Download 1.5. Matrix Equations

1.5. Matrix Equations
If we have an m × n system of linear equations
a11x1 + a12x2 + · · · + a1nxn = b1
a21x1 + a22x2 + · · · + a2nxn = b2
..
am1x1 + am2x2 + · · · + amnxn = bm
with coefficient matrix



A=

and we define
x
a11 a12
a21 a22
..
..
am1 am2

x1
 x

=  ..2

xn





· · · a1n
· · · a2n
..
...
· · · amn
and b

b1
 b

=  ..2

bm








,

then the linear system can be expressed simply as the
matrix equation
Ax = b.
Solving the Matrix Equation When A is Invertible
If the coefficient matrix, A, in the matrix equation Ax =
b is an n × n invertible matrix, then we can solve the
equation for x as follows:
−1 (Ax) = A−1b
A
Ax = b =⇒
³
´
−1
=⇒ A A x = A−1b
=⇒ Inx = A−1b
=⇒ x = A−1b.
Example
Write the linear system
−2x − 3y + z = 1
−x + y + 2z = 12
−3x + z = 3
as a matrix equation and then solve the equation by computing A−1.
Theorem
If A is an n×n invertible matrix and b is an n—dimensional
vector, then the matrix equation Ax = b has the unique
solution x = A−1b.
Homogeneous Linear Systems
An m × n system of linear equations of the form
a11x1 + a12x2 + · · · + a1nxn = 0
a21x1 + a22x2 + · · · + a2nxn = 0
..
am1x1 + am2x2 + · · · + amnxn = 0
is called a homogeneous system. It is obvious that
any homogeneous linear system has at least one solution
which is x1 = x2 = · · · = xn = 0. This solution is
called the trivial solution.
We can write the above homogeneous system as the matrix equation
Ax = 0m.
If A is an n×n invertible matrix, then the unique solution
of this matrix equation is
x = A−10m = 0m.
Thus, if A is invertible, then the trivial solution is the
only solution of the homogeneous system.
Example
Find the solution set of the homogeneous system
x + 2y + z = 0
x + 3y = 0
x + y + 2z = 0.
Comparison of Homogeneous and Nonhomogeneous n × n Systems
1. Suppose that A is an n × n (square) matrix.
(a) If A is invertible, then Ax = 0 has a unique
solution.
(b) if A is not invertible then Ax = 0 has infinitely
many solutions.
2. Suppose that A is an n × n (square) matrix.
(a) If A is invertible then Ax = b has a unique
solution (no matter what b is).
(b) If A is not invertible then Ax = b has either no
solution or infinitely many solutions. Which of
these is the case depends on what b is.
Example
Explain (without doing any computations) why the system
x − 5y − 15z = 0
−3x + 13y + 39z = 0
2x − 9y − 27z = 0
has at least one solution. Then (by doing computations)
decide whether this system has exactly one solution or
infinitely many solutions.
Example
By referring to the previous example, explain why the
system
x − 5y − 15z = b1
−3x + 13y + 39z = b2
2x − 9y − 27z = b3
cannot have a unique solution (no matter what b1, b2,
and b3 are). Then:
1) Find b1, b2, and b3 such that the above system has
infinitely many solutions.
2) Find b1, b2, and b3 such that the above system has
no solutions.
Homework
In Section 1.5 (page 51), do problems 1—31 (odd)