Symbolic substitution - MATLAB subs (2024)

Replace a with 4 in this expression.

syms a bsubs(a + b,a,4)

ans =$$b+4$$

Replace a*b with 5 in this expression.

subs(a*b^2,a*b,5)

ans =$$5\hspace{0.17em}b$$

Substitute the default symbolic scalar variable in this expression with a. If you do not specify the scalar variable or expression to replace, subs uses symvar to find the default variable. For x + y, the default variable is x.

syms x y asymvar(x + y,1)

ans =$$x$$

Therefore, subs replaces x with a.

subs(x + y,a)

ans =$$a+y$$

When you assign a new value to a symbolic scalar variable, expressions containing the variable are not automatically evaluated. Instead, evaluate expressions by using subs.

Define the expression y = x^2.

syms xy = x^2;

Assign 2 to x. The value of y is still x^2 instead of 4.

x = 2;y

y =$$x^2$$

Evaluate y with the new value of x by using subs.

subs(y)

ans =$$4$$

Make multiple substitutions by specifying the old and new values as vectors.

syms a bsubs(cos(a) + sin(b), [a,b], [sym('alpha'),2])

ans =$$\sin \left(2\right)+\cos \left(\alpha \right)$$

Alternatively, for multiple substitutions, use cell arrays.

subs(cos(a) + sin(b), {a,b}, {sym('alpha'),2})

ans =$$\sin \left(2\right)+\cos \left(\alpha \right)$$

Replace the symbolic scalar variable a in this expression with the 3-by-3 magic square matrix. Note that the constant 1 expands to the 3-by-3 matrix with all its elements equal to 1.

syms a tsubs(exp(a*t) + 1, a, -magic(3))

ans =
$$\left(\begin{array}{ccc}{\mathrm{e}}^{-8\hspace{0.17em}t}+1& {\mathrm{e}}^{-t}+1& {\mathrm{e}}^{-6\hspace{0.17em}t}+1\\ {\mathrm{e}}^{-3\hspace{0.17em}t}+1& {\mathrm{e}}^{-5\hspace{0.17em}t}+1& {\mathrm{e}}^{-7\hspace{0.17em}t}+1\\ {\mathrm{e}}^{-4\hspace{0.17em}t}+1& {\mathrm{e}}^{-9\hspace{0.17em}t}+1& {\mathrm{e}}^{-2\hspace{0.17em}t}+1\end{array}\right)$$

You can also replace an element of a vector, matrix, or array with a nonscalar value. For example, create these 2-by-2 matrices.

A = sym('A',[2,2])

A =
$$\left(\begin{array}{cc}{A}_{1,1}& A_{1,2}\\ A_{2,1}& A_{2,2}\end{array}\right)$$

B = sym('B',[2,2])

B =
$$\left(\begin{array}{cc}{B}_{1,1}& B_{1,2}\\ B_{2,1}& B_{2,2}\end{array}\right)$$

Replace the first element of the matrix A with the matrix B. While making this substitution, subs expands the 2-by-2 matrix A into this 4-by-4 matrix.

A44 = subs(A, A(1,1), B)

A44 =
$$\left(\begin{array}{cccc}{B}_{1,1}& B_{1,2}& A_{1,2}& A_{1,2}\\ B_{2,1}& B_{2,2}& A_{1,2}& A_{1,2}\\ A_{2,1}& A_{2,1}& A_{2,2}& A_{2,2}\\ A_{2,1}& A_{2,1}& A_{2,2}& A_{2,2}\end{array}\right)$$

subs does not let you replace a nonscalar or matrix with a scalar that shrinks the matrix size.

Create a structure array with symbolic expressions as the field values.

syms x y zS = struct('f1',x*y,'f2',y + z,'f3',y^2)

S = struct with fields: f1: x*y f2: y + z f3: y^2

Replace the symbolic scalar variables x, y, and z with numeric values.

