Question

I solved the following system by Cramer's rule and found 3.2 and 11. But he says the right is 1.2 and 3. I do not understand. http://migre.me/5uUDj

Answer 1

Do you have to solve with cramer's rule?

Answer 2

yes!

Answer 3

\[\left(
\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 1 \\
1 & 1 & 1
\end{array}
\right)\left(
\begin{array}{c}
2 \\
3 \\
6
\end{array}
\right)\]


we have this .

Answer 4

So first find Determinant of
\[\left(
\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 1 \\
1 & 1 & 1
\end{array}
\right)\]

I got -7, is that what you got?

Answer 5

wait,let me see...

Answer 6

in the first time,yes.but next I realized that I was wrong and found 3

Answer 7

Actually it is -7

Answer 8

really?I don't think so '-' I did the determinant of X and I found 9,how can 9:-7 be 1?

Answer 9

let me recheck

Answer 10

wait,I'm gonna see if I made some mistake in this count...

Answer 11

ok

Answer 12

determinant of x= (-2+12-3)-(6+2+6) > 7 right?

Answer 13

\[1 \left(
\begin{array}{cc}
-1 & 1 \\
1 & 1
\end{array}
\right)-2\left(
\begin{array}{cc}
2 & 1 \\
1 & 1
\end{array}
\right)-1\left(
\begin{array}{cc}
2 & -1 \\
1 & 1
\end{array}
\right)\]

Answer 14

1(-2) -2(1)-1(3)
-2-2-3
-7

Answer 15

ok,but is not jus that.Are you doing the determinant of X,Y and Z?

Answer 16

Now we have D, we have to Find \[D_x,D_y,D_z\]

Answer 17

yes

Answer 18

\[D_x=\text{} \text{}\left(
\begin{array}{ccc}
2 & 2 & -1 \\
3 & -1 & 1 \\
6 & 1 & 1
\end{array}
\right)\]

Answer 19

Take determinant to find Dx

Answer 20

did you get -7

Answer 21

its a question?

Answer 22

\[D_x=-7\]
\[x=\frac{D_x}{D}\]
\[x=\frac{-7}{-7}\]
x=1

Answer 23

hmmmm...Ok,I saw what I did wrong.I was multiplying "-(-etc)" instead of doying the count in the parentheses to next subtract.tnx