Examples of Linear Systems as Augmented Matrices .

> restart;with(linalg):

Warning, new definition for norm

Warning, new definition for trace

A Homogeneous system

Definition : A LS is called homogeneous if the constant terms are all zero. Clearly, every homogeneous LS has at least one solution, namely = ... (called the trivial solution ). Therefore, homogeneous LS's always have either just the trivial solution or infinitely many solutions.

> m2:=matrix([[2, 3, 5, 0], [-1, 7, -1, 0], [4, -11, k, 0]]);

Write out the homogeneous LS represented by m2 and determine k so that the LS has infinitely many solutions.

> m3:=gausselim(m2);gaussjord(m2);

Clearly, m2 will have infinitely many solutions if the last row of m3 becomes a zero row (why?). Therefore, take k = 7.

> m3[3,3];

> solve(%);

Theorem : A homogeneous LS with more unknowns (variables) than equations has infinitely many solutions.

Why?

Conditions on the constants of a LS.

Write out the LS represented by the augmented m4 matrix and determine the conditions on a, b and c so that the LS has infinitely many solutions.

> m4:=matrix(3,3,[3,2,-4,-4,1,-1,7,12,-22]);

> augment(m4,matrix(3,1,[a,b,c]));

> gausselim(%);

The result above shows that c - 5a - 2b must be zero for infinitely many sol.'s (why?).

> augment(m4,matrix(3,1,[1,-3,-1])); # check with concrete numbers that obey the conditions.

The system represented by this matrix obeys c - 5a - 2b = 0.

> rref(%);

Conditions on Coeff's & Constants of a LS.

Write out the LS represented by m5 and determine conditions on a so that the LS has:

a) infinitely many solutions b) no solution c) exactly one solution.

> m5:=matrix(3,4,[1,1,2,-1,1,2,1,a-1,-1,0,a^2-12,2*a+4]);

> gausselim(m5);rref(m5);

Look at the near REF above and analyze. Clearly there will be:

a) infinitely many solutions if = 0 = . I.E. if a = -3 .

b) no solution if = 0 but . I.E. if a = 3 .

c) exactly one solution for all other choices of a (why?)