Cramer's rule

Proof

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 to this column means that this function has the form

${\displaystyle D_j(a_{1,j},\dots ,a_{n,j})=C_{1,j}{a}_{1,j}+\cdots ,C_{n,j}{a}_{n,j},}$

where the ${\displaystyle C_{i,j}}$ are coefficients that depend on the entries of A that are not in column j. So, one has

${\displaystyle det(A)=D_j(a_{1,j},\dots ,a_{n,j})=C_{1,j}{a}_{1,j}+\cdots ,C_{n,j}{a}_{n,j}}$

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

If the function ${\displaystyle D_j}$ 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 ${\displaystyle x_1,\dots ,x_n}$, whose coefficient matrix is A, with det(A) assumed to be nonzero:

${\displaystyle \begin{array}{ccc}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n& =& b_1\\ a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n& =& b_2\\ & ⋮& \\ a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{nn}{x}_n& =& b_n.\end{array}}$

If one combines these equations by taking C_1,j times the first equation, plus C_2,j times the second, and so forth until C_n,j times the last, then for every k the resulting coefficient of x_k becomes

${\displaystyle D_j(a_{1,k},\dots ,a_{n,k}).}$

So, all coefficients become zero, except the coefficient of ${\displaystyle x_j}$ that becomes ${\displaystyle det(A).}$ Similarly, the constant coefficient becomes ${\displaystyle D_j(b_1,\dots ,b_n),}$ and the resulting equation is thus

${\displaystyle det(A)x_j=D_j(b_1,\dots ,b_n),}$

which gives the value of ${\displaystyle x_j}$ as

${\displaystyle x_j=\frac{1}{det(A)}{D}_j(b_1,\dots ,b_n).}$

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 ${\displaystyle D_j}$ as jth row, for ${\displaystyle j=1,\dots ,n}$ (this is the adjugate matrix for A). Expressed in matrix terms, we have thus to prove that

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

is a solution; that is, that

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

For that, it suffices to prove that

${\displaystyle A\left(\frac{1}{det(A)}M\right)=I_n,}$

where ${\displaystyle I_n}$ is the identity matrix.

The above properties of the functions ${\displaystyle D_j}$ show that one has MA = det(A)I_n, and therefore,

${\displaystyle \left(\frac{1}{det(A)}M\right)A=I_n.}$

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.

Finding inverse matrix

Let A be an n × n matrix with entries in a field F. Then

${\displaystyle A\mathrm{adj}(A)=\mathrm{adj}(A)A=det(A)I}$

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

${\displaystyle A^{-1}=\frac{1}{det(A)}\mathrm{adj}(A).}$

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).

Applications

Explicit formulas for small systems

Consider the linear system

${\displaystyle \{\begin{array}{cc}{a}_{1}x+b_{1}y& =c_1\\ a_{2}x+b_{2}y& =c_2\end{array}}$

which in matrix format is

${\displaystyle \left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right].}$

Assume a₁b₂ − b₁a₂ is nonzero. Then, with the help of determinants, x and y can be found with Cramer's rule as

${\displaystyle \begin{array}{cc}x& =\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}=\frac{c_1{b}_2-b_1{c}_2}{a_1{b}_2-b_1{a}_2},y=\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}=\frac{a_1{c}_2-c_1{a}_2}{a_1{b}_2-b_1{a}_2}\end{array}.}$

The rules for 3 × 3 matrices are similar. Given

${\displaystyle \{\begin{array}{cc}{a}_{1}x+b_{1}y+c_{1}z& =d_1\\ a_{2}x+b_{2}y+c_{2}z& =d_2\\ a_{3}x+b_{3}y+c_{3}z& =d_3\end{array}}$

which in matrix format is

${\displaystyle \left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right].}$

Then the values of x, y and z can be found as follows:

${\displaystyle x=\frac{\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|},y=\frac{\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|},z=\frac{\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|}.}$

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.

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 ${\displaystyle (M,g)}$ be a Riemannian manifold equipped with local coordinates ${\displaystyle (x^1,x^2,\dots ,x^n)}$. Let ${\displaystyle A=A^i\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^i}}$ be a vector field. We use the summation convention throughout.

Theorem.

The divergence of ${\displaystyle A}$,

${\displaystyle \mathrm{div}A=\frac{1}{\sqrt{detg}}\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^i}\left(A^i\sqrt{detg}\right),}$

is invariant under change of coordinates.

