Question
Mathematics Question on Applications of Determinants and Matrices
Solve the system of equations x+2y+z=4, −x+y+z=0 and x−3y+z=4.
A
x=2 , y=0, z=2
B
x=2,y=0, z=−2
C
x=−2, y=2, z=0
D
x=−2, y−0, z=2
Answer
x=2 , y=0, z=2
Explanation
Solution
Given system of equations is x+2y+z=4, −x+y+z=0 and x−3y+z=4 In matrix form, it can be written as 1 −1 121−3111x y z=4 0 4 or AX=B where, A=1 −1 121−3111,B=4 0 4andX=x y z Now, ∣A∣=1(1+3)−2(−1−1)+1(3−1)=4+4+2=10 ∵∣A∣=0, hence unique solution exists. Cofactor matrix of A=4 −5 120−2253 ∴adj(A)=4 −5 120−2253T=4 2 2−5051−23 A−1=∣A∣1adj(A)=1014 2 2−5051−23Now,X=A−1B=1014 2 2−5051−234 0 4 x y z=10116+0+4 8+0−8 8+0+12=10120 0 20=2 0 2 On comparing the corresponding elements, we get x=2, y=0 and z=2