Given the eqns
x + 3y +z = 1
2x - y -z =6
5x + y + z =1
let ....A=
1 3 1 1
2 -1 -6 = 6
5 1 1 1
Det(A) =1(0) - 3(7) +1(7)
=-14
- By substituting the values of x by the column 1 6 1
det(A1) = 1(0) -3(7)+1(7)
=-14
X=Det(A1)/Det(A)
= -14/-14
X = 1
By substituting the values of y by the column 1 6 1
det(A2)=1(7) -1(7) +1(-28)
=-28
Y=Det(A2)/Det(A)
=-28/-14
=2
- By substituting the values of z by the column 1 6 1
Det(A3)=1(-7) -3(-28) +1(7)
=84
Z=det(A3)/det(A)
=84/-14
=-6
Therefore the values (x,y,z)=(1,2,-6)