Solution of system of linear equation in MATLAB (original) (raw)

Last Updated : 15 Jul, 2025

Let us see how to solve a system of linear equations in MATLAB. Here are the various operators that we will be deploying to execute our task :

\ operator : A \ B is the matrix division of A into B, which is roughly the same as INV(A) * B. If A is an NXN matrix and B is a column vector with N components or a matrix with several such columns, then X = A \ B is the solution to the equation **A * X = B**. A warning message is printed if A is badly scaled or nearly singular. A\EYE(SIZE(A)) produces the inverse of A.
linsolve operator : **X = LINSOLVE(A, B)** solves the linear system A * X = B using LU factorization with partial pivoting when A is square, and QR factorization with column pivoting. A warning is given if A is ill conditioned for square matrices and rank deficient for rectangular matrices.

Example 1 : Non-homogeneous System Ax = b, where A is a square and is invertible. In our example we will consider the following equations :

2x + y - z = 7 x -2y + 5z = -13 3x + 5y - 4z = 18

We will convert these equations into matrices A and b :

MATLAB `

% declaring the matrices based on the equations A = [2 1 -1; 1 -2 5; 3 5 -4] b = [7; -13; 18]

Output :

A =

2 1 -1 1 -2 5 3 5 -4

b =

-13 18

Now we will create an augmented matrix Ab. We will compare the ranks of Ab and A, if the ranks are equal then a unique solution exists.

MATLAB `

% creating augmented matrix Ab = [A b]

% checking the ranks if rank(A) == rank(Ab) display("Unique solution exists") else display("Unique solution does not exist")
end

Output :

Ab =

2    1   -1    7
1   -2    5  -13
3    5   -4   18

Unique solution exists

Now we can find the solution to this system of equations by using 3 methods:

conventional way : inv(A) * b
using mid-divide routine : A \ b
using linsolve routine : linsolve(A, b) MATLAB `

% conventional way of finding solution x_inv = inv(A) * b

% using mid-divide routine of MATLAB x_bslash = A \ b

% using linsolve routine of MATLAB x_linsolve = linsolve(A, b)

Output :

x_inv =

2.0000e+00 8.8818e-16 -3.0000e+00

x_bslash =

2.0000e+00 9.6892e-16 -3.0000e+00

x_linsolve =

2.0000e+00 9.6892e-16 -3.0000e+00

We can verify the correctness of the solution by finding the error using A * x - b. The error should be 0.

MATLAB `

% check for errors Er1 = A * x_inv - b Er2 = A * x_bslash - b Er3 = A * x_linsolve - b

Output :

Er1 =

-8.8818e-16 -3.5527e-15 0.0000e+00

Er2 =

-1.7764e-15 -1.7764e-15 0.0000e+00

Er3 =

-1.7764e-15 -1.7764e-15 0.0000e+00

As all the errors are close to 0, we can say that the solution is correct.Example 2 : Non-homogeneous system Ax = b, where A is a square and it is not invertible. In our example we will consider the following equations :

2x + 4y + 6z = 7 3x -2y + 1z = 2 1x + 2y + 3z = 5

MATLAB `

% declaring the matrices based on the equations A = [2 4 6; 3 -2 1; 1 2 3] b = [7; 2; 5]

% creating augmented matrix Ab = [A b]

% checking the ranks if rank(A) == rank(Ab) display("Unique solution exists") else display("Unique solution does not exist")
end

% conventional way of finding solution % gives warning and wrong answer. x_inv = inv(A) * b

% using mid-divide routine of MATLAB % this too gives warning and wrong answer. x_bslash = A \ b

% check for errors Er1 = A * x_inv - b Er2 = A * x_bslash - b

Output :

A =

2 4 6 3 -2 1 1 2 3

b =

7 2 5

Ab =

2 4 6 7 3 -2 1 2 1 2 3 5

Unique solution does not exist warning: matrix singular to machine precision warning: called from testing at line 17 column 7 x_inv =

Inf Inf Inf

warning: matrix singular to machine precision warning: called from testing at line 21 column 10
x_bslash =

-Inf -Inf Inf

Er1 =

Inf NaN Inf

Er2 =

NaN NaN NaN

Example 3 : Non-homogeneous system Ax = b where A is not a square. In our example we will consider the following equations :

2a + c - d + e = 2 a + c - d + e = 1 12a + 2b + 8c + 2e = 12

MATLAB `

% declaring the matrices based on the equations A = [2 0 1 -1 1; 1 0 1 -1 1; 12 2 8 0 2] b = [2; 1; 12]

% creating augmented matrix Ab = [A b]

% checking the ranks if rank(A) == rank(Ab) display("Solution exists") else display("Solution does not exist")
end

% checking for unique solution if rank(A) == 5 display("Unique solution exists") else display("Unique solution does not exist")
end

Output :

A =

2    0    1   -1    1
1    0    1   -1    1

12 2 8 0 2

b =

2
1

Ab =

2    0    1   -1    1    2
1    0    1   -1    1    1

12 2 8 0 2 12

Solution exists Unique solution does not exist

Example 4 : Homogeneous system Ax = 0 where A is a square and is invertible. In our example we will consider the following equations :

6x + 2y + 3z = 0 4x - y + 2z = 0 2x + y + 5z = 0

MATLAB `

% declaring the matrices based on the equations A = [6 2 3; 4 -1 2; 2 1 5] b = [0; 0; 0]

% checking for unique solution if rank(A) == 3 display("Unique solution exists") else display("Unique solution does not exist")
endif

% trivial solution x = A \ b

% getting a null set. % this is obvious as A is invertible. % so its null space contains only zero vector x = null(A)

Output :

A =

6 2 3 4 -1 2 2 1 5

b =

0 0 0

Unique solution exists x =

0 0 0

x =

Example 5 : Homogeneous system Ax = 0 where A is a square and is not invertible. In our example we will consider the following equations :

1x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 9z = 0

MATLAB `

% declaring the matrices based on the equations A = [1 2 3; 4 5 6; 7 8 9] b = [0; 0; 0]

% checking for unique solution if rank(A) == 3 display("Unique solution exists") else display("Unique solution does not exist")
endif

% trivial solution with warning x = A \ b

% this will return a set containing % only one basis vector of null space of A % the null space of A is spanned by this vector % hence this vector or its scalar multiple % is the solution of the given homogeneous system x = null(A)

% finding the errors Err = A*x - b

Output :

A =

1 2 3 4 5 6 7 8 9

b =

0 0 0

Unique solution does not exist warning: matrix singular to machine precision, rcond = 1.54198e-18 warning: called from testing at line 13 column 3 x =

0 0 0

x =

0.40825 -0.81650 0.40825

Err =

-1.3323e-15 -4.4409e-16 4.4409e-16