Sval = subs(S,[x y z],[0.5 1 1.5])

Sval = struct with fields: f1: 1/2 f2: 5/2 f3: 1

Replace the symbolic scalar variables x and y with these 2-by-2 matrices. When you make multiple substitutions involving vectors or matrices, use cell arrays to specify the old and new values.

syms x ysubs(x*y, {x,y}, {[0 1; -1 0], [1 -1; -2 1]})

ans =
$$\left(\begin{array}{cc}0& -1\\ 2& 0\end{array}\right)$$

Note that because x and y are scalars, these substitutions are element-wise.

[0 1; -1 0].*[1 -1; -2 1]

ans = 2×2 0 -1 2 0

Eliminate scalar variables from an equation by using the variable's value from another equation. In the second equation, isolate the variable on the left side using isolate, and then substitute the right side with the variable in the first equation.

First, declare the equations eqn1 and eqn2.

syms x yeqn1 = sin(x)+y == x^2 + y^2;eqn2 = y*x == cos(x);

Isolate y in eqn2 by using isolate.

eqn2 = isolate(eqn2,y)

eqn2 =
$$y=\frac{\cos \left(x\right)}{x}$$

Eliminate y from eqn1 by substituting the left side of eqn2 with the right side of eqn2.

eqn1 = subs(eqn1,lhs(eqn2),rhs(eqn2))

eqn1 =
$$\sin \left(x\right)+\frac{\cos \left(x\right)}{x}=\frac{{\cos \left(x\right)}^2}{x^2}+x^2$$

Replace x with a in this symbolic function.

syms x y asyms f(x,y)f(x,y) = x + y;f = subs(f,x,a)

f(x, y) =$$a+y$$

subs replaces the values in the symbolic function formula, but it does not replace input arguments of the function.

formula(f)

ans =$$a+y$$

argnames(f)

ans =$$\left(\begin{array}{cc}x& y\end{array}\right)$$

Replace the arguments of a symbolic function explicitly.

syms x yf(x,y) = x + y;f(a,y) = subs(f,x,a);f

f(a, y) =$$a+y$$

Suppose you want to verify the solutions of this system of equations.

syms x yeqs = [x^2 + y^2 == 1, x == y];S = solve(eqs,[x y]);S.x

ans =
$$\left(\begin{array}{c}-\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)$$

S.y

ans =
$$\left(\begin{array}{c}-\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)$$
See Also
Symbolic Toolbox and Related Functions

Verify the solutions by substituting the solutions into the original system.

isAlways(subs(eqs,S))

ans = 2x2 logical array 1 1 1 1

Since R2021b

Define the product of two 2-by-2 matrices. Declare the matrices as symbolic matrix variables with the symmatrix data type.

syms X Y [2 2] matrixsM = X*Y

sM =$$X\hspace{0.17em}Y$$

Replace the matrix variables $$X$$ and $$Y$$ with 2-by-2 symbolic matrices. When you make multiple substitutions involving vectors or matrices, use cell arrays to specify the matrix variables to be substituted and their new values. The new values must have the same size as the matrix variables to be substituted.

S = subs(sM,{X,Y},{[0 sqrt(sym(2)); sqrt(sym(2)) 0], [1 -1; -2 1]})

S =
$$\begin{array}{l}{\Sigma }_1\\ \\ \mathrm{where}\\ \\ \mathrm{}{\Sigma }_1=\left(\begin{array}{cc}-2\hspace{0.17em}\sqrt{2}& \sqrt{2}\\ \sqrt{2}& -\sqrt{2}\end{array}\right)\end{array}$$

Convert the expression S to the sym data type to show the result of the substituted matrix multiplication.

Ssym = symmatrix2sym(S)

Ssym =
$$\left(\begin{array}{cc}-2\hspace{0.17em}\sqrt{2}& \sqrt{2}\\ \sqrt{2}& -\sqrt{2}\end{array}\right)$$

Since R2021b

Create a matrix of symbolic numbers.

