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This section shows you how to solve a system of linear equations using the Symbolic Math Toolbox™.

A system of linear equations

$$\begin{array}{l}{a}_{11}{x}_{1}+{a}_{12}{x}_{2}+\dots +{a}_{1n}{x}_{n}={b}_{1}\\ {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+\dots +{a}_{2n}{x}_{n}={b}_{2}\\ \cdots \\ {a}_{m1}{x}_{1}+{a}_{m2}{x}_{2}+\dots +{a}_{mn}{x}_{n}={b}_{m}\end{array}$$

can be represented as the matrix equation $$A\cdot \overrightarrow{x}=\overrightarrow{b}$$, where *A* is
the coefficient matrix,

$$A=\left(\begin{array}{ccc}{a}_{11}& \dots & {a}_{1n}\\ \vdots & \ddots & \vdots \\ {a}_{m1}& \cdots & {a}_{mn}\end{array}\right)$$

and $$\overrightarrow{b}$$ is the vector containing the right sides of equations,

$$\overrightarrow{b}=\left(\begin{array}{c}{b}_{1}\\ \vdots \\ {b}_{m}\end{array}\right)$$

If you do not have the system of linear equations in the form ```
AX
= B
```

, use `equationsToMatrix`

to convert
the equations into this form. Consider the following system.

$$\begin{array}{l}2x+y+z=2\\ -x+y-z=3\\ x+2y+3z=-10\end{array}$$

Declare the system of equations.

syms x y z eqn1 = 2*x + y + z == 2; eqn2 = -x + y - z == 3; eqn3 = x + 2*y + 3*z == -10;

Use `equationsToMatrix`

to convert the equations
into the form `AX = B`

. The second input to `equationsToMatrix`

specifies
the independent variables in the equations.

[A,B] = equationsToMatrix([eqn1, eqn2, eqn3], [x, y, z])

A = [ 2, 1, 1] [ -1, 1, -1] [ 1, 2, 3] B = 2 3 -10

Use `linsolve`

to solve `AX = B`

for
the vector of unknowns `X`

.

X = linsolve(A,B)

X = 3 1 -5

From `X`

, *x* = 3, *y* = 1 and *z* = -5.

Use `solve`

instead of `linsolve`

if
you have the equations in the form of expressions and not a matrix
of coefficients. Consider the same system of linear equations.

$$\begin{array}{l}2x+y+z=2\\ -x+y-z=3\\ x+2y+3z=-10\end{array}$$

Declare the system of equations.

syms x y z eqn1 = 2*x + y + z == 2; eqn2 = -x + y - z == 3; eqn3 = x + 2*y + 3*z == -10;

Solve the system of equations using `solve`

.
The inputs to `solve`

are a vector of equations,
and a vector of variables to solve the equations for.

sol = solve([eqn1, eqn2, eqn3], [x, y, z]); xSol = sol.x ySol = sol.y zSol = sol.z

xSol = 3 ySol = 1 zSol = -5

`solve`

returns the solutions in a structure
array. To access the solutions, index into the array.