Question

# 12, x-y+z=42x+y-3z = 0"x + y + z = 2

Open in App
Solution

## 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.

Suggest Corrections
0