Solution
A homogeneous system has the form AX=0.
(i)
⎩⎨⎧x+4y−2z=02x+y+5z=05x+2y+8z=0
Row-reduction gives rank 3, so only the trivial solution exists:
x=0,y=0,z=0
(ii)
⎩⎨⎧x1+4x2+2x3=02x1+x2−3x3=03x1+2x2−4x3=0
RREF gives:
x1−2x3x2+x3=0=0
Let x3=t. Then x1=2t and x2=−t.
(x1,x2,x3)=t(2,−1,1)
(iii)
⎩⎨⎧x1+2x2−x3=0x1−x2+5x3=02x1+x2+4x3=0
RREF gives:
x1+3x3x2−2x3=0=0
Let x3=t. Then x1=−3t and x2=2t.
(x1,x2,x3)=t(−3,2,1)