What is the correct way to solve a matrix equation with variables x and y?

[Ry122]

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.

$$\begin{bmatrix}<br /> 1 & -2 \\ <br /> 3 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 & 0 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> <br /> <br />$$
new r1= r1-(r2 x -2)
$$\begin{bmatrix}<br /> 7 & 0 \\ <br /> 3 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 & -2 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> <br /> <br />$$
new r1=r1/7
$$\begin{bmatrix}<br /> 1 & 0 \\ <br /> 3 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1/7 & -2/7 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> <br /> <br />$$
new r2 = r2-(r1 x 3)
$$\begin{bmatrix}<br /> 1 & 0 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1/7 & -2/7 \\ <br /> -3/7 & 13/7<br /> \end{bmatrix} <br /> <br /> <br />$$

$$<br /> b=\begin{bmatrix}<br /> 1 \\ <br /> -2<br /> \end{bmatrix}$$
so
$$<br /> \begin{bmatrix}<br /> 1/7 & -2/7 \\ <br /> -3/7 & 13/7<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 \\ <br /> -2<br /> \end{bmatrix} <br /> <br /> <br />$$

$$4/7 + 1/7 = 5/7$$
$$-26/7 + 3/7 = -23/7$$

 

[Defennder]

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.

 

[Ry122]

yes its negative 5/7 actually.

 

[HallsofIvy]

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
$$\begin{bmatrix}1 & -2 \\3 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ -2\end{bmatrix}$$
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.

$$<br /> <br /> \begin{bmatrix}<br /> 1 & -2 \\ <br /> 3 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 & 0 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> <br /> <br />$$
new r1= r1-(r2 x -2)
$$<br /> <br /> \begin{bmatrix}<br /> 7 & 0 \\ <br /> 3 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 & -2 \\ <br /> 0 & 1<br /> \end{bmatrix}$$

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
$$<br /> <br /> \begin{bmatrix}<br /> 1 & 0 \\ <br /> 3 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1/7 & -2/7 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> <br /> <br />$$
new r2 = r2-(r1 x 3)
$$<br /> <br /> \begin{bmatrix}<br /> 1 & 0 \\ <br /> 0 & 1<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1/7 & -2/7 \\ <br /> -3/7 & 13/7<br /> \end{bmatrix} <br /> <br /> <br />$$

$$<br /> b=\begin{bmatrix}<br /> 1 \\ <br /> -2<br /> \end{bmatrix}$$
so
$$<br /> \begin{bmatrix}<br /> 1/7 & -2/7 \\ <br /> -3/7 & 13/7<br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> 1 \\ <br /> -2<br /> \end{bmatrix} <br /> <br /> <br />$$

$$4/7 + 1/7 = 5/7$$
$$-26/7 + 3/7 = -23/7$$