In linear algebra, Cramer's rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the column vector of right-sides of the equations. It is named after Gabriel Cramer, who published the rule for an arbitrary number of unknowns in 1750,[1][2] although Colin 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 of n equations in n unknowns, it requires computation of n + 1 determinants, while Gaussian elimination produces the result with the same computational complexity as the computation of a single determinant.[8][9][verification needed] Cramer's rule can also be numerically unstable even for 2×2 systems.[10] However, Cramer's rule can be implemented with the same complexity as Gaussian elimination,[11][12] (consistently requires twice as many arithmetic operations and has the same numerical stability when the same permutation matrices are applied).
Consider a system of n linear equations for n unknowns, represented in matrix multiplication form as follows:
![{\displaystyle A\mathbf {x} =\mathbf {b} }](//wikimedia.org/api/rest_v1/media/math/render/svg/45d894430af69e29d6dda5aacbf4bb19336226a0)
where the n × n matrix A has a nonzero determinant, and the vector
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:
![{\displaystyle x_{i}={\frac {\det(A_{i})}{\det(A)}}\qquad i=1,\ldots ,n}](//wikimedia.org/api/rest_v1/media/math/render/svg/39364326c745c93758b2a762e1e25e6e185e2fa7)
where
is the matrix formed by replacing the i-th column of A by the column vector b.
A more general version of Cramer's rule[13] considers the matrix equation
![{\displaystyle AX=B}](//wikimedia.org/api/rest_v1/media/math/render/svg/e51b626e9793be3e577dc5157905281ec0af0d32)
where the n × n matrix A has a nonzero determinant, and X, B are n × m matrices. Given sequences
and
, let
be the k × k submatrix of X with rows in
and columns in
. Let
be the n × n matrix formed by replacing the
column of A by the
column of B, for all
. Then
![{\displaystyle \det X_{I,J}={\frac {\det(A_{B}(I,J))}{\det(A)}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/09d1c90167bf04aa9368d3cbbb5e5f17cb3144e2)
In the case
, this reduces to the normal Cramer's rule.
The rule holds for systems of equations with coefficients and unknowns in any field, not just in the real numbers.
The proof for Cramer's rule uses the following properties 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 index j 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 the jth column. Linearity with respect of this column means that this function has the form
![{\displaystyle D_{j}(a_{1,j},\ldots ,a_{n,j})=C_{1,j}a_{1,j}+\cdots ,C_{n,j}a_{n,j},}](//wikimedia.org/api/rest_v1/media/math/render/svg/c8042fededdee012c05a9b9610733616a2ebd1a9)
where the
are coefficients that depend on the entries of A that are not in column j. So, one has
![{\displaystyle \det(A)=D_{j}(a_{1,j},\ldots ,a_{n,j})=C_{1,j}a_{1,j}+\cdots ,C_{n,j}a_{n,j}}](//wikimedia.org/api/rest_v1/media/math/render/svg/3aae0dcaeae9af10bc4b1376ecf1c5aa7d794cb7)
(Laplace expansion provides a formula for computing the
but their expression is not important here.)
If the function
is applied to any other column k of A, then the result is the determinant of the matrix obtained from A by replacing column j by a copy of column k, so the resulting determinant is 0 (the case of two equal columns).
Now consider a system of n linear equations in n unknowns
, whose coefficient matrix is A, with det(A) assumed to be nonzero:
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/0e5b07174955a83288cf8a4c004cc8e5af2b216e)
If one combines these equations by taking C1,j times the first equation, plus C2,j times the second, and so forth until Cn,j times the last, then for every k the resulting coefficient of xk becomes
![{\displaystyle D_{j}(a_{1,k},\ldots ,a_{n,k}).}](//wikimedia.org/api/rest_v1/media/math/render/svg/745f5196096af73d8b33df6d09d5f57497769b74)
So, all coefficients become zero, except the coefficient of
that becomes
Similarly, the constant coefficient becomes
and the resulting equation is thus
![{\displaystyle \det(A)x_{j}=D_{j}(b_{1},\ldots ,b_{n}),}](//wikimedia.org/api/rest_v1/media/math/render/svg/08bf3e6ac987c0ff0d0648796d3f3760d3908572)
which gives the value of
as
![{\displaystyle x_{j}={\frac {1}{\det(A)}}D_{j}(b_{1},\ldots ,b_{n}).}](//wikimedia.org/api/rest_v1/media/math/render/svg/79310f1c0fc09d5c13261b5dbe1be40d74901f8a)
As, by construction, the numerator is the determinant of the matrix obtained from A by replacing column j by b, 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. Let M be the n × n matrix that has the coefficients of
as jth row, for
(this is the adjugate matrix for A). Expressed in matrix terms, we have thus to prove that
![{\displaystyle \mathbf {x} ={\frac {1}{\det(A)}}M\mathbf {b} }](//wikimedia.org/api/rest_v1/media/math/render/svg/ac765f17a9a6a9fbbaf605d6adb982e557117c62)
is a solution; that is, that
![{\displaystyle A\left({\frac {1}{\det(A)}}M\right)\mathbf {b} =\mathbf {b} .}](//wikimedia.org/api/rest_v1/media/math/render/svg/18b7b30ed0d41119b388ac2c142cf47bdc791ecd)
For that, it suffices to prove that
![{\displaystyle A\,\left({\frac {1}{\det(A)}}M\right)=I_{n},}](//wikimedia.org/api/rest_v1/media/math/render/svg/84d0938bf006d47042e9bcab1ef48116d8ea7675)
where
is the identity matrix.
