Cambridge Additional Mathematics

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