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labwork [276]
4 years ago
5

Solve the system, please show each step.

Mathematics
2 answers:
Alekssandra [29.7K]4 years ago
8 0

Answer:

The given equations are

x + y + z = -4


-x + 2 y + 3 z = 3


x - 4 y - 2 z = -15

Writing in matrix form

A=   1     1      1             X=   x       B=   -4

     -1     2     3                    y               3            ⇒A,X,B are in matrix form.

       1     -4    -2                   z              -15

i.e  Ax=B

x =A^{-1}B

but ,A^{-1}=Adj.(A)/Determinant A

Determinant of A= 1(-4+12) -1(2-3)+1(4-2)=8+1+2=11

To find Adjoint of matrix A, we will find the cofactor of A and then it's transpose.

a_{11}=-4+12=8, a_{12}=-[2-3]=1,

,a_{13}=4-2=2,\\,a_{21}=-[-2+4]=-2\\,a_{22}=-2-1=-3,\\a_{23}=-[-4-1]=5,\\a_{31}=[3-2]=1\\,a_{32}=-[3+1]=-4\\,a_{33}=2+1=3

Now taking cofactor, and getting the adjoint

Adjoint (A)= 8     -2       1

                     1      - 3       -4

                     2       5         3

Adjoint(A). B=  -53

                        47

                         -38


\frac{Adjoint (A)\times B}{Determinant A} =   -53/11

                                                                                        47/11

                                                                                         -38/11

So, solution set is , x=-53/11, y=47/11, z=-38/11



Arlecino [84]4 years ago
3 0

Answer: The solution of the system of equations is x=\frac{-53}{11}, y=\frac{47}{11} and z=\frac{-38}{11}.

Explanation:

The given equations are,

x+y+z=-4            ..... (1)

-x+2y+3z=3         ..... (2)

x-4y-2z=-15        ...... (3)

From equation we get,

x=-4-y-z

Put this value in equation (4) and (5).

-(-4-y-z)+2y+3z=3

3y+4z=-1     .... (4)

(-4-y-z)-4y-2z=-15

-5y-3z=-11  .....(5)

Use elimination method to solve the equations (4) and (5).

Multiply equation (4) by 3 and equation (5) by 4, then add both equations as shown in figure,

-11y=-47

y=\frac{47}{11}

Put this value in equation (4).

3(\frac{47}{11})+4z=-1

4z=-1-\frac{141}{11}

4z=\frac{-11-141}{11}

z=\frac{-152}{11\times 4}

z=\frac{-38}{11}

Put y=\frac{47}{11} and z=\frac{-38}{11} in equation (1).

x+\frac{47}{11}+\frac{-38}{11}=-4

x+\frac{9}{11}=-4

x=\frac{-44-9}{11}

x=\frac{-53}{11}

Therefore, the The solution of the system of equations is x=\frac{-53}{11}, y=\frac{47}{11} and z=\frac{-38}{11}.

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