The above properties of the functions
show that one has MA = det(A)In, and therefore,
![{\displaystyle \left({\frac {1}{\det(A)}}M\right)\,A=I_{n}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/0dc9784861bac61ef7065b9a17c680baa751c891)
This completes the proof, since a left inverse of a square matrix is also a right-inverse (see Invertible matrix theorem).
For other proofs, see below.
Let A be an n × n matrix with entries in a field F. Then
![{\displaystyle A\,\operatorname {adj} (A)=\operatorname {adj} (A)\,A=\det(A)I}](//wikimedia.org/api/rest_v1/media/math/render/svg/772e1a41d38d7da3e0deea206a6b6422950a40d0)
where adj(A) denotes the adjugate matrix, det(A) is the determinant, and I is the identity matrix. If det(A) is nonzero, then the inverse matrix of A is
![{\displaystyle A^{-1}={\frac {1}{\det(A)}}\operatorname {adj} (A).}](//wikimedia.org/api/rest_v1/media/math/render/svg/be8419966809179aab9ca75dc761b582ffc0b569)
This gives a formula for the inverse of A, provided det(A) ≠ 0. In fact, this formula works whenever F is a commutative ring, provided that det(A) is a unit. If det(A) is not a unit, then A is not invertible over the ring (it may be invertible over a larger ring in which some non-unit elements of F may be invertible).
Consider the linear system
![{\displaystyle \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.}](//wikimedia.org/api/rest_v1/media/math/render/svg/4877b2a677a47973c8703266eeddc5c695653edd)
which in matrix format is
![{\displaystyle {\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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/bb5d9569703337ad94d4badcfc4013af8cdfd6ac)
Assume a1b2 − b1a2 is nonzero. Then, with the help of determinants, x and y can be found with Cramer's rule as
![{\displaystyle {\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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/567da479f71b8d15a62e17388c0d1fff4550434a)
The rules for 3 × 3 matrices are similar. Given
![{\displaystyle \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.}](//wikimedia.org/api/rest_v1/media/math/render/svg/84885104605b22cb330496b879fcdb0796bfac60)
which in matrix format is
![{\displaystyle {\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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/bf640e00506027189b4a7bf0668c16ab42c095bc)
Then the values of x, y and z can be found as follows:
![{\displaystyle 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}}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/c53430c6efce025a9bb8fc7cd6e4983c0f3b10ad)
Differential geometry
Ricci calculus
Cramer's rule is used in the Ricci calculus in various calculations involving the Christoffel symbols of the first and second kind.[14]
In particular, Cramer's rule can be used to prove that the divergence operator on a Riemannian manifold is invariant with respect to change of coordinates. We give a direct proof, suppressing the role of the Christoffel symbols.
Let
be a Riemannian manifold equipped with local coordinates
. Let
be a vector field. We use the summation convention throughout.
- Theorem.
- The divergence of
,
![{\displaystyle \operatorname {div} A={\frac {1}{\sqrt {\det g}}}{\frac {\partial }{\partial x^{i}}}\left(A^{i}{\sqrt {\det g}}\right),}](//wikimedia.org/api/rest_v1/media/math/render/svg/a747d3b5d0a40e5c6da6819f25ec22f1bd67a2cf)
- is invariant under change of coordinates.
More information Let
...
Proof |
Let be a coordinate transformation with non-singular Jacobian. Then the classical transformation laws imply that where . Similarly, if , then .
Writing this transformation law in terms of matrices yields , which implies .
Now one computes
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/c0bbb78374fe5bc8322cb4c64248bed44748296b)
In order to show that this equals
,
it is necessary and sufficient to show that
![{\displaystyle {\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 ,}](//wikimedia.org/api/rest_v1/media/math/render/svg/c89832b94bc831005750c3077e8b4a8eed6b2361)
which is equivalent to
![{\displaystyle {\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 }}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/ae03a7009e20a94371677960c981e4b1264a7ba8)
Carrying out the differentiation on the left-hand side, we get:
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/663b93468c64515dbf6c3a262cb2334517ac4886)
where denotes the matrix obtained from by deleting the th row and th column.
