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Question

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

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Solution

The given system of equations is,

xy+z=4

2x+y3z=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( 21 ) =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 [ 21 ] =1

A 21 = ( 1 ) 2+1 [ 11 ] =( 2 ) =2

A 22 = ( 1 ) 2+2 [ 11 ] =0

A 23 = ( 1 ) 2+3 [ 1+1 ] =2

A 31 = ( 1 ) 3+1 [ 31 ] =2

A 32 = ( 1 ) 3+2 [ 32 ] =( 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.


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