1. ## Clarifification on matricies

Hey guys, i just need clarification on how to start this problem.

$\begin{bmatrix}
4 & -1\\
3 & 1
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
3\\
4
\end{bmatrix}$

Basically how do i get this to look like a regular system of equations?

2. Originally Posted by Mp5xm8
Hey guys, i just need clarification on how to start this problem.

$\begin{bmatrix}
4 & -1\\
3 & 1
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
3\\
4
\end{bmatrix}$

Basically how do i get this to look like a regular system of equations?
to get "regular" system of equations from that

$4x -y = 3$

$3x+y= 4$

it's just matter of multiplying those matrices do you know how to do that ?

3. this hasn't come up in a while, i kinda forgot how to do it

4. Originally Posted by Mp5xm8
this hasn't come up in a while, i kinda forgot how to do it

to multiply two matrices AB=C

$A= \begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}$

$B = \begin{bmatrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33}
\end{bmatrix}$

$\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix} \cdot \begin{bmatrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33}
\end{bmatrix} = \begin{bmatrix}
a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31} & a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32} & a_{11}b_{13}+a_{12}b_{23}+a_{13}b_{33}\\

a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31} & a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32} & a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33}\\

a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31} & a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32} & a_{31}b_{13}+a_{32}b_{23}+a_{13}b_{33}\\
\end{bmatrix}
$

than

$C= \begin{bmatrix}
a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31} & a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32} & a_{11}b_{13}+a_{12}b_{23}+a_{13}b_{33}\\

a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31} & a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32} & a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33}\\

a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31} & a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32} & a_{31}b_{13}+a_{32}b_{23}+a_{13}b_{33}\\
\end{bmatrix}
$

two matrices are equal if their elements are equal :

$A= \begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}$

$B = \begin{bmatrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33}
\end{bmatrix}$

$a_{ij} = b_{ij}$ where i,j = 1,2,3... depending on the size of the matrix

if that is true than A=B

5. Originally Posted by Mp5xm8
Hey guys, i just need clarification on how to start this problem.

$\begin{bmatrix}
4 & -1\\
3 & 1
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
3\\
4
\end{bmatrix}$

Basically how do i get this to look like a regular system of equations?

although when given system like that, perhaps it's point do solve it using matrix not write them as....
but if you need to solve it like that ...

$A= \begin{bmatrix}
4 & -1\\
3 & 1
\end{bmatrix}$

$C= \begin{bmatrix}
3\\
4
\end{bmatrix}$

$B = \begin{bmatrix}
x\\
y
\end{bmatrix}$

to solve it you need inverse matrix A ... so when multiply A from the left you get

$AB=C$
$A^{-1}AB=A^{-1}C$
$IB =A^{-1}C$
$B =A^{-1}C$

P.S. you shouldn't write so many lines... i did so you could remember what's that all about (i hope you know to find inverse)