# Solve a Matrix Equation Algebraically#

Use SymPy to solve a matrix (linear) equation. For example, solving $$\left[\begin{array}{cc} c & d\\1 & -e\end{array}\right] \left[\begin{array}{cc} x\\y\end{array}\right] = \left[\begin{array}{cc} 2\\0\end{array}\right]$$ yields $$\left[\begin{array}{cc} x\\y\end{array}\right] = \left[\begin{array}{cc} \frac{2e}{ce+d}\\\frac{2}{ce+d}\end{array}\right]$$.

## Alternatives to Consider#

• If your matrix and constant vector contain only numbers, not symbols, for example $$\left[\begin{array}{cc} 1 & 2\\3 & 4\end{array}\right] \left[\begin{array}{cc} x\\y\end{array}\right] = \left[\begin{array}{cc} 2\\0\end{array}\right]$$, you can use one of these other free and open-source packages instead of SymPy:

• Solving a matrix equation is equivalent to solving a system of linear equations, so if you prefer you can Solve a System of Equations Algebraically

• If you formulated your problem as a system of linear equations, and want to convert it to matrix form, you can use linear_eq_to_matrix() and then follow the procedures in this guide.

## Solve a Matrix Equation#

Here is an example of solving a matrix equation with SymPy’s sympy.matrices.matrices.MatrixBase.solve(). We use the standard matrix equation formulation $$Ax=b$$ where

• $$A$$ is the matrix representing the coefficients in the linear equations

• $$x$$ is the column vector of unknowns to be solved for

• $$b$$ is the column vector of constants, where each row is the value of an equation

>>> from sympy import init_printing
>>> init_printing(use_unicode=True)

>>> from sympy import symbols
>>> from sympy.matrices import Matrix
>>> c, d, e = symbols("c, d, e")
>>> A = Matrix([[c,d], [1, -e]])
>>> A
⎡c  d ⎤
⎢     ⎥
⎣1  -e⎦
>>> b = Matrix([2, 0])
>>> b
⎡2⎤
⎢ ⎥
⎣0⎦
>>> A.solve(b)
⎡  2⋅e  ⎤
⎢───────⎥
⎢c⋅e + d⎥
⎢       ⎥
⎢   2   ⎥
⎢───────⎥
⎣c⋅e + d⎦


## Guidance#

### Matrix Usually Must Be Square#

The matrix $$A$$ usually must be square to represent a system of linear equations with the same number of unknowns as equations. If not, SymPy will give the error ShapeError: self and rhs must have the same number of rows.

The exception to the requirement that a matrix be square comes from SymPy’s use of the Moore-Penrose pseudoinverse.

### Methods for Solving Matrix Equations#

SymPy’s matrix solving method, sympy.matrices.matrices.MatrixBase.solve(), can use several different methods, which are listed at that API reference link. Depending on the nature of the matrix, a given method may be more efficient. By default, Gauss-Jordan elimination will be used.

Specifying a method in solve is equivalent to using a specialized solving function. For example, using solve with method='LU' calls LUsolve().

### Solving Several Matrix Equations With the Same Matrix#

If you need to repeatedly solve matrix equations with the same matrix $$A$$ but different constant vectors $$b$$, it is more efficient to use one of the following methods.

You can use LU decomposition via LUsolve():

>>> from sympy import symbols, Matrix, eye, simplify
>>> c, d, e = symbols("c, d, e")
>>> A = Matrix([[c,d], [1, -e]])
>>> A
⎡c  d ⎤
⎢     ⎥
⎣1  -e⎦
>>> b = Matrix([2, 0])
>>> b
⎡2⎤
⎢ ⎥
⎣0⎦
>>> solution = A.LUsolve(b)
>>> solution
⎡  2⋅e  ⎤
⎢───────⎥
⎢c⋅e + d⎥
⎢       ⎥
⎢   2   ⎥
⎢───────⎥
⎣c⋅e + d⎦
>>> # Demonstrate that solution is correct
>>> simplify(A * solution)
⎡2⎤
⎢ ⎥
⎣0⎦
>>> b2 = Matrix([4, 0])
>>> b2
⎡4⎤
⎢ ⎥
⎣0⎦
>>> solution2 = A.LUsolve(b2)
>>> solution2
⎡  4⋅e  ⎤
⎢───────⎥
⎢c⋅e + d⎥
⎢       ⎥
⎢   4   ⎥
⎢───────⎥
⎣c⋅e + d⎦
>>> # Demonstrate that solution2 is correct
>>> simplify(A * solution2)
⎡4⎤
⎢ ⎥
⎣0⎦


