\partial }{\partial x^{i}}}\left(A^{i}{\sqrt {\det g}}\right)\\&=\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right){\frac {1}{\sqrt {\det {\bar {g}}}}}{\frac {\partial {\bar {x}}^{k}}{\partial x^{i}}}{\frac {\partial }{\partial {\bar {x}}^{k}}}\left({\frac {\partial x^{i}}{\partial {\bar {x}}^{\ell }}}{\bar {A}}^{\ell }\det \!\left({\frac {\partial x}{\partial {\bar {x}}}}\right)^{\!\!-1}\!{\sqrt {\det {\bar {g}}}}\right).\end{aligned}}

In order to show that this equals ${\frac {1}{\sqrt {\det {\bar {g}}}}}{\frac {\partial }{\partial {\bar {x}}^{k}}}\left({\bar {A}}^{k}{\sqrt {\det {\bar {g}}}}\right)$ ,it is necessary and sufficient to show that

{\frac {\partial {\bar {x}}^{k}}{\partial x^{i}}}{\frac {\partial }{\partial {\bar {x}}^{k}}}\left({\frac {\partial x^{i}}{\partial {\bar {x}}^{\ell }}}\det \!\left({\frac {\partial x}{\partial {\bar {x}}}}\right)^{\!\!\!-1}\right)=0\qquad {\text{for all }}\ell ,

which is equivalent to

{\frac {\partial }{\partial {\bar {x}}^{\ell }}}\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right)=\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right){\frac {\partial {\bar {x}}^{k}}{\partial x^{i}}}{\frac {\partial ^{2}x^{i}}{\partial {\bar {x}}^{k}\partial {\bar {x}}^{\ell }}}.

Carrying out the differentiation on the left-hand side, we get:

{\begin{aligned}{\frac {\partial }{\partial {\bar {x}}^{\ell }}}\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right)&=(-1)^{i+j}{\frac {\partial ^{2}x^{i}}{\partial {\bar {x}}^{\ell }\partial {\bar {x}}^{j}}}\det M(i|j)\\&={\frac {\partial ^{2}x^{i}}{\partial {\bar {x}}^{\ell }\partial {\bar {x}}^{j}}}\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right){\frac {(-1)^{i+j}}{\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right)}}\det M(i|j)=(\ast ),\end{aligned}}

where $M(i|j)$ denotes the matrix obtained from $\left({\frac {\partial x}{\partial {\bar {x}}}}\right)$ by deleting the $i {\displaystyle i}$ th row and $j {\displaystyle j}$ th column.But Cramer's Rule says that

{\frac {(-1)^{i+j}}{\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right)}}\det M(i|j)

is the $(j,i)$ th entry of the matrix $\left({\frac {\partial {\bar {x}}}{\partial x}}\right)$ .Thus

(\ast )=\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right){\frac {\partial ^{2}x^{i}}{\partial {\bar {x}}^{\ell }\partial {\bar {x}}^{j}}}{\frac {\partial {\bar {x}}^{j}}{\partial x^{i}}},

completing the proof.

Computing derivatives implicitly

Consider the two equations $F(x,y,u,v)=0$ and $G(x,y,u,v)=0$ . Whenu andv are independent variables, we can define $x=X(u,v)$ and $y=Y(u,v).$

An equation for ${\dfrac {\partial x}{\partial u}}$ can be found by applying Cramer's rule.