Computing derivatives implicitly

Consider the two equations ${\displaystyle F(x,y,u,v)=0}$ and ${\displaystyle G(x,y,u,v)=0}$. When u and v are independent variables, we can define ${\displaystyle x=X(u,v)}$ and ${\displaystyle y=Y(u,v).}$

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

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.

Examples

2x2 System

Consider the linear system

${\displaystyle \begin{array}{cc}12x+3y& =15\\ 2x-3y& =13\end{array}}$

Applying Cramer's Rule gives

${\displaystyle \begin{array}{cc}x& =\frac{\left|\begin{array}{cc}15& 3\\ 13& -3\end{array}\right|}{\left|\begin{array}{cc}12& 3\\ 2& -3\end{array}\right|}=\frac{-84}{-42}=2,y=\frac{\left|\begin{array}{cc}12& 15\\ 2& 13\end{array}\right|}{\left|\begin{array}{cc}12& 3\\ 2& -3\end{array}\right|}=-\frac{126}{42}=-3\end{array}.}$

These values can be verified by substituting back into the original equations: $${\displaystyle 12x+3y=(12\times 2)+(3\times (-3))=24-9=15}$$ and $${\displaystyle 2x-3y=(2\times 2)-(3\times (-3))=4-(-9)=13,}$$

as required.

3x3 System

Consider the linear system

${\displaystyle \begin{array}{cc}3x-2y+5z& =2\\ 4x-7y-z& =19\\ 5x-6y+4z& =13\end{array}}$

To simplify notation, define $${\displaystyle A=\left[\begin{array}{ccc}3& -2& 5\\ 4& -7& -1\\ 5& -6& 4\end{array}\right]}$$ as the matrix of coefficients, and $${\displaystyle A_1=\left[\begin{array}{ccc}2& -2& 5\\ 19& -7& -1\\ 13& -6& 4\end{array}\right],A_2=\left[\begin{array}{ccc}3& 2& 5\\ 4& 19& -1\\ 5& 13& 4\end{array}\right],A_3=\left[\begin{array}{ccc}3& -2& 2\\ 4& -7& 19\\ 5& -6& 13\end{array}\right].}$$ Calculating the required determinants gives $${\displaystyle |A|=-5,|A_1|=-5,|A_2|=10,|A_3|=5.}$$

Applying Cramer's Rule gives

${\displaystyle \begin{array}{cc}x& =\frac{\left|\begin{array}{c}{A}_1\end{array}\right|}{\left|\begin{array}{c}A\end{array}\right|}=\frac{-5}{-5}=1,y=\frac{\left|\begin{array}{c}{A}_2\end{array}\right|}{\left|\begin{array}{c}A\end{array}\right|}=\frac{10}{-5}=-2,z=\frac{\left|\begin{array}{c}{A}_3\end{array}\right|}{\left|\begin{array}{c}A\end{array}\right|}=\frac{5}{-5}=-1\end{array}.}$

These values can be verified by substituting back into the original equations: $${\displaystyle 3x-2y+5z=(3\times 1)-(2\times (-2))+(5\times (-1))=3+4-5=2}$$ $${\displaystyle 4x-7y-z=(4\times 1)-(7\times (-2))-(1\times (-1))=4+14+1=19}$$ $${\displaystyle 5x-6y+4z=(5\times 1)-(6\times (-2))+(4\times (-1))=5+12-4=13}$$

as required.

Algorithm

The following pseudocode describes Cramer's rule for calculating the solution of a linear system ${\displaystyle A\mathbf{x}=\mathbf{b}}$, provided a unique solution exists.

<b id="mwAVU">input:</b> Coefficient matrix <i id="mwAVY">A</i>, Column vector <i id="mwAVc">b</i>, Number of variables <i id="mwAVg">n</i> <b id="mwAVk">output:</b> Column vector <i id="mwAVo">x</i> representing the solution of <i id="mwAVs">Ax</i> = <i id="mwAVw">b</i> <i id="mwAV0">D</i> ← <i id="mwAV4">det</i>(<i id="mwAV8">A</i>) <b id="mwAWA">if</b> <i id="mwAWE">D</i> = 0 <b id="mwAWI">then</b> print("<i id="mwAWM">Unique solution does not exist.</i>") return <b id="mwAWQ">end if</b> <b id="mwAWU">for</b> <i id="mwAWY">i</i> ← <i id="mwAWc">1</i> to <i id="mwAWg">n</i> <b id="mwAWk">do</b> <i id="mwAWo">Aᵢ</i> ← <i id="mwAWs">A</i> Replace column <i id="mwAWw">i</i> of <i id="mwAW0">Aᵢ</i> with column vector <i id="mwAW4">b</i> <i id="mwAW8">xᵢ</i> ← <i id="mwAXA">det</i>(<i id="mwAXE">Aᵢ</i>) / <i id="mwAXI">D</i> <b id="mwAXM">end if</b> <b id="mwAXQ">return</b> <i id="mwAXU">x</i>