A = sym([1 4 2; 4 1 2; 2 2 3])

A =
$$\left(\begin{array}{ccc}1& 4& 2\\ 4& 1& 2\\ 2& 2& 3\end{array}\right)$$

Compute the coefficients of the characteristic polynomial of A using the charpoly function.

c = charpoly(A)

c =$$\left(\begin{array}{cccc}1& -5& -17& 21\end{array}\right)$$

Next, define $$X$$ as a 3-by-3 symbolic matrix variable. Use the coefficients c to create the polynomial $$p(X)=c_1{X}^3+c_2{X}^2+c_{3}X+c_4{I}_3$$, where $$X$$ is an indeterminate that represents a 3-by-3 matrix.

syms X [3 3] matrixp = c(1)*X^3 + c(2)*X^2 + c(3)*X + c(4)*X^0

p =$$21\hspace{0.17em}{\mathrm{I}}_3-17\hspace{0.17em}X-5\hspace{0.17em}{X}^2+X^3$$

Substitute $$X$$ in the polynomial $$p(X)$$ with A using the subs function. According to the Cayley-Hamilton theorem, this substitution results in a 3-by-3 zero matrix because the coefficients c are the characteristic polynomial of A. Use symmatrix2sym to convert the substituted expression to a matrix of symbolic numbers.

Y = subs(p,A)

Y =
$$\begin{array}{l}-17\hspace{0.17em}{\Sigma }_1-5\hspace{0.17em}{{\Sigma }_1}^2+{{\Sigma }_1}^3+21\hspace{0.17em}{\mathrm{I}}_3\\ \\ \mathrm{where}\\ \\ \mathrm{}{\Sigma }_1=\left(\begin{array}{ccc}1& 4& 2\\ 4& 1& 2\\ 2& 2& 3\end{array}\right)\end{array}$$

Z = symmatrix2sym(Y)

Z =
$$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$$

Since R2022a

Define the function $\mathit{f}\left(\mathit{A}\right)={\mathit{A}}^2-2\mathit{A}+{\mathrm{I}}_2$, where $\mathit{A}$ is a 2-by-2 matrix and ${\mathrm{I}}_2$ is a 2-by-2 identity matrix. Substitute the variable $\mathit{A}$ with another expression and evaluate the new function.

Create a 2-by-2 symbolic matrix variable $\mathit{A}$. Create a symbolic matrix function $\mathit{f}\left(\mathit{A}\right)$, keeping the existing definition of $\mathit{A}$ in the workspace. Assign the polynomial expression of $\mathit{f}\left(\mathit{A}\right)$.

syms A 2 matrixsyms f(A) 2 matrix keepargsf(A) = A*A - 2*A + eye(2)

f(A) =$${\mathrm{I}}_2-2\hspace{0.17em}A+A^2$$

Next, create new symbolic matrix variables $\mathit{B}$ and $\mathit{C}$. Create a new symbolic matrix function $\mathit{g}\left(\mathit{B},\mathit{C}\right)$, keeping the existing definitions of $\mathit{B}$ and $\mathit{C}$ in the workspace.

syms B C 2 matrixsyms g(B,C) 2 matrix keepargs

Substitute the variable $\mathit{A}$ in $\mathit{f}\left(\mathit{A}\right)$ with $\mathit{B}+\mathit{C}$. Assign the substituted result to the new function $\mathit{g}\left(\mathit{B},\mathit{C}\right)$.

g(B,C) = subs(f,A,B+C)

g(B, C) =$${\left(B+C\right)}^2+{\mathrm{I}}_2-2\hspace{0.17em}B-2\hspace{0.17em}C$$

Evaluate $\mathit{g}\left(\mathit{B},\mathit{C}\right)$ for the matrix values $\mathit{B}=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ and $\mathit{C}=\left[\begin{array}{cc}1& -1\\ -2& 1\end{array}\right]$ using subs.

S = subs(g(B,C),{B,C},{[0 1; -1 0],[1 -1; -2 1]})

S =
$$\begin{array}{l}-2\hspace{0.17em}{\Sigma }_1-2\hspace{0.17em}{\Sigma }_2+{\left({\Sigma }_1+{\Sigma }_2\right)}^2+{\mathrm{I}}_2\\ \\ \mathrm{where}\\ \\ \mathrm{}{\Sigma }_1=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ \\ \mathrm{}{\Sigma }_2=\left(\begin{array}{cc}1& -1\\ -2& 1\end{array}\right)\end{array}$$

Convert the expression S from the symmatrix data type to the sym data type to show the result of the substituted polynomial.

