Question

Solve the system, please show each step.

Answer 1

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

Answer 2

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 $z=\frac{-38}{11}$.