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 =
B
but ,
=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.
=-4+12=8,
![a_(12)=-[2-3]=1,](https://img.qammunity.org/2019/formulas/mathematics/high-school/ns7jv20w9qrzj04qygc429muvsae1jo5ww.png)
![,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](https://img.qammunity.org/2019/formulas/mathematics/high-school/oh4u4tlj66mdcb9t9acacur50l4otrri2a.png)
Now taking cofactor, and getting the adjoint
Adjoint (A)= 8 -2 1
1 - 3 -4
2 5 3
Adjoint(A). B= -53
47
-38
= -53/11
47/11
-38/11
So, solution set is , x=-53/11, y=47/11, z=-38/11