Question

14. x-y+2z = 73x + 4y-5z =-52x-y+3z = 12

Answer 1

The given system of equations is,

x−y+2z=7

3x+4y−5z=−5

2x−y+3z=12

Write the system of equations in the form of AX=B.

[ 1 −1 2 3 4 −5 2 −1 3 ][ x y z ]=[ 7 −5 12 ]

Now, the determinant of A is,

| A |=1( 12−5 )+1( 9+10 )+2( −3−8 ) =7+19−22 =4

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 [ 12−5 ] =7

A 12 = ( −1 ) 1+2 [ 9+10 ] =−19

A 13 = ( −1 ) 1+3 [ −3−8 ] =−11

A 21 = ( −1 ) 2+1 [ −3+2 ] =−( −1 ) =1

A 22 = ( −1 ) 2+2 [ 3−4 ] =−1

A 23 = ( −1 ) 2+3 [ −1+2 ] =−1

A 31 = ( −1 ) 3+1 [ 5−8 ] =−3

A 32 = ( −1 ) 3+2 [ −5−6 ] =−( −11 ) =11

A 33 = ( −1 ) 3+3 [ 4+3 ] =7

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 ] =[ 7 1 −3 −19 −1 11 −11 −1 7 ]

Since | A |=4, thus,

A −1 = 1 4 [ 7 1 −3 −19 −1 11 −11 −1 7 ]

Now,

X= A −1 B [ x y z ]= 1 4 [ 7 1 −3 −19 −1 11 −11 −1 7 ][ 7 −5 12 ] [ x y z ]= 1 4 [ 8 4 12 ]

Thus,

[ x y z ]=[ 2 1 3 ]

Hence,

x=2, y=1 and z=3.