Another approach is to compute the inverse matrix, but this is almost always slower, and significantly slower for larger matrices. If efficient computation is not a priority, you can use inv():

>>> from sympy import symbols, Matrix, simplify
>>> c, d, e = symbols("c, d, e")
>>> A = Matrix([[c,d], [1, -e]])
>>> b = Matrix([2, 0])
>>> b
⎡2⎤
⎢ ⎥
⎣0⎦
>>> b2 = Matrix([4, 0])
>>> b2
⎡4⎤
⎢ ⎥
⎣0⎦
>>> inv = A.inv()
>>> inv
⎡   e        d   ⎤
⎢───────  ───────⎥
⎢c⋅e + d  c⋅e + d⎥
⎢                ⎥
⎢   1       -c   ⎥
⎢───────  ───────⎥
⎣c⋅e + d  c⋅e + d⎦
>>> # Solves Ax = b for x
>>> solution = inv * b
>>> solution
⎡  2⋅e  ⎤
⎢───────⎥
⎢c⋅e + d⎥
⎢       ⎥
⎢   2   ⎥
⎢───────⎥
⎣c⋅e + d⎦
>>> # Demonstrate that solution is correct
>>> simplify(A * solution)
⎡2⎤
⎢ ⎥
⎣0⎦
>>> # Solves Ax = b2 for x
>>> solution2 = inv * b2
>>> solution2
⎡  4⋅e  ⎤
⎢───────⎥
⎢c⋅e + d⎥
⎢       ⎥
⎢   4   ⎥
⎢───────⎥
⎣c⋅e + d⎦
>>> # Demonstrate that solution2 is correct
>>> simplify(A * solution2)
⎡4⎤
⎢ ⎥
⎣0⎦


Determining the inverse of a large symbolic matrix may not be computationally tractable.

### Work With Symbolic Matrices#

The computational complexity of manipulating symbolic matrices can increase rapidly with matrix size. For example, the number of terms in the determinant of a symbolic matrix increases with the factorial of the matrix dimension. As a result, the maximum dimensionality of matrices that can be solved is more limited than for numerical matrices. For example, the determinant of this 4x4 symbolic matrix has 24 terms with four elements in each term:

>>> from sympy import MatrixSymbol
>>> A = MatrixSymbol('A', 4, 4).as_explicit()
>>> A
⎡A₀₀  A₀₁  A₀₂  A₀₃⎤
⎢                  ⎥
⎢A₁₀  A₁₁  A₁₂  A₁₃⎥
⎢                  ⎥
⎢A₂₀  A₂₁  A₂₂  A₂₃⎥
⎢                  ⎥
⎣A₃₀  A₃₁  A₃₂  A₃₃⎦
>>> A.det()
A₀₀⋅A₁₁⋅A₂₂⋅A₃₃ - A₀₀⋅A₁₁⋅A₂₃⋅A₃₂ - A₀₀⋅A₁₂⋅A₂₁⋅A₃₃ + A₀₀⋅A₁₂⋅A₂₃⋅A₃₁ + A₀₀⋅A₁
₃⋅A₂₁⋅A₃₂ - A₀₀⋅A₁₃⋅A₂₂⋅A₃₁ - A₀₁⋅A₁₀⋅A₂₂⋅A₃₃ + A₀₁⋅A₁₀⋅A₂₃⋅A₃₂ + A₀₁⋅A₁₂⋅A₂₀⋅
A₃₃ - A₀₁⋅A₁₂⋅A₂₃⋅A₃₀ - A₀₁⋅A₁₃⋅A₂₀⋅A₃₂ + A₀₁⋅A₁₃⋅A₂₂⋅A₃₀ + A₀₂⋅A₁₀⋅A₂₁⋅A₃₃ -
A₀₂⋅A₁₀⋅A₂₃⋅A₃₁ - A₀₂⋅A₁₁⋅A₂₀⋅A₃₃ + A₀₂⋅A₁₁⋅A₂₃⋅A₃₀ + A₀₂⋅A₁₃⋅A₂₀⋅A₃₁ - A₀₂⋅A₁
₃⋅A₂₁⋅A₃₀ - A₀₃⋅A₁₀⋅A₂₁⋅A₃₂ + A₀₃⋅A₁₀⋅A₂₂⋅A₃₁ + A₀₃⋅A₁₁⋅A₂₀⋅A₃₂ - A₀₃⋅A₁₁⋅A₂₂⋅
A₃₀ - A₀₃⋅A₁₂⋅A₂₀⋅A₃₁ + A₀₃⋅A₁₂⋅A₂₁⋅A₃₀