Calculation of ${\dfrac {\partial x}{\partial u}}$

First, calculate the first derivatives ofF,G,x, andy:

{\begin{aligned}dF&={\frac {\partial F}{\partial x}}dx+{\frac {\partial F}{\partial y}}dy+{\frac {\partial F}{\partial u}}du+{\frac {\partial F}{\partial v}}dv=0\\[6pt]dG&={\frac {\partial G}{\partial x}}dx+{\frac {\partial G}{\partial y}}dy+{\frac {\partial G}{\partial u}}du+{\frac {\partial G}{\partial v}}dv=0\\[6pt]dx&={\frac {\partial X}{\partial u}}du+{\frac {\partial X}{\partial v}}dv\\[6pt]dy&={\frac {\partial Y}{\partial u}}du+{\frac {\partial Y}{\partial v}}dv.\end{aligned}}

Substitutingdx,dy intodF anddG, we have:

{\begin{aligned}dF&=\left({\frac {\partial F}{\partial x}}{\frac {\partial x}{\partial u}}+{\frac {\partial F}{\partial y}}{\frac {\partial y}{\partial u}}+{\frac {\partial F}{\partial u}}\right)du+\left({\frac {\partial F}{\partial x}}{\frac {\partial x}{\partial v}}+{\frac {\partial F}{\partial y}}{\frac {\partial y}{\partial v}}+{\frac {\partial F}{\partial v}}\right)dv=0\\[6pt]dG&=\left({\frac {\partial G}{\partial x}}{\frac {\partial x}{\partial u}}+{\frac {\partial G}{\partial y}}{\frac {\partial y}{\partial u}}+{\frac {\partial G}{\partial u}}\right)du+\left({\frac {\partial G}{\partial x}}{\frac {\partial x}{\partial v}}+{\frac {\partial G}{\partial y}}{\frac {\partial y}{\partial v}}+{\frac {\partial G}{\partial v}}\right)dv=0.\end{aligned}}

Sinceu,v are both independent, the coefficients ofdu,dv must be zero. So we can write out equations for the coefficients:

{\begin{aligned}{\frac {\partial F}{\partial x}}{\frac {\partial x}{\partial u}}+{\frac {\partial F}{\partial y}}{\frac {\partial y}{\partial u}}&=-{\frac {\partial F}{\partial u}}\\[6pt]{\frac {\partial G}{\partial x}}{\frac {\partial x}{\partial u}}+{\frac {\partial G}{\partial y}}{\frac {\partial y}{\partial u}}&=-{\frac {\partial G}{\partial u}}\\[6pt]{\frac {\partial F}{\partial x}}{\frac {\partial x}{\partial v}}+{\frac {\partial F}{\partial y}}{\frac {\partial y}{\partial v}}&=-{\frac {\partial F}{\partial v}}\\[6pt]{\frac {\partial G}{\partial x}}{\frac {\partial x}{\partial v}}+{\frac {\partial G}{\partial y}}{\frac {\partial y}{\partial v}}&=-{\frac {\partial G}{\partial v}}.\end{aligned}}

Now, by Cramer's rule, we see that:

{\frac {\partial x}{\partial u}}={\frac {\begin{vmatrix}-{\frac {\partial F}{\partial u}}&{\frac {\partial F}{\partial y}}\\-{\frac {\partial G}{\partial u}}&{\frac {\partial G}{\partial y}}\end{vmatrix}}{\begin{vmatrix}{\frac {\partial F}{\partial x}}&{\frac {\partial F}{\partial y}}\\{\frac {\partial G}{\partial x}}&{\frac {\partial G}{\partial y}}\end{vmatrix}}}.

This is now a formula in terms of twoJacobians:

{\frac {\partial x}{\partial u}}=-{\frac {\left({\frac {\partial (F,G)}{\partial (u,y)}}\right)}{\left({\frac {\partial (F,G)}{\partial (x,y)}}\right)}}.

Similar formulas can be derived for ${\frac {\partial x}{\partial v}},{\frac {\partial y}{\partial u}},{\frac {\partial y}{\partial v}}.$

Integer programming

Cramer's rule can be used to prove that aninteger programming problem whose constraint matrix istotally unimodular and whose right-hand side is integer, has integer basic solutions. This makes the integer program substantially easier to solve.

Ordinary differential equations

Cramer's rule is used to derive the general solution to an inhomogeneous linear differential equation by the method ofvariation of parameters.