But Cramer's Rule says that
![{\displaystyle {\frac {(-1)^{i+j}}{\det \left({\frac {\partial x}{\partial {\bar {x}}}}\right)}}\det M(i|j)}](//wikimedia.org/api/rest_v1/media/math/render/svg/d6bb4966e095ea29c537a72a1d638bb8c3fded3e)
is the th entry of the matrix .
Thus
![{\displaystyle (\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}}},}](//wikimedia.org/api/rest_v1/media/math/render/svg/6f6e00c66051e74b640bb6ad440914283d8d18cc)
completing the proof. |
Close
Computing derivatives implicitly
Consider the two equations
and
. When u and v are independent variables, we can define
and ![{\displaystyle y=Y(u,v).}](//wikimedia.org/api/rest_v1/media/math/render/svg/ca30c1576f21e64d2c5c91bdaf0f57c96c3ab2e5)
An equation for
can be found by applying Cramer's rule.
More information Calculation of
...
Calculation of ![{\displaystyle {\dfrac {\partial x}{\partial u}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/3f661a2acc487ec3c1f0b5cb7a60c768f1e03a73) |
First, calculate the first derivatives of F, G, x, and y:
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/717cfe9b94cd928ab12a788deedfa99a2ea41e09)
Substituting dx, dy into dF and dG, we have:
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/db9780c618030244fe08ce9e6f53ac6ad0cd9a49)
Since u, v are both independent, the coefficients of du, dv must be zero. So we can write out equations for the coefficients:
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/faa8362cce15633e1c0bc856b3d6be35b8cf9231)
Now, by Cramer's rule, we see that:
![{\displaystyle {\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}}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/f37ad34982b4d7ef027bd3aae6b7d0fade25ec2e)
This is now a formula in terms of two Jacobians:
![{\displaystyle {\frac {\partial x}{\partial u}}=-{\frac {\left({\frac {\partial (F,G)}{\partial (u,y)}}\right)}{\left({\frac {\partial (F,G)}{\partial (x,y)}}\right)}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/5d4db738a2f7334725e1bbd98061b8ba4f207190)
Similar formulas can be derived for ![{\displaystyle {\frac {\partial x}{\partial v}},{\frac {\partial y}{\partial u}},{\frac {\partial y}{\partial v}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/a4a552943a352a244c330efdde84be8b035ba93c) |
Close
Integer programming
Cramer's rule can be used to prove that an integer programming problem whose constraint matrix is totally 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 of variation of parameters.
A proof by abstract linear algebra
This is a restatement of the proof above in abstract language.
Consider the map
where
is the matrix
with
substituted in the
th column, as in Cramer's rule. Because of linearity of determinant in every column, this map is linear. Observe that it sends the
th column of
to the
th basis vector
(with 1 in the
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
on the column space; hence it agrees with
on the span of the column space. Since
is invertible, the column vectors span all of
, so our map really is the inverse of
. Cramer's rule follows.
A short proof
A short proof of Cramer's rule [15] can be given by noticing that
is the determinant of the matrix
![{\displaystyle 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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/6f6076678f2a1cfed489fe0462c5269301793eb0)
On the other hand, assuming that our original matrix A is invertible, this matrix
has columns
, where
is the n-th column of the matrix A. Recall that the matrix
has columns
, and therefore
. Hence, by using that the determinant of the product of two matrices is the product of the determinants, we have
![{\displaystyle x_{1}=\det(X_{1})=\det(A^{-1})\det(A_{1})={\frac {\det(A_{1})}{\det(A)}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/db8f1cf8b140c99cd47f899df35deb29c82c70d0)
The proof for other
is similar.
A system of equations is said to be inconsistent when there are no solutions and it is called indeterminate 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 incompatible 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 incompatible is the 3×3 system x+y+z=1, x+y+z=2, x+y+z=3.
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.
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. Nicholas J. Higham (2002). Accuracy and Stability of Numerical Algorithms: Second Edition. SIAM. p. 13. ISBN 978-0-89871-521-7. G.I.Malaschonok (1983). "Solution of a System of Linear Equations in an Integral Ring". USSR J. Of Comput. Math. And Math. Phys. 23: 1497–1500. arXiv:1711.09452. Levi-Civita, Tullio (1926). The Absolute Differential Calculus (Calculus of Tensors). Dover. pp. 111–112. ISBN 9780486634012.