082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

284 Chapter 8 Matrix algebra
To understand this expression it is necessary that matrix multiplication has been fully
mastered, for, by multiplying outthe l.h.s.,wefind
( ( ) ( )
2 4 x 2x +4y
=
1 −3)
y 1x−3y
and the form (8.7)follows immediately.
We can writeEquation (8.7) as
AX=B (8.8)
( )
( ( )
2 4 x 14
whereAisthe matrix ,X isthe matrix andBis the matrix .
1 −3 y)
−8
( x
In order to findX = it is now necessary to makeX the subject of the equation
y)
AX =B. We can premultiply Equation (8.8) byA −1 , the inverse ofA, provided such an
inverse exists,togive
A −1 AX =A −1 B
Then, noting thatA −1 A =I, we find
that is
IX =A −1 B
X =A −1 B
usingthepropertiesoftheidentitymatrix.WehavenowmadeX thesubjectoftheequationasrequiredandweseethattofindX
wemustpremultiplyther.h.s.ofEquation(8.8)
by the inverse ofA.
Inthiscase
A −1 = 1 ( ) −3 −4
−10 −1 2
( ) 3/10 2/5
=
1/10 −1/5
and
A −1 B =
=
( ) ( ) 3/10 2/5 14
1/10 −1/5 −8
( 1
3)
( x
that is,X = =
y)
( 1
3)
, so thatx = 1 andy = 3 isthe required solution.
IfAX =BthenX =A −1 BprovidedA −1 exists.
Thistechniquecanbeappliedtothreeequationsinthreeunknownsinananalogousway.