Geometric interpretation

Geometric interpretation of Cramer's rule. The areas of the second and third shaded parallelograms are the same and the second is $x_{1}$ times the first. From this equality Cramer's rule follows.

{\displaystyle x_{1}} — Geometric interpretation of Cramer's rule. The areas of the second and third shaded parallelograms are the same and the second is $x_{1}$ times the first. From this equality Cramer's rule follows.

Cramer's rule has a geometric interpretation that can be considered also a proof or simply giving insight about its geometric nature. These geometric arguments work in general and not only in the case of two equations with two unknowns presented here.

Given the system of equations

{\begin{matrix}a_{11}x_{1}+a_{12}x_{2}&=b_{1}\\a_{21}x_{1}+a_{22}x_{2}&=b_{2}\end{matrix}}

it can be considered as an equation between vectors

x_{1}{\binom {a_{11}}{a_{21}}}+x_{2}{\binom {a_{12}}{a_{22}}}={\binom {b_{1}}{b_{2}}}.

The area of the parallelogram determined by ${\binom {a_{11}}{a_{21}}}$ and ${\binom {a_{12}}{a_{22}}}$ is given by the determinant of the system of equations:

{\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}.

In general, when there are more variables and equations, the determinant ofn vectors of lengthn will give thevolume of theparallelepiped determined by those vectors in then-th dimensionalEuclidean space.

Therefore, the area of the parallelogram determined by $x_{1}{\binom {a_{11}}{a_{21}}}$ and ${\binom {a_{12}}{a_{22}}}$ has to be $x_{1}$ times the area of the first one since one of the sides has been multiplied by this factor. Now, this last parallelogram, byCavalieri's principle, has the same area as the parallelogram determined by ${\binom {b_{1}}{b_{2}}}=x_{1}{\binom {a_{11}}{a_{21}}}+x_{2}{\binom {a_{12}}{a_{22}}}$ and ${\binom {a_{12}}{a_{22}}}.$

Equating the areas of this last and the second parallelogram gives the equation

{\begin{vmatrix}b_{1}&a_{12}\\b_{2}&a_{22}\end{vmatrix}}={\begin{vmatrix}a_{11}x_{1}&a_{12}\\a_{21}x_{1}&a_{22}\end{vmatrix}}=x_{1}{\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}

from which Cramer's rule follows.

Other proofs

A proof by abstract linear algebra

This is a restatement of the proof above in abstract language.

Consider the map $\mathbf {x} =(x_{1},\ldots ,x_{n})\mapsto {\frac {1}{\det A}}\left(\det(A_{1}),\ldots ,\det(A_{n})\right),$ where $A_{i}$ is the matrix $A {\displaystyle A}$ with $\mathbf {x}$ substituted in the $i {\displaystyle i}$ th column, as in Cramer's rule. Because of linearity of determinant in every column, this map is linear. Observe that it sends the $i {\displaystyle i}$ th column of $A {\displaystyle A}$ to the $i {\displaystyle i}$ th basis vector $\mathbf {e} _{i}=(0,\ldots ,1,\ldots ,0)$ (with 1 in the $i {\displaystyle i}$ th place), because determinant of a matrix with a repeated column is 0. So we have a linear map which agrees with the inverse of $A {\displaystyle A}$ on the column space; hence it agrees with $A^{-1}$ on the span of the column space. Since $A {\displaystyle A}$ is invertible, the column vectors span all of $\mathbb {R} ^{n}$ , so our map really is the inverse of $A {\displaystyle A}$ . Cramer's rule follows.

