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Problems #1 and #7, pg.47 are graded.
Problem #1,pg.47:Determine if the system has a nontrivial solution or not:
{2x1−5x2+8x3=0−2x1−7x2+x3=04x1+2x2+7x3=0.
Solution: We form the augmented matrix associated with this sytem of equations and compute
[2−580−2−7104270]∼[10178001−3400000].
Now interpret the reduced echelon form as a system to get
{x1+178x3=0x2−34x3=00=0
or equivalently
{x1=−178x3x2=34x30=0.
Thus we see the solution vector →x has the form
→x=[x1x2x3]=[−178x334x3x3]=x3[−178341],
where x3 is a free variable. Hence the system has nontrivial solutions (one for each nonzero value that x3 can take).
Problem #7,pg.47: Describe all solutions of A→x=→0 in parametric vector form, where
A=[13−3701−45].
Solution: We write the matrix equation A→x=→0 as a vector equation in the following way:
x1[10]+x2[31]+x3[−3−4]+x4[−3−4]=→0.
We solve this vector equation by writing the augmented matrix and compute
[13−37001−450]∼[109−8001−450].
Hence we see that this is equivalent to the system of equations
{x1+9x3−8x4=0x2−4x3+5x4=0,
or equivalently
{x1=−9x3+8x4x2=4x3−5x4.
Therefore we may write the solution vector →x as
→x=[x1x2x3x4]=[−9x3+8x44x3−5x4x3x4]=x3[−9410]+x4[8−501],
which is the solution expressed in parametric vector form, with parameters (i.e. free variables) x3 and x4.
