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Question
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)]`
A11 = 1 × 1-3×(-2)=1+6=7
A12 = - [0×1-3×1]=-(-3)=3
A13 =- 0×(-2) -1×1=-1
A21 = [1×1-(-2)×1]=-[1+2]=-3
A22=1×1-1×1=1-1=0
A23 = [1×(-2)-1×1]=-[-2-1]=-(-3)=3
A31 = 1×3-1×1=3-1=2
A32 =-[1×3-0×1]=-[3-0]=-3
A33=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
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