#### Question

The sum of three numbers is 6. When second number is subtracted from thrice the sum of first and third number, we get number 10. Four times the sum of third number is subtracted from five times the sum of first and second number, the result is 3. Using above information, find these three numbers by matrix method.

#### Solution

Given that the sum of three numbers, x, y and z is 6.

From the given statement, we have,

3 (x+z)- y=10

5 (x+y)- 4z= 3

Thus, the system of equations are:

x+y+z=6

3x-y+3z=10

5x+5y-4z=3

Let us write the above equations in the matrix form as:

`[[1,1,1],[3,-1,3],[5,5,-4]][[x],[y],[z]]=[[6],[10],[3]]`

AX=B

`A=[[1,1,1],[3,-1,3],[5,5,-4]],X=[[x],[y],[z]],B=[[6],[10],[3]]`

Thus ,`[[1,1,1],[3,-1,3],[5,5,-4]]A^-1=[[1,0,0],[0,1,0],[0,0,1]]`

Thus ,`[[1,1,1],[3,-1,3],[5,5,-4]]A^-1=[[1,0,0],[0,1,0],[0,0,1]]`

Thus ,`[[1,1,1],[3,-1,3],[5,5,-4]]A^-1=[[1,0,0],[0,1,0],[0,0,1]]`

where `A=[[1,1,1],[3,-1,3],[5,5,-4]],X=[[x],[y],[z]] and B=[[6],[10],[3]]`

Thus, A^{-1} exists

We know that AA^{-1} = I

`"Thus " ,[[1,1,1],[3,-1,3],[5,5,-4]]A^-1=[[1,0,0],[0,1,0],[0,0,1]]`

Applying R_{2}→ R_{2} - 3R1 , we have

`[[1,1,1],[0,-4,0],[5,5,-4]] A^-1=[[1,0,0],[-3,1,0],[0,0,1]]`

Applying R_{3→} R_{3}- 5R_{1} , we have

`[[1,1,1],[0,-4,0],[0,0,-9]] A^-1=[[1,0,0],[-3,1,0],[-5,0,1]]`

Applying R3 →R_{3}/-9, we have

`[[1,1,1],[0,-4,0],[0,0,1]] A^-1=[[1,0,0],[-3,1,0],[-5/9,0,-1/9]]`

Applying R_{2→}R_{2}/4 , we have

`[[1,1,1],[0,1,0],[0,0,1]] A^-1=[[1/4,1,0],[3/4,-1/4,0],[5/9,0,-1/9]]`

Applying R_{1}→ R_{1}- R_{2} , we have

`[[1,0,1],[0,1,0],[0,0,1]] A^-1=[[1/4,1/4,0],[3/4,-1/4,0],[5/9,0,-1/9]]`

Applying R_{1}→R_{1}-R_{3} , we have

`[[1,0,0],[0,1,0],[0,0,1]] A^-1=[[11/36,1/4,1/9],[3/4,-1/4,0],[5/9,0,-1/9]]`

AX=B

A^{-1} AX=A^{-1} B

IX=A^{-1} B

X=A^{-1} B

`X=[[11/36,1/4,1/9],[3/4,-1/4,0],[5/9,0,-1/9]][[6],[10],[3]]`

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

Thus, the numbers are 1, 2 and 3.