Symbolic Linear Algebra
You can perform operations with symbolic matrices in much the same way as .with numeric matrices. Here we give examples of finding matrix products, the matrix inverse, eigenvalues, and the characteristic polynomial of a matrix  Remember that using symbolic matrices avoids numerical imprecision in subsequent operations. You can create a symbolic matrix from a numeric matrix  in several ways, as shown in the following session:
»A = sym([3, 5; 2, 7]); % the most direct method
»B = [3, 5; 2, 7];
»C = sym(S);%B is preserved as a numeric matrix
»D = subs (A, t 3, 5; 2, 7]);
The first method is the most direct. Use the second method when you want to keep a numeric version of the matrix. The matrices A and C are symbolic and identical. The matrices Band Dlook like A and C but are numeric of class double. You can create a symbolic matrix consisting of functions. For example, the relationship between the coordinates (X2. Y2) of a coordinate system rotated

counterclockwise through an angle a relative to the (x l s YI) coodinate system is
X2 = XI cosa +YI sina
Y2 = YI cosa – XI sina
These equations can be expressed in matrix form as
The symbolic matrix R.can be defined in MATLAB as follows:
»syrns a
»R = [cos(a), sin(a); -sin(a), cos(a) 1
R
cos (a),
-sin (a),
sin(a) 1
cos(a) 1
If we rotate the coordinate system twice by the same angle to produce a third coordinate system (X3, Y3), the result is the same as a single rotation with twice the angle. Let us see if MATLAB gives that result. The vector-matrix equation is = R = RR
Thus R(a)R(a) should be the same as R(2a). Continue the previous session as
follows:
»Q = R*R
Q
cos(a)A2-sin(a)A2, 2*cos(a)*sin(a)
[ -2*cos(a)*sin(a), cos(a)A2-sin(a)A2
»Q = sirnple(Q)
Q =
cos(2*a), sin(2*a)
-sin(2*a), cos(2*a)
The matrix Q is the same as R(2a), as we suspected. To evaluate a matrix numerically, use the subs and double functions. For example, for a rotation of a = 1r/4 rad (45°),
»R = subs(R,a,p

Characteristic polynom1 and Roots
Sets of first-order differential equations can be expressed in vector-matrix notation as
10.6 Symbolic Linear Algebra 633 x=Ax+Bf(t)
whereX is the vector of dependent variables and r(t) is a vectorccntaining the forcing functions. For example, the equation set
XI = X2
X2 = -kxl – 2X2 + f(t)
comes from the equation of motion of a mass connected to a spring and sliding on a surface having viscous friction. The term f(t) is the applied force acting on the mass. For this set the vector x and the matrices A and B are
B =  The equation 151- AI = 0 is the characteristic equation of the model.where 5 represents the characteristic roots of the model. Use the poly (A)function to find this polynomial, and note that MATLAB uses the default symbolic variable x to represent the roots. For example, to find the characteristic equation and solve
for the roots in terms of the spring constant k, use the following session:

»syms k
»A = [0 ,1i-k, -z r •
»poly(A)
ans =
x”2+2*x+k
»solve(ans)
ans
[ -1+(1-k)”(1/2)
[ -1- (l-k) A (1/2)
Thus the roots are 5 = -1 ± .;r-=k.
Use the eig (A) function to find the roots directly without finding the char-‘ acteristic equation (eig stands for “eigenvalue,” which is another term for “characteristic
root”). For example,
»syms k
»A = [0 ,1i-k, -2]i
»eig(A)
ans =
-1+(1-k)”(1/2)
-1- (l-k)” (1/2) 1

You can use the inv (A) and det; (A) functions to invert and find the tenninant of a matrix symbolically. For example, using the same matrix A the previous session,
>::>inv(A)
ans =
[-2/k, -11k J
[ I, a J
»A*ans % verify that the inverse is correct
ans =
[ I, a
[. a I 1
»det(A)
ans =
k
Solving Linear Algebraic Equations
You can use matrix methods in M~TLAB to solve linear algebraic equa•.••.a•• sym . By. You can use the matrix inverse method, if the inverse exis t-division method (see Chapter 6 for a discussion of these methods .
example, to solve the set
2x – 3y = 3
5x +cy = 19
using both methods, the session is
»syms c
»A = sym( [2, -3; 5, cJ);
»b = sym( [3; 19J);
»x = inv(A)*b % the matrix inverse method
x =
[3*c/(2*c+15)+57/(?*c+15)J
[231 (2*c+15) J
»x = A\b % the left-division method
. x =
[3*(19+c)/(2*c+15)J
[231 (2-*c+15) J
.Although the results appear to be different, they both reduce to the same sol
x = 3(19 + c)/(2c + 15), y = 23/(2c + 15).
Table 10.6-1 summarizes the functions used in this section. Note that 