var('a')
A = matrix ([[a,3],[4,6]])

A.add_multiple_of_row(1,0,-4/a)

B=matrix([[-3],[6]])
X=A.solve_right(B)
print 'soln is ::', X

print 'for the system to  break down permanently-'

a=A[0][1]*A[1][0]/A[1][1]

print 'a=',a
print 'for non permanent breakdown a != 2'