Geometric interpretation

Cramer's rule has a geometric interpretation that can also be considered 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

${\displaystyle \begin{array}{cc}{a}_{11}{x}_1+a_{12}{x}_2& =b_1\\ a_{21}{x}_1+a_{22}{x}_2& =b_2\end{array}}$

it can be considered as an equation between vectors

${\displaystyle x_1(\genfrac{}{}{0ex}{}{a_{11}}{a_{21}})+x_2(\genfrac{}{}{0ex}{}{a_{12}}{a_{22}})=(\genfrac{}{}{0ex}{}{b_1}{b_2}).}$

The area of the parallelogram determined by ${\displaystyle (\genfrac{}{}{0ex}{}{a_{11}}{a_{21}})}$ and ${\displaystyle (\genfrac{}{}{0ex}{}{a_{12}}{a_{22}})}$ is given by the determinant of the system of equations:

${\displaystyle \left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|.}$

In general, when there are more variables and equations, the determinant of n vectors of length n will give the volume of the parallelepiped determined by those vectors in the n-th dimensional Euclidean space.

Therefore, the area of the parallelogram determined by ${\displaystyle x_1(\genfrac{}{}{0ex}{}{a_{11}}{a_{21}})}$ and ${\displaystyle (\genfrac{}{}{0ex}{}{a_{12}}{a_{22}})}$ has to be ${\displaystyle x_1}$ times the area of the first one since one of the sides has been multiplied by this factor. Now, this last parallelogram, by Cavalieri's principle, has the same area as the parallelogram determined by ${\displaystyle (\genfrac{}{}{0ex}{}{b_1}{b_2})=x_1(\genfrac{}{}{0ex}{}{a_{11}}{a_{21}})+x_2(\genfrac{}{}{0ex}{}{a_{12}}{a_{22}})}$ and ${\displaystyle (\genfrac{}{}{0ex}{}{a_{12}}{a_{22}}).}$

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

${\displaystyle \left|\begin{array}{cc}{b}_1& a_{12}\\ b_2& a_{22}\end{array}\right|=\left|\begin{array}{cc}{a}_{11}{x}_1& a_{12}\\ a_{21}{x}_1& a_{22}\end{array}\right|=x_1\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|}$

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 ${\displaystyle \mathbf{x}=(x_1,\dots ,x_n)\mapsto \frac{1}{detA}\left(det(A_1),\dots ,det(A_n)\right),}$ where ${\displaystyle A_i}$ is the matrix ${\displaystyle A}$ with ${\displaystyle \mathbf{x}}$ substituted in the ${\displaystyle i}$th column, as in Cramer's rule. Because of linearity of determinant in every column, this map is linear. Observe that it sends the ${\displaystyle i}$th column of ${\displaystyle A}$ to the ${\displaystyle i}$th basis vector ${\displaystyle {\mathbf{e}}_i=(0,\dots ,1,\dots ,0)}$ (with 1 in the ${\displaystyle i}$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 ${\displaystyle A}$ on the column space; hence it agrees with ${\displaystyle A^{-1}}$ on the span of the column space. Since ${\displaystyle A}$ is invertible, the column vectors span all of ${\displaystyle {\mathbb{R}}^n}$, so our map really is the inverse of ${\displaystyle A}$. Cramer's rule follows.