Ssym = symmatrix2sym(S)

Ssym =
$$\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$$

Since R2022b

Define the equation $\frac{\partial }{\partial X^{\mathrm{T}}}\mathrm{}\frac{\partial }{\partial X}\mathrm{}f\left(X,A\right)=2\hspace{0.17em}A$, where $\mathit{A}$ is a 3-by-3 matrix and $\mathit{X}$ is a 3-by-1 matrix. Substitute $f(\mathit{X},\mathit{A})$ with another symbolic expression and $\mathit{A}$ with symbolic values. Check if the equation is true for these values.

Create two symbolic matrix variables $\mathit{A}$ and $\mathit{X}$. Create a symbolic matrix function $f(\mathit{X},\mathit{A})$, keeping the existing definitions of $\mathit{A}$ and $\mathit{X}$ in the workspace. Create the equation.

syms A [3 3] matrixsyms X [3 1] matrixsyms f(X,A) [1 1] matrix keepargseq = diff(diff(f,X),X.') == 2*A

eq(X, A) =
$$\frac{\partial }{\partial X^{\mathrm{T}}}\mathrm{}\frac{\partial }{\partial X}\mathrm{}f\left(X,A\right)=2\hspace{0.17em}A$$

Substitute $f(\mathit{X},\mathit{A})$ with ${\mathit{X}}^{\mathit{T}}\mathit{AX}$ and evaluate the second-order differential function in the equation for this expression.

eq = subs(eq,f,X.'*A*X)

eq(X, A) =$$A^{\mathrm{T}}+A=2\hspace{0.17em}A$$

Substitute $\mathit{A}$ with the Hilbert matrix of order 3.

eq = subs(eq,A,hilb(3))

eq(X, A) =
$$\begin{array}{l}{\Sigma }_1+{{\Sigma }_1}^{\mathrm{T}}=2\hspace{0.17em}{\Sigma }_1\\ \\ \mathrm{where}\\ \\ \mathrm{}{\Sigma }_1=\left(\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{2}& \frac{1}{3}& \frac{1}{4}\\ \frac{1}{3}& \frac{1}{4}& \frac{1}{5}\end{array}\right)\end{array}$$

Check if the equation is true for these values by using isAlways. Because isAlways accepts only a symbolic input of type symfun or sym, convert eq from type symfunmatrix to type symfun before using isAlways.

tf = isAlways(symfunmatrix2symfun(eq))

tf = 3x3 logical array 1 1 1 1 1 1 1 1 1

Since R2023b

Define the expression $\mathit{X}{\mathit{Y}}^2-\mathit{Y}{\mathit{X}}^2$, where $\mathit{X}$ and $\mathit{Y}$ are 3-by-3 matrices. Create the matrices as symbolic matrix variables.

syms X Y [3 3] matrixC = X*Y^2 - Y*X^2

C =$$X\hspace{0.17em}{Y}^2-Y\hspace{0.17em}{X}^2$$

Assign values to the matrices X and Y.

X = [-1 2 pi; 0 1/2 2; 2 1 0];Y = [3 2 2; -1 2 1; 1 2 -1];

Evaluate the expression C with the assigned values of X and Y by using subs.

Cnew = subs(C)

Cnew =
$$\begin{array}{l}-{\Sigma }_1\hspace{0.17em}{{\Sigma }_2}^2+{\Sigma }_2\hspace{0.17em}{{\Sigma }_1}^2\\ \\ \mathrm{where}\\ \\ \mathrm{}{\Sigma }_1=\left(\begin{array}{ccc}3& 2& 2\\ -1& 2& 1\\ 1& 2& -1\end{array}\right)\\ \\ \mathrm{}{\Sigma }_2=\left(\begin{array}{ccc}-1& 2& \pi \\ 0& \frac{1}{2}& 2\\ 2& 1& 0\end{array}\right)\end{array}$$

Convert the result from the symmatrix data type to the double data type.

Cnum = double(Cnew)

Cnum = 3×3 -42.8496 -13.3584 -13.4336 -0.7168 3.1416 0.0752 -3.2832 29.8584 16.4248