Question

Solve the system of the following equations

Answer 1

The equations of the system are given as,

2 x + 3 y + 10 z =4 4 x − 6 y + 5 z =1 6 x + 9 y − 20 z =2

Take 1 x =a, 1 y =b and 1 z =c.

The system of equation can be written as,

2a+3b+10c=4 4a−6b+5c=1 6a+9b−20c=2

The equation is given as CX=D,

[ 2 3 10 4 −6 5 6 9 −20 ][ a b c ]=[ 4 1 2 ]

The value of determinant can be calculated as,

| C |=| 2 3 10 4 −6 5 6 9 −20 | =2( 120−45 )−3( −80−30 )+10( 36+36 ) =1200

Here, | C | is determinant of matrix C and | C |≠0, then system of equation is consistent and has unique solution.

It can be written as,

X= C −1 D(1)

The inverse of matrix C can be written as,

C −1 = 1 | C | adj( C ), it exists if | C |≠0.

Further we have to calculate adj C,

C=[ B 11 B 21 B 31 B 12 B 22 B 32 B 13 B 23 B 33 ]

The minors can be calculated as,

R 11 =| −6 5 9 −20 |=75 R 12 =| 4 5 6 −20 |=−110 R 13 =| 4 −6 6 9 |=72 R 21 =| 3 10 9 −20 |=−150

Further calculate the minors as,

R 22 =| 2 10 6 −20 |=−100 R 23 =| 2 3 6 9 |=0 R 31 =| 3 10 −6 5 |=75 R 32 =| 2 10 4 5 |=−30

Further calculate the minors as,

R 33 =| 2 3 4 −6 |=−24

Calculate the cofactors as,

B 11 = ( −1 ) 1+1 R 11 = ( −1 ) 2 ×( 75 )=75 B 12 = ( −1 ) 1+2 R 12 = ( −1 ) 3 ×( −110 )=110 B 13 = ( −1 ) 1+3 R 13 = ( −1 ) 4 ×( 72 )=72 B 21 = ( −1 ) 2+1 R 21 = ( −1 ) 3 ×( −150 )=150

Further calculate the factors as,

B 22 = ( −1 ) 2+2 R 22 = ( −1 ) 4 ×( −100 )=−100 B 23 = ( −1 ) 2+3 R 23 = ( −1 ) 5 ×( 0 )=0 B 31 = ( −1 ) 3+1 R 31 = ( −1 ) 4 ×( 75 )=75 B 32 = ( −1 ) 3+2 R 32 = ( −1 ) 5 ×( −30 )=30

Further calculate the cofactors as,

B 33 = ( −1 ) 3+3 R 33 = ( −1 ) 6 ×( −24 )=−24

By substituting all values in the matrix we get,

adj( C )=[ 75 150 75 110 −100 30 72 0 −24 ]

By substituting the value of adj( A ) in the formula of inverse of matrix A we get,

C −1 = 1 1200 [ 75 150 75 110 −110 30 72 0 −24 ]

We have to put the value of C −1 and D in the equation (1).

[ a b c ]= 1 1200 [ 75 150 75 110 −110 30 72 0 −24 ][ 4 1 2 ] [ a b c ]= 1 1200 [ 600 400 240 ] [ a b c ]=[ 1 2 1 3 1 5 ]

Thus, the values of x=2, y=3 and z=5.