Math and matrix?

Question

I need help with this math equation and i can only solve it using the matrix method.

x + y + z = 3
3x + 3y + z = 0
x + 2y + z = 0

Please help me and explain how you did it as well, Thank you.
- solve it using matrix only.

Answer 1

first, write out the augmented matrix

| 1 1 1 3 |
| 3 3 1 0 |
| 1 2 1 0 |

then row reduce. first i'll let R2 = R2 - 3R1. then i'll let R3=R3-R1
we get:

| 1 1 1 3 |
| 0 0 -2 -9 |
| 0 1 0 -3 |

we can solve it here. this says y = -3 and -2z = -9

so y = -3, z = 9/2 and x + y + z = 3
thus x = 3/2

(3/2, -3, 9/2)

Answer 2

let A= 1 1 1
3 3 1
1 2 1
X= x
y
z
B=3
0
0
the given system of equations can be put in the form of the matrix equation AX=B
!A!= 1(3-2)-3(1-2)+1(1-3)=2 is non zero
A11=1
A12=-2
A13=3
A21=-1
A22=0
!23=-1
A31=-2
A32=2
A33=0
Adjoint of A=1 -1 -2
-2 0 2
3 -1 0
A^-1=1/!A! adjA
=2 1 -1 -2
-2 0 -2
3 -1 0

=2 1*3+0+0
-2*3+0+0
3*3+0+0
so x=6
y=-12
z=18
is the same as 1,-2 and 3 taking out 3 as the common factor

Answer 3

x + y + z = 3
3x + 3y + z = 0
x + 2y + z = 0

x y z

R1 1 1 1 3 R3-R1
R2 3 3 1 0 --------->
R3 1 2 1 0

x y z
1 1 1 3 (1/3)*R2-R1
3 3 1 0 -------------->
0 1 0 -3

x y z
1 1 1 3
0 0 -2/3 -3
0 1 0 -3
therefore... x+y+z=3
(-2/3)z=-3 ==> z=9/2
y=-3

therefore...x=-y-z+3 ===> y= 3-9/2+3=(6-9+6)/2 = 3/2

thus, x=3/2
y =-3
z=9/2

Answer 4

Matrix multiplication is purely taking the dot made of row i of matrix A with column j of matrix B, to type the ij get right of entry to of the matrix (AB). think of "row dot column". The order of matrix is (rows)x(columns). If there are 3 rows in A and 3 columns in B, then there could be 3 rows and 3 columns in (AB) for the above dot product to artwork out. additionally, the dimensions of each row of A (the style of columns of A) and the top of each row of B (the style of rows of B) could be equivalent for the dot product to be defined. In all, right this is the rule governing matrix multiplication orders: (nxm)x(mxp) = (nxp)

Answer 5

my answer is at this link
http://img154.imageshack.us/img154/6685/untitled404hw6.jpg
(1.5 , -3 , 4.5 )