mathmatrix equations
posted by Jennifer .
I have a problem with the following homework question:
Which product represents the solution to the system: y + 7x=14
x + 4y=1
a.  (1/27) (4/27)
 (7/27) (1/27)
b.  (7/3) (4/3) 
 (1/3) (1/3) 
c.  (7/3) (4/3) 
 (1/3) (1/3) 
d.  (1/27) (4/27) 
 (4/27) (1/27) 
in every answer above, the matrices are multiplied by
1 
14
I'm totally confused about how they got those answers...I came up with this as a solution: 1/27 mult. by
4 1 
1 7  mult. by
14 
1 
i'm not sure what i'm doing wrong...
i really need help with this question. thanks in advance!

7 1 14
1 4 1
divide first row by 7 to get 1 at A11
+1 1/7 + 2
1 + 4 + 1
add them to get 0 in A21
+1 1/7 + 2
+0 27/7 + 3
multiply second row by 7/27 to get 1 in A22
+1 1/7 + 2
+0 + 1 + 7/9
that means y = 7/9. to get rid of the 1/7 in A12, add 1/7 times row 2 to row 1
1 0 19/9
0 1 7/9
so
x = 19/9
y = 7/9
check
y + 7 x = 7/9 +133/9 = 126/9 = 14 check
x+4y = 19/9 + 28/9 = 9/9 = 1 check
I think you may have a typo in the problem statement. 
no, the problem statement is as it is on my sheet.
do you know how to answer the question how they want it (a, b, c, or d)? 
trying inverse at the same time
7 1 14 1 0
1 4 1 0 1
divide first row by 7 to get 1 at A11
+1 1/7 + 2 +1/7 0
1 + 4 + 1 0 1
add them to get 0 in A21
+1 1/7 + 2 1/7 0
+0 27/7 + 3 1/7 1
multiply second row by 7/27 to get 1 in A22
+1 1/7 + 2 1/7 0
+0 + 1 + 7/9 1/27 7/27
that means y = 7/9. to get rid of the 1/7 in A12, add 1/7 times row 2 to row 1
1 0 19/9 4/27 1/27
0 1 7/9 1/27 7/27
That means the inverse is
4/27 1/27
1/27 7/27
that inverse times the column 14 1
should solve it
4/27 1/27  14
1/27 7/27  1
19/9
7/9
caramba 
so answer a does it

but I had to multiply by {14 1} not {1 14}

In other words, we agree.

thanks, this question is very strange but your replies are veryvery helpful =)
much appreciated!