epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 161
This system can be separated into two parts:
{
ax-\-bz=\ . {ay-\-bt — 0
ex + Jz = 0 \cy -\- dt =1'
The solutions can be found because det A = ad — bc^Q. Performing
the calculation we obtain:
So A~^ is the inverse matrix of A.
Part 3: Statement 2 impHes statement 3. The system is formed by the
two equations:
If we solve it, we obtain:
ax^-hy = Q and cx-\- dy = 0.
{ad — bc)x = 0 and (ad — bc)y = 0.
Obviously, x = 0 and }; = 0 is a solution. Because det ^ = ad - fee / 0,
each one of these two equations has only one solution. Therefore, the
system's only solution is x = 0 and y = 0.
Part 4: Statement 3 implies statement 2. The system is formed by the
two equations:
ax + by = 0 and cx~\- dy — 0.
Because the system has a unique solution, either a^^O or c^^O.
Indeed, if a = 0 and c = 0 the system would have an infinite number
of solutions of the form (x,0), where x is any real number. We assume
that a/0. Then x = —by/a. Substituting this expression into the
second equation, we obtain:
(ad - bc)y ^ ^
a
This means that
(ad — bc)y = 0.
In order ior y = 0 to be the only solution of this equation, we must
have ad~bc^ 0. Therefore, det v4 / 0.