Question

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

Answer 1

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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.