and solving a matrix equation of it takes about a minute, whereas the analogous 3x3 matrix takes less than one second. The more unrelated, symbolic entries in a matrix, the more likely it is to be slow to manipulate. This example, finding a general solution to a matrix where all elements are independent symbols, is the extreme case and thus the slowest for a matrix of its size.

### Speed up Solving Matrix Equations#

Here are some suggestions:

• If matrix elements are zero, ensure that they are recognized as zero. You can do this by either making them zero or by applying assumptions.

• Selecting a solve method suited to the properties of the matrix, for example hermitian, symmetric, or triangular. Refer to Methods for Solving Matrix Equations.

• Use the DomainMatrix class, which can be faster to operate on because it limits the domain of matrix elements.

## Use the Solution Result#

### Use the Solution as a Vector#

You can use the solution result as a vector. For example, to prove that the solution $$x$$ is correct, you can multiply it the matrix $$A$$ and verify that it produces the constants vector $$b$$:

>>> from sympy import symbols, simplify
>>> from sympy.matrices import Matrix
>>> c, d, e = symbols("c, d, e")
>>> A = Matrix([[c,d], [1, -e]])
>>> b = Matrix([2, 0])
>>> solution = A.solve(b)
>>> solution
⎡  2⋅e  ⎤
⎢───────⎥
⎢c⋅e + d⎥
⎢       ⎥
⎢   2   ⎥
⎢───────⎥
⎣c⋅e + d⎦
>>> # Not immediately obvious whether this result is a zeroes vector
>>> (A * solution) - b
⎡ 2⋅c⋅e      2⋅d      ⎤
⎢─────── + ─────── - 2⎥
⎢c⋅e + d   c⋅e + d    ⎥
⎢                     ⎥
⎣          0          ⎦
>>> # simplify reveals that this result is a zeroes vector
>>> simplify((A * solution) - b)
⎡0⎤
⎢ ⎥
⎣0⎦


Note that we had to use simplify() to make SymPy simplify the expression in a matrix element to make it immediately obvious that the solution is correct.

### Extract Elements From the Solution#

Because you can iterate through the elements in a column vector, you can extract its elements using standard Python techniques. For example, you can create a list of the elements using list comprehension

>>> [element for element in solution]
⎡  2⋅e       2   ⎤
⎢───────, ───────⎥
⎣c⋅e + d  c⋅e + d⎦


or you can extract individual elements by subscripting

>>> solution[0]
2⋅e
───────
c⋅e + d


## Equations With No Solution#

If the determinant of a matrix is zero, matrix equations with it have no solution:

>>> from sympy import symbols
>>> from sympy.matrices import Matrix
>>> c, d, e = symbols("c, d, e")
>>> A = Matrix([[c*e**2, d*e], [c*e, d]])
>>> A
⎡   2     ⎤
⎢c⋅e   d⋅e⎥
⎢         ⎥
⎣c⋅e    d ⎦
>>> b = Matrix([2, 0])
>>> A.LUsolve(b)
Traceback (most recent call last):
...
NonInvertibleMatrixError: Matrix det == 0; not invertible.


## Report a Bug#

If you find a bug with matrix-solving functions, please post the problem on the SymPy mailing list. Until the issue is resolved, you can use a different method listed in Alternatives to Consider.