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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 Numbe - Mathematics and Statistics

Sum

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.

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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

 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

Step 3 

Calculating X= A-1

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 

` [(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 

Concept: Inverse of Matrix
  Is there an error in this question or solution?
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