Using matrices to solve the following simultaneous equation

Hi undertheshade,

Welcome to Web2.0calc  forum     

y=x+1

1x-1y=-1

y=7-2x

2x+1y=7

\(\begin{bmatrix} 1 &-1 \\ 2&1 \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix}= \begin{bmatrix} -1 \\ 7 \end{bmatrix}\\ \)

There are different ways to do this - I shall use reduced row escelon form.

\(\begin{bmatrix} 1 &-1\:\:|\:-1\\ 2&1\:|\:7 \end{bmatrix} \\~\\\mbox{Top row (row 1) }\times 2\\~\\ \begin{bmatrix} 2 &-2\:\:|\:-2\\ 2&1\:|\:7 \end{bmatrix}\\~\\ \mbox{row 2 - row 1}\\~\\ \begin{bmatrix} 2 &-2\:\:|\:-2\\ 0&3\:|\:9 \end{bmatrix}\\~\\ \mbox{row 1 times 3}\\ \mbox{row 2 times 2} \\~\\ \begin{bmatrix} 6 &-6\:\:|\:-6\\ 0&6\:|\:18 \end{bmatrix}\\~\\ \mbox{row 1 plus row 2} \\~\\ \begin{bmatrix} 6 &0\:\:|\:12\\ 0&6\:|\:18 \end{bmatrix}\\~\\ \mbox{row 1 divided by 6}\\\mbox{row 2 divided by 6} \\~\\ \begin{bmatrix} 1 &0\:\:|\:2\\ 0&1\:|\:3 \end{bmatrix}\\~\\ x=2 \qquad y=3\)

first re-arrange to get

x - y  = -1  and

2x + y  = 7

Now form your matrices

1   -1         x      =     -1 

2     1        y              7   

Now pre-multiply both sides by the inverse of  matrix    1 -1  

                                                                                       2   1           and just read off  for x  and y .

(Assumes that you know how to find the inverse of a 2 x 2 matrix.)  

Thanks guest, i'm just practicing using Latex  :)

ok now I will try it using the inverse method.

\( \begin{bmatrix} 1 &-1 \\ 2&1 \end{bmatrix} \begin{bmatrix}x  \\y \end{bmatrix}= \begin{bmatrix} -1 \\ 7 \end{bmatrix}\\ \frac{1}{1*1--1*2} \begin{bmatrix} 1 && 1 \\ -2 &&1 \end{bmatrix} \begin{bmatrix} 1 &-1 \\ 2&1 \end{bmatrix} \begin{bmatrix}x  \\y \end{bmatrix}= \frac{1}{1*1--1*2} \begin{bmatrix} 1 && 1 \\ -2 &&1 \end{bmatrix} \begin{bmatrix} -1 \\ 7 \end{bmatrix}\\~\\ \frac{1}{3} \begin{bmatrix} 1+2 && -1+1 \\ -2+2 &&2+1 \end{bmatrix} \begin{bmatrix}x  \\y \end{bmatrix}= \frac{1}{3} \begin{bmatrix} -1+7 \\ 2+7 \end{bmatrix}\\~\\ \frac{1}{3} \begin{bmatrix} 3 &&0 \\ 0 &&3 \end{bmatrix} \begin{bmatrix}x  \\y \end{bmatrix}= \frac{1}{3} \begin{bmatrix} 6 \\ 9 \end{bmatrix}\\~\\ \begin{bmatrix} 1 &&0 \\ 0 &&1 \end{bmatrix} \begin{bmatrix}x  \\y \end{bmatrix}= \begin{bmatrix} 2 \\ 3 \end{bmatrix}\\~\\ \begin{bmatrix}x  \\y \end{bmatrix}= \begin{bmatrix} 2 \\ 3 \end{bmatrix}\\~\\ x=2 \qquad and \qquad y=3 \)

If you are confused try watching this video and the one that follows it.

https://www.khanacademy.org/math/precalculus/precalc-matrices/determinant-of-2x2-matrix/v/finding-the-determinant-of-a-2x2-matrix