||'''Book'''|| Getting started with MATLAB ||
 ||'''Author'''|| Rudra Pratap ||
 ||'''Edition'''||  ||

/* code_begins */

{{{
var('r')
A=matrix([[5,2*r,r],[3,6,2*r-1],[2,r-1,3*r]])
print "Matrix A is given as:\n",A
b=matrix([[2],[3],[5]])
print "Matrix b is given as:\n",b
var('x1,x2,x3')
X=matrix([[x1],[x2],[x3]])
print "Matrix X is given as:\n",X
print "To solve matrix AX=b ,find inverse for A"
print "Inverse is only possible if the matrix is non-singular i.e determinant of matrix is not zero."
x(r)=A.det()
print "Determinant of A \nx(r)=",x(r)
print "To solve this problem consider any valid value for r.For instance consider r=1.So x(1)=",x(1)
K1=A(1)

Result:

Matrix A is given as:
[      5     2*r       r]
[      3       6 2*r - 1]
[      2   r - 1     3*r]
Matrix b is given as:
[2]
[3]
[5]
Matrix X is given as:
[x1]
[x2]
[x3]
To solve matrix AX=b ,find inverse for A
Inverse is only possible if the matrix is non-singular i.e determinant
of matrix is not zero.
Determinant of A 
x(r)= -5*(r - 1)*(2*r - 1) + 3*(r - 1)*r + 4*(2*r - 1)*r - 18*r^2 + 78*r
To solve this problem consider any valid value for r.For instance
consider r=1.So x(1)= 64
A(1)=
[5 2 1]
[3 6 1]
[2 0 3]
X=(A(1).inverse)*b
Resultant matrix X=
[-1/32]
[15/64]
[27/16]
Value of x1= (-1/32)
Value of x2= (15/64)
Value of x3= (27/16)
print "A(1)=\n" ,K1
K=A(1).inverse()
print "X=(A(1).inverse)*b"
print "Resultant matrix X=\n",K*b
X=K*b
print "Value of x1=",X[0]
print "Value of x2=",X[1]
print "Value of x3=",X[2]
}}}

/* code_ends */

 * '''Solution by''': 
   * <T.R.N.Karthik>, <Student>, <Gitam University>