The sum of three numbers is 6. If we multiply the third number by 3 and add it to the second number we get 11. By adding first and third numbers we get a number, which is double than the second number. Use this information and find a system of linear equations. Find these three numbers using matrices.

#### Solution

Let the first , second & third number be x, y, z respectively

Given,

∴ x + y + z = 6

y + 3z = 11

x + z = 2y or x - 2y + z = 0

Step 1

write equation as AX = B

A = `[(1,1,1),(0,1,3),(1,-2,1)] , "X" = [(x),(y),(z)] , B = [(6),(11),(0)]`

Hence A = ` [(1,1,1),(0,1,3),(1,-2,1)] ,X = [(x),(y),(z)] &B= [(6),(11),(0)]`

Step 2

calculate |A|

|A| = `[(1,1,1),(0,1,3),(1,-2,1)]`

= 1(1 + 6) - 0 (1 + 2) + 1(3 + 1)

= 7 + 2

= 9

So, |A| ≠0

∴ The system of equation is consistent & has a unique solutions

Now , AX = B

X = A^{-1} B

Hence A =` [(1,1,1),(0,1,3),(1,-2,1)] ,X = [(x),(y),(z)] &B= [(6),(11),(0)]`

= 1 (1+6)-0(1+2)+1(3-1)

=7+2

=9≠0

Since determinant is not equal to O , A^{-1 }exists

Now find adj (A)

now AX = B

X = A^{-1} B

Step 3

Calculating X= A^{-1} B

Calculating A^{-1}

Now A^{-1 }= `1/|A| `adj (A)

adj A = `[(A_11,A_12,A_ 13) ,(A_21,A_22,A_23),(A_31 ,A_32,A_33)]^'=[(A_11,A_21,A_31) , (A_12,A_22,A_32), (A_13,A_23,A_33)]`

A = `[(1,-1,2) ,(3,4,-5), (2,-1,3)]`

A_{11} = 1 × 1-3×(-2)=1+6=7

A_{12} = - [0×1-3×1]=-(-3)=3

A_{13} =- 0×(-2) -1×1=-1

A_{21} = [1×1-(-2)×1]=-[1+2]=-3

A_{22}=1×1-1×1=1-1=0

A_{23} = [1×(-2)-1×1]=-[-2-1]=-(-3)=3

A_{31} = 1×3-1×1=3-1=2

A_{32} =-[1×3-0×1]=-[3-0]=-3

A_{33}=1×1-1×0=1-0=1

Hence , adj (A) = `[(7,-3,2),(3,0,-3),(-1,3,1)]`

Now ,

A^{-1} = `1/|A|` adj (A)

A^{-1 }=`1/9[(7,-3,2),(3,0,-3),(-1,3,1)]`

Solution of given system of equations is

X = A^{-1 } B

` [(x),(y),(z)] = 1/9 [(7,-3,2),(3,0,-3),(-1,3,1)] [(6),(11),(0)]`

` [(x),(y),(z)] = 1/9 [(42,-33,+0),(18,+0,+0),(-6,+33,+0)]`

` [(x),(y),(z)] = 1/9 [(9),(18),(27)]`

` [(x),(y),(z)] = [(1),(2),(3)]`

**∴ x = 1, y = 2, z = 3 **