A short proof

A short proof of Cramer's rule can be given by noticing that ${\displaystyle x_1}$ is the determinant of the matrix

${\displaystyle X_1=\left[\begin{array}{ccccc}{x}_1& 0& 0& \cdots & 0\\ x_2& 1& 0& \cdots & 0\\ x_3& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ x_n& 0& 0& \cdots & 1\end{array}\right]}$

On the other hand, assuming that our original matrix A is invertible, this matrix ${\displaystyle X_1}$ has columns ${\displaystyle A^{-1}\mathbf{b},A^{-1}{\mathbf{v}}_2,\dots ,A^{-1}{\mathbf{v}}_n}$, where ${\displaystyle {\mathbf{v}}_n}$ is the n-th column of the matrix A. Recall that the matrix ${\displaystyle A_1}$ has columns ${\displaystyle \mathbf{b},{\mathbf{v}}_2,\dots ,{\mathbf{v}}_n}$, and therefore ${\displaystyle X_1=A^{-1}{A}_1}$. Hence, by using that the determinant of the product of two matrices is the product of the determinants, we have

${\displaystyle x_1=det(X_1)=det(A^{-1})det(A_1)=\frac{det(A_1)}{det(A)}.}$

The proof for other ${\displaystyle x_j}$ is similar.

Using Geometric Algebra

Inconsistent and indeterminate cases

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 inconsistent if the numerator determinants are nonzero, or indeterminate if the numerator determinants are zero.

Example 1: An Indeterminate System

Consider the linear system

${\displaystyle \begin{array}{cc}2x-y& =8\\ 4x-2y& =16\end{array}}$

By inspection, the second equation is a multiple of the first, so there are infinitely many solutions. Applying Cramer's Rule gives

${\displaystyle \begin{array}{cc}x& =\frac{\left|\begin{array}{cc}8& -1\\ 16& -2\end{array}\right|}{\left|\begin{array}{cc}2& -1\\ 4& -2\end{array}\right|}=\frac{0}{0},y=\frac{\left|\begin{array}{cc}2& 8\\ 4& 16\end{array}\right|}{\left|\begin{array}{cc}2& -1\\ 4& -2\end{array}\right|}=\frac{0}{0}\end{array}}$

which both have zero numerators and denominators, meaning the system has infinitely many solutions and is therefore indeterminate.

Example 2: An Inconsistent System

Consider the linear system

${\displaystyle \begin{array}{cc}3x-4y& =1\\ 3x-4y& =2\end{array}}$

By inspection, the left hand side of both equations are identical, with different right hand sides. Therefore, the system is inconsistent, meaning there are no solutions. Applying Cramer's Rule gives

${\displaystyle \begin{array}{cc}x& =\frac{\left|\begin{array}{cc}1& -4\\ 2& -4\end{array}\right|}{\left|\begin{array}{cc}3& -4\\ 3& -4\end{array}\right|}=\frac{4}{0},y=\frac{\left|\begin{array}{cc}3& 1\\ 3& 2\end{array}\right|}{\left|\begin{array}{cc}3& -4\\ 3& -4\end{array}\right|}=\frac{3}{0}\end{array}}$

which are both undefined due to division by zero, however the non-zero numerators means the system has no solutions and is therefore inconsistent.

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 system x+y+z=1, x+y+z=2, x+y+z=3.

Practical Considerations

Cramer's rule, implemented in a naive way, is computationally inefficient for systems of more than two or three equations. 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 (up to a constant factor independent of ⁠${\displaystyle n}$⁠) computational complexity as the computation of a single determinant. Moreover, the 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. In 1983, an algorithm for solving the system using Cramer's rule in ${\displaystyle n^3+O(n^2)}$ operations was proposed. This algorithm can use permutations exactly like those in the Gaussian method. Therefore, if approximate calculation methods are used, the solution will be more stable than the solution using the Gaussian method, since the first steps are calculated without error, before rounding off the numbers. In 1991, an algorithm with complexity of ${\displaystyle n^3/2+O(n^2)}$ operations was published. . In 1997, an algorithm for solving a system using Cramer's rule with complexity equal to that of matrix multiplication was proposed.. For example, for the Strassen's Multiplication Algorithm, this algorithm computes the solution in ${\displaystyle (7/15)n^{{\log }_2(7)}+O(n^2)}$ operations.

Despite the computational inefficiency of Cramer's Rule, it still remains an important theoretical tool in linear algebra for studying the mathematical properties of a solution without the need for solving the system, and for engineers and mathematicians working on low dimensional problems by hand with messy decimals. For example, faced with a problem like

${\displaystyle \begin{array}{cc}2.87761x-3.16534y& =7.82234\\ -5.91823x+2.81859y& =11.34308\end{array}}$

with no tools other than a hand calculator, applying Cramer's Rule would provide the solution quite efficiently.