Question

Please help!!! (algebra)
"Given the system of equations, what is the value of the system determinant?
2x + y = 8
x - y = 10"

Answer 1

Okay, so I know that to find the system determinant I need to make a matrix...but I'm confused. Don't I need at least 3 variables to do that? What exactly is a "system determinant"...how could I use it in a problem? Most importantly, how do I solve for the system determinant?

Answer 2

ok... so if you add your 2 equations y is eliminated... and you will be able to solve for x.

when you have x, substitute it into either equation... and find y

Answer 3

allright
x = 4
y = -6

Answer 4

not quite add the equations

2x + y = 8
x - y = 10
----------
3x = 18

solve for x...

Answer 5

ok,
x = 6
y = -4

Answer 6

that looks a bit better

Answer 7

is it correct?

Answer 8

I'd say yes...

Answer 9

you can check by substituting beck into each equation.

Answer 10

ok, now how do i find the system determinant?

Answer 11

you have a perfect guy here. let him help

Answer 12

@campbell_st "using determinants"

Answer 13

when you have equations\[a_1x+b_1y=c_1\qquad\text{and}\\a_2x+b_2y=c_2\]
then, the system can be represented in a matrix form as
\[\left[\begin{matrix}
a_1&b_1\\a_2&b_2
\end{matrix}\right]\left[\begin{matrix}
x\\y
\end{matrix}\right]=\left[\begin{matrix}
c_1\\c_2
\end{matrix}\right]
\]

Answer 14

now, write your system of equations in this format

Answer 15

@ScienceQs

Answer 16

I only have x and y...how do I get "b" and "a"?

Answer 17

\[(2)x+(1)y=8\\(1)x+(-1)y=10\]

Answer 18

get it?

Answer 19

ok, but why do I do that? will I multiply 2 times x like that every time I solve a problem tp find the system determinant?

Answer 20

nobody said to multiply.. just write them down

Answer 21

this matrix determinant is the easiest and ONLY method to solve large problems.

Answer 22

smaller ones you can getaway with the method shown by campbell above

Answer 23

now, lets get back to creating our matrices

Answer 24

ok, so i wrote it down. now what? oh, and should I substitute x and y for their real values in the matrix?

Answer 25

nope.. hold your horses

Answer 26

so, it would look like

Answer 27

\[\left[\begin{matrix}
2&1\\1&-1
\end{matrix}\right]\left[\begin{matrix}
x\\y
\end{matrix}\right]=\left[\begin{matrix}
8\\10
\end{matrix}\right]
\]

Answer 28

now, think of matrices like numbers
\[AX=B\]
so,
\[{1\over A}AX={1\over A}B\]

Answer 29

we cool?

Answer 30

@ScienceQs ola

Answer 31

I see how you change a linear equation into a matrix now, but I am still confused...
does A = 2 and B = 1, always?
or does A and B represent a matrix? If so, which one?

Answer 32

no. A represents the entire first block with numbers (2 by 2)
X represents the second block (with variables in it -> we need to find this)
C represents the last block (again of numbers but smaller)

Answer 33

i meant to say B, not C

Answer 34

k, so
A =[211−1]?

Answer 35

bingo

Answer 36

now look at that "A" dividin.. how can we divide with a "block" of numbers?

Answer 37

yay!
and X = [xy]?

Answer 38

yep. but the format is important...
coming back to our topic. did you understand my recent comments?

Answer 39

so, we are wanting to divide A? How do we do that?

Answer 40

yes. getting there

Answer 41

instead of dividing, we convert it to multiplication... like this:
\[{1\over A}=A^{-1}\]
the \(A^{-1}\) is called the inverse of the matrix \(A\)

Answer 42

so, what we now have is:
\[A^{-1}AX=A^{-1}B\]
but, \(A^{-1}A=I\)
I is the identity matrix.. it has similar properties like our regular number "1" but using a bulk (matrix)

Answer 43

\[[12 -11]\] = A^-1 then?

Answer 44

so, what we now have is \[X=A^{-1}B\]
yay!

Answer 45

nope.. again, keep holding your horses.. dont let em loose yet

Answer 46

understand that since A is a group of numbers, so is its inverse.

Answer 47

ok! i don't really understand much of that, but let's move on. what's next?

Answer 48

if\[A=\left[\begin{matrix}
a_1&b_1\\a_2&b_2
\end{matrix}\right]\]
then \[A^{-1}=\frac{1}{a_1b_2-a_2b_1}
\left[\begin{matrix}
b_2&-b_1\\-a_2&b_1
\end{matrix}\right]\]

Answer 49

we good? keep the horses ready

Answer 50

ok, i think I got that about A

Answer 51

now, plug in the numbers..
\[
A^{-1}=\frac{1}{(2)(-1)-(1)(1)}
\left[\begin{matrix}
-1&-1\\-1&1
\end{matrix}\right]\\
\therefore A^{-1}=\frac{1}{-3}
\left[\begin{matrix}
-1&-1\\-1&1
\end{matrix}\right]\\\therefore
X=\frac{1}{-3}
\left[\begin{matrix}
-1&-1\\-1&1
\end{matrix}\right]
\left[\begin{matrix}
8\\10
\end{matrix}\right]
\]

Answer 52

and now you let em loose

Answer 53

the three dots, those mean "therefore" right?
and the -3 and the -1s...how do I know when to put those in? Like, would it work the same if I put in a -2 where there is a -1? How do I know when to use -1...I am very confused/curious about that.

Answer 54

yes..(therefore)

each line is a step... did you see how I put the numbers for find the inverse?

Answer 55

compare the original A with the formula for inverse

Answer 56

you do not get to change the signs... just plug the numbers with their signs and the numbers will reveal themselves

Answer 57

do you think you could refer me to a good webpage about matrices and how they work? i have a lesson on it but maybe there's another way to explain it that I would get a lot better

Answer 58

all hail wikipeda

Answer 59

sincerely, only a one-on-one tutor can teach you any better than this!