Math, asked by shraddhaprajapat, 5 months ago

5x-6y+4z=15, 7x+4y-3z=19, 2x+y+6z=46 solve by elimination method​

Answers

Answered by Vinetkumar
2

1. 5x – 6y + 4z = 15

7x + 4y – 3z = 19

2x + y + 6z = 46

Hint : 1 2 3 x ,y ,z

∆ ∆ ∆ === ∆∆∆

Sol. i) Cramer’s rule :

5 64

74 3

21 6

∆= −

5(24 3) 6(42 6) 4(7 8)

135 288 4 419

= ++ ++ −

= + −=

1

15 6 4

19 4 3

46 1 6

∆= −

15(24 3) 6(114 138) 4(19 184)

405 1512 660 1917 660 1257

= ++ + + −

=+ −= −=

2

5 15 4

7 19 3

2 46 6

∆= −

5(114 138) 15(42 6) 4(322 38)

1260 720 1136 1676

= + − ++ −

= −+ =

3

5 6 15

7 4 19

2 1 46

∆ =

5(184 19) 6(322 38) 15(7 8)

825 1074 15 2529 15 2514x 3

419

1676 y 4 419

2514 z 6

419

∆ == = ∆

∆ == = ∆

∆ == = ∆

Solution is x = 3, y = 4, z = 6.

ii) Matrix inversion method :

Hint : 1 AdjA A

det A

− =

5 64

A 74 3

21 6

  −

  = −  

   

1

1

1

4 3

A 24 3 27

1 6

7 3

B (42 6) 48 1 6

7 4

C 78 1

2 1

− = = +=

− =− =− + =−

= = − =−

2

2

2

6 4

A ( 36 4) 40 1 6

5 4

B 30 8 22

2 6

5 6

C (5 12) 17 2 1

− =− =− − − =

= = −=

− =− =− + =−

3

6 4

A 18 16 2

4 3

− = =−= −

3

5 4

B ( 15 28) 43 7 3 =− =− − − = −

3

5 6

C 20 42 62

7 4

− = =+=

Adj A =

123

123

123

A A A 27 40 2

B B B 48 22 43

C C C 1 17 62

  

   = −

   − −

Det A = ∆ = 419A 48 22 43

DetA 419

1 17 62

 

  = =  

    − −

1

27 40 2 15

1 x A D 48 22 43 19

419

1 17 62 46

405 760 92

1 720 418 1978

419

15 323 2852

1257 3

1 1676 4

419

2514 6

  

   = =−

   − −

  ++ +

  = −+ +

  −− +

  

   = =   

     

∴ Solution is x = 3, y = 4, z = 6.

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