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# Solution of system of linear equation in MATLAB

• Last Updated : 28 Jul, 2020

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 :

 `% 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 =

7
-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.

 `% 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)`

 `% 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.

 `% 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
```

 `% 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
```

 `% 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
12

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
```

 `% 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 = [](3x0)
```

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
```

 `% 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
```

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