A short proof

A short proof of Cramer's rule^[12] can be given by noticing that $x_{1}$ is the determinant of the matrix

X_{1}={\begin{bmatrix}x_{1}&0&0&\cdots &0\\x_{2}&1&0&\cdots &0\\x_{3}&0&1&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\x_{n}&0&0&\cdots &1\end{bmatrix}}

On the other hand, assuming that our original matrixA is invertible, this matrix $X_{1}$ has columns $A^{-1}\mathbf {b} ,A^{-1}\mathbf {v} _{2},\ldots ,A^{-1}\mathbf {v} _{n}$ , where $\mathbf {v} _{n}$ is then-th column of the matrixA. Recall that the matrix $A_{1}$ has columns $\mathbf {b} ,\mathbf {v} _{2},\ldots ,\mathbf {v} _{n}$ , and therefore $X_{1}=A^{-1}A_{1}$ . Hence, by using that the determinant of the product of two matrices is the product of the determinants, we have

x_{1}=\det(X_{1})=\det(A^{-1})\det(A_{1})={\frac {\det(A_{1})}{\det(A)}}.

The proof for other $x_{j}$ is similar.

Using Geometric Algebra

Main article:Comparison of vector algebra and geometric algebra § Matrix related

Inconsistent and indeterminate cases

A system of equations is said to beinconsistent when there are no solutions and it is calledindeterminate when there is more than one solution. For linear equations, an indeterminate system will have infinitely many solutions (if it is over an infinite field), since the solutions can be expressed in terms of one or more parameters that can take arbitrary values.

Cramer's rule applies to the case where the coefficient determinant is nonzero. In the 2×2 case, if the coefficient determinant is zero, then the system is inconsistent if the numerator determinants are nonzero, or indeterminate if the numerator determinants are zero.

For 3×3 or higher systems, the only thing one can say when the coefficient determinant equals zero is that if any of the numerator determinants are nonzero, then the system must be inconsistent. However, having all determinants zero does not imply that the system is indeterminate. A simple example where all determinants vanish (equal zero) but the system is still inconsistent is the 3×3 systemx+y+z=1,x+y+z=2,x+y+z=3.

See also

References

^Cramer, Gabriel (1750)."Introduction à l'Analyse des lignes Courbes algébriques" (in French). Geneva: Europeana. pp. 656–659. Retrieved2012-05-18.
^Kosinski, A. A. (2001). "Cramer's Rule is due to Cramer".Mathematics Magazine.74 (4):310–312.doi:10.2307/2691101.JSTOR 2691101.
^MacLaurin, Colin (1748).A Treatise of Algebra, in Three Parts. Printed for A. Millar & J. Nourse.
^Boyer, Carl B. (1968).A History of Mathematics (2nd ed.). Wiley. p. 431.
^Katz, Victor (2004).A History of Mathematics (Brief ed.). Pearson Education. pp. 378–379.
^Hedman, Bruce A. (1999)."An Earlier Date for "Cramer's Rule""(PDF).Historia Mathematica.26 (4):365–368.doi:10.1006/hmat.1999.2247.S2CID 121056843.
^David Poole (2014).Linear Algebra: A Modern Introduction. Cengage Learning. p. 276.ISBN 978-1-285-98283-0.
^Joe D. Hoffman; Steven Frankel (2001).Numerical Methods for Engineers and Scientists, Second Edition. CRC Press. p. 30.ISBN 978-0-8247-0443-8.
^Thomas S. Shores (2007).Applied Linear Algebra and Matrix Analysis. Springer Science & Business Media. p. 132.ISBN 978-0-387-48947-6.
^Zhiming Gong; M. Aldeen; L. Elsner (2002)."A note on a generalized Cramer's rule".Linear Algebra and Its Applications.340 (1–3):253–254.doi:10.1016/S0024-3795(01)00469-4.
^Levi-Civita, Tullio (1926).The Absolute Differential Calculus (Calculus of Tensors). Dover. pp. 111–112.ISBN 9780486634012.
^Robinson, Stephen M. (1970). "A Short Proof of Cramer's Rule".Mathematics Magazine.43 (2):94–95.doi:10.1080/0025570X.1970.11976018.

External links

Wikibooks has a book on the topic of:Linear Algebra/Cramer's Rule

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