Question

h

Answer 1

Linear algebra uses matrices and vectors
to represent systems of equations like this

A x = b

A: coefficient matrix
x: variables, in this case x and y
b: result vector, in this case -16 and 108

Answer 2

\[\left[\begin{matrix}-8 & -8 \\ 6 & -9\end{matrix}\right] \left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}-16 \\ -108\end{matrix}\right)\]

Answer 3

basically you would divide by the A matrix to get x,y alone right?

instead of dividing by a matrix (which is not possible),
you multiply its inverse

Answer 4

you multiply both sides with the inverse, which has the same effect
as dividing both sides by the A matrix

Answer 5

\[\left[\begin{matrix}-8 & -8 \\ 6 & -9\end{matrix}\right] \left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}-16 \\ -108\end{matrix}\right)\]the first matrix is A, then the vector on LHS is x = the vector on RHS is b

Answer 6

I'm still very confused.. So I multiply -8 - 8 by 8 and 8? thats the inverse? and then what?

Answer 7

finding the inverse matrix of a square matrix is a distinct process

Answer 8

if you're not familiar with this kind of stuff (determinants, inverse matrices)
then it doesn't make much sense to write a syst. of eq in matrix form.

Answer 9

do you know how to multiply a matrix with a vector?

Answer 10

no :(

Answer 11

OK I can show the steps and I guess you don't have to remember it ;)

first I read off the coefficients of x,y in the 2 equations to get A
in our case A:
\[\left[\begin{matrix}-8 & -8 \\ 6 & -9\end{matrix}\right]\]The inverse involves matrix operations and is:
\[A^{-1}=\left[\begin{matrix}-0.075 & 0.0667 \\ -0.05 & -0.0667\end{matrix}\right]\]

Answer 12

we want x = and had

A x = b

we would want to divide by A on both sides, which is not possible.
instead of dividing by A on both sides, multiply A^-1 on both sides

A^-1*A x = A^-1 * b

which is the same as

x = A^-1*b

Answer 13

so we need to make a matrix multiplication of

A^-1*b

to get the x vector (the value for variables x and y)

Answer 14

\[A^{-1} * b=\left[\begin{matrix}-0.075 & 0.0667 \\ -0.05 & -0.0667\end{matrix}\right] * \left(\begin{matrix}-16 \\ -108\end{matrix}\right)\]which gives
\[x=\left(\begin{matrix}-6 \\ 8\end{matrix}\right).\]
The first entry corresponds to the variable in the first column - x
the 2nd entry corresponds to the variable in the second column - y

Answer 15

Thank you so much! I understand now, And you totally didnt have to give me the answer i was just wondering how to solve it for future problems thank you :)