The sum of three number is 6 if we multiply the third number by 300 second number to it we get 2 by adding first and third number we get double the second second number represent the information and difficulty and find the numbers using matrix method
Answers
Need homework help? Click here.
Need revision notes? Click here.
Toolbox:
If |A|≠0,then the matrix is non-singular and hence inverse exists.
A−1=1|A|(adjA)
AX=B⇒x=A−1B.
Step 1:Let the numbers be x,y and z.
From the given condition we get,
x+y+z=6
x+2z=7
3x+y+z=12
This is of the form AX=B.
(i.e)⎡⎣⎢113101121⎤⎦⎥⎡⎣⎢xyz⎤⎦⎥=⎡⎣⎢6712⎤⎦⎥
Where A=⎡⎣⎢113101121⎤⎦⎥,X=⎡⎣⎢xyz⎤⎦⎥,B=⎡⎣⎢6712⎤⎦⎥
Now by expanding along R1 let us find |A|
|A|=1(0-2)-1(1-6)+1(1-0)
=−2+5+1
=4≠0
Hence it is non-singular and A−1 exists.
Step 2:Now let us find the adj A.
A11=(−1)1+1∣∣∣0121∣∣∣=0-2=-2
A12=(−1)1+2∣∣∣1321∣∣∣=-(1-6)=5
A13=(−1)1+3∣∣∣1301∣∣∣=1-0=1
A21=(−1)2+1∣∣∣1111∣∣∣=-(1-1)=0
A22=(−1)2+2∣∣∣1311∣∣∣=1-3=-2
A23=(−1)2+3∣∣∣1311∣∣∣=-(1-3)=2
A31=(−1)3+1∣∣∣1012∣∣∣=2-0=2
A32=(−1)3+2∣∣∣1112∣∣∣=-(2-1)=-1
A33=(−1)3+3∣∣∣1110∣∣∣=0-1=-1
Step 3:Hence adj A=⎡⎣⎢A11A12A13A21A22A23A31A32A33⎤⎦⎥
=⎡⎣⎢−2510−222−1−1⎤⎦⎥
A−1=1|A|adj(A) we know |A|=4.
A−1=14⎡⎣⎢−2510−222−1−1⎤⎦⎥
Step 4:We know X=A−1B
Now substituting for X,A−1,B we get
⎡⎣⎢xyz⎤⎦⎥=14⎡⎣⎢−2510−222−1−1⎤⎦⎥
⎡⎣⎢xyz⎤⎦⎥=14⎡⎣⎢−12+0+2430−14−1216+14−12⎤⎦⎥
⎡⎣⎢xyz⎤⎦⎥=⎡⎣⎢⎢⎢1244484⎤⎦⎥⎥⎥=⎡⎣⎢312⎤⎦⎥
Therefore x=3,y=1,z=2.