Cambridge Additional Mathematics

(singke) #1
parallel
(no solution)

coincident
(infinitely many solutions)

unique solution

328 Matrices (Chapter 12)

We can solve

½
2 x+3y=4
5 x+4y=17

algebraically to get x=5, y=¡ 2.

Notice that this system can be written as a matrix equation

μ
23
54

¶μ
x
y


=

μ
4
17


The solution x=5, y=¡ 2 is easily checked as
μ
23
54

¶μ
5
¡ 2


=

μ
2(5) + 3(¡2)
5(5) + 4(¡2)


=

μ
4
17


X

In general, a system of linear equations can be written in the form AX=B whereAis the matrix of
coefficients,Xis the unknown column matrix, andBis a column matrix of constants.
We can use inverses to solve the matrix equation AX=B forX.

If we premultiply each side of AX=B byA¡^1 , we get
A¡^1 (AX)=A¡^1 B
) (A¡^1 A)X=A¡^1 B
) IX=A¡^1 B
and so X=A¡^1 B

If the matrix of coefficients A is invertible, then calculating
X =A¡^1 B will give a unique solution to the pair of linear
equations. This indicates that the lines intersect at a single point.

If the matrix of coefficientsAis singular, then we cannot calculate X=A¡^1 B. This indicates that either
the lines are parallel and there are no solutions, or that the lines are coincident and there are infinitely many
solutions.

Example 13 Self Tutor


a If A=

μ
23
54


, find A¡^1.

b Write the system

½
2 x+3y=4
5 x+4y=17

in matrix form.

c Hence, solve the simultaneous linear equations.

a detA= 2(4)¡3(5)
=¡ 7

) A¡^1 =¡^17

μ
4 ¡ 3
¡ 52


E Simultaneous linear equations


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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_12\328CamAdd_12.cdr Thursday, 27 February 2014 2:40:00 PM GR8GREG

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