wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

The given system of equations is,

xy+2z=7

3x+4y5z=5

2xy+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( 125 )+1( 9+10 )+2( 38 ) =7+1922 =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 [ 125 ] =7

A 12 = ( 1 ) 1+2 [ 9+10 ] =19

A 13 = ( 1 ) 1+3 [ 38 ] =11

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

A 22 = ( 1 ) 2+2 [ 34 ] =1

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

A 31 = ( 1 ) 3+1 [ 58 ] =3

A 32 = ( 1 ) 3+2 [ 56 ] =( 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.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Solving Linear Equation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon