Finding x and y in matrix

  • Thread starter Ry122
  • Start date
  • #1
565
2
Can someone tell me what i'm doing wrong. the answer for x is supposed to be -3/7 and y is supposed to be 5/7 but i have 5/7 and -23/7.

[tex]


\begin{bmatrix}
1 & -2 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}



[/tex]
new r1= r1-(r2 x -2)
[tex]


\begin{bmatrix}
7 & 0 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
1 & -2 \\
0 & 1
\end{bmatrix}



[/tex]
new r1=r1/7
[tex]


\begin{bmatrix}
1 & 0 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
1/7 & -2/7 \\
0 & 1
\end{bmatrix}



[/tex]
new r2 = r2-(r1 x 3)
[tex]


\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1/7 & -2/7 \\
-3/7 & 13/7
\end{bmatrix}



[/tex]

[tex]

b=\begin{bmatrix}
1 \\
-2
\end{bmatrix}
[/tex]
so
[tex]

\begin{bmatrix}
1/7 & -2/7 \\
-3/7 & 13/7
\end{bmatrix}
\begin{bmatrix}
1 \\
-2
\end{bmatrix}



[/tex]

[tex]
4/7 + 1/7 = 5/7
[/tex]
[tex]
-26/7 + 3/7 = -23/7
[/tex]
 
Last edited:

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
Are you sure y is supposed to be 5/7? I got -5/7. As for your error, check your first step. The matrix on the left hand side should be 1,2 not 1,-2.
 
  • #3
565
2
yes its negative 5/7 actually.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
961
It would help if you would tell us right at the beginning what the problem is!
I can guess, as Defennnder apparently did, that you are trying to solve
[tex]\begin{bmatrix}1 & -2 \\3 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ -2\end{bmatrix}[/tex]
and that you found the inverse matrix then mutliplied the inverse by b.

Can someone tell me what i'm doing wrong. the answer for x is supposed to be -3/7 and y is supposed to be 5/7 but i have 5/7 and -23/7.

[tex]


\begin{bmatrix}
1 & -2 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}



[/tex]
new r1= r1-(r2 x -2)
[tex]


\begin{bmatrix}
7 & 0 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
1 & -2 \\
0 & 1
\end{bmatrix}[/tex]
Here's your error. In the last column "r1- (r2*-2)" would be 0- (1(-2))= 2, not -2. it would have been simpler to think "r1= r1+ (r2*2)" rather than having the two negatives.




new r1=r1/7
[tex]


\begin{bmatrix}
1 & 0 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
1/7 & -2/7 \\
0 & 1
\end{bmatrix}



[/tex]
new r2 = r2-(r1 x 3)
[tex]


\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1/7 & -2/7 \\
-3/7 & 13/7
\end{bmatrix}



[/tex]

[tex]

b=\begin{bmatrix}
1 \\
-2
\end{bmatrix}
[/tex]
so
[tex]

\begin{bmatrix}
1/7 & -2/7 \\
-3/7 & 13/7
\end{bmatrix}
\begin{bmatrix}
1 \\
-2
\end{bmatrix}



[/tex]

[tex]
4/7 + 1/7 = 5/7
[/tex]
[tex]
-26/7 + 3/7 = -23/7
[/tex]
 

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