The given system of equations is,
x−y+z=4
2x+y−3z=0
x+y+z=2
Write the system of equations in the form of AX=B.
[ 1 −1 1 2 1 −3 1 1 1 ][ x y z ]=[ 4 0 2 ]
Now, the determinant of A is,
| A |=1( 1+3 )+1( 2+3 )+1( 2−1 ) =4+5+1 =10
Since | A |≠0, thus A is non-singular, therefore, its inverse exists.
Since AX=B, thus, X= A -1 B.
It is known that,
A −1 = adjA | A |
The co-factors of each elements of the matrix are,
A 11 = ( −1 ) 1+1 [ 1+3 ] =4
A 12 = ( −1 ) 1+2 [ 2+3 ] =−5
A 13 = ( −1 ) 1+3 [ 2−1 ] =1
A 21 = ( −1 ) 2+1 [ −1−1 ] =−( −2 ) =2
A 22 = ( −1 ) 2+2 [ 1−1 ] =0
A 23 = ( −1 ) 2+3 [ 1+1 ] =−2
A 31 = ( −1 ) 3+1 [ 3−1 ] =2
A 32 = ( −1 ) 3+2 [ −3−2 ] =−( −5 ) =5
A 33 = ( −1 ) 3+3 [ 1+2 ] =3
So, the value of adjA is,
adjA=[ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ] =[ 4 2 2 −5 0 5 1 −2 3 ]
Since | A |=10, thus,
A −1 = 1 10 [ 4 2 2 −5 0 5 1 −2 3 ]
Now,
X= A −1 B [ x y z ]= 1 10 [ 4 2 2 −5 0 5 1 −2 3 ][ 4 0 2 ] [ x y z ]= 1 10 [ 20 −10 10 ]
Thus,
[ x y z ]=[ 2 −1 1 ]
Hence,
x=2, y=−1 and z=1.