Question

Quick help: Explain how to solve the following system of equations. What is the solution to the system?

2x+2y+z=-5
3x+4y+2z=0
x+3y+2z=1

Answer 1

I would use matrix

Answer 2

I used matrix a wile ago I cannot remember how use it :/

Answer 3

check thi out and follow that pattern https://www.youtube.com/watch?v=TtxVGMWXMSE

Answer 4

\(\large\color{black}{ \left[\begin{matrix}? & ? & ? & ?\\ ? & ? & ? & ? \\ ? & ? & ? &?\end{matrix}\right] }\) lets start filling in numbers

Answer 5

this*

Answer 6

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 3 & 4 & 2 &~~~~~~ 0 \\ 1 & 3 & 2 &~~~~~~1\end{matrix}\right] }\)

Answer 7

like this

Answer 8

ok

Answer 9

you need to it to:

\(\large\color{black}{ \left[\begin{matrix}1 & 0 & 0 &~~~~~~ x\\ 0 & 1 & 0 &~~~~~~ y \\ 0 & 0 & 1 &~~~~~~z\end{matrix}\right] }\)

Answer 10

woah how'd you get that

Answer 11

so lets try to zero out the 1st row 2nd column

Answer 12

I would multiply the 3rd row, times -3, and add it to the second row.

Answer 13

wait. How did you get the above matrix

Answer 14

well, x y and z in the last column are just going to be the numbers that they (x y z) are equal to

Answer 15

yeah, that part makes sense

Answer 16

How did you get with the 1s and 0s

Answer 17

that is what the set up should be at the end

Answer 18

ok gotcha

Answer 19

so how would I get there?

Answer 20

step by step, lol

Answer 21

ok...

Answer 22

Now,

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 3 & 4 & 2 &~~~~~~ 0 \\ 1 & 3 & 2 &~~~~~~1\end{matrix}\right] }\)


\(\large\color{black}{ \left[\begin{matrix} 1 & 3 & 2 &~~~~~~1\end{matrix}\right] }\) times -3, \(\large\color{black}{ \left[\begin{matrix} -3 & -9 & -6 &~~~~~~-3\end{matrix}\right] }\)

adding to the 2nd row,
\(\large\color{black}{ ~~~\left[\begin{matrix} -3 & -9 & -6 &~~~~~~-3\end{matrix}\right] }\)
\(\large\color{black}{^+ \left[\begin{matrix}3 &~~~~~ 4 & ~~~~2 &~~~~~~~~~~~ 0\end{matrix}\right] }\)

Answer 23

what will your second row become?

Answer 24

Is every row times -3? always?

Answer 25

I multiplied only the last row times -3, so that I can add (I am allowed to do this)

Answer 26

what will the indicated sum (for second row) be?

Answer 27

So am I multiplying row 2 and 1 by -3 as well?

Answer 28

or is that -2 then -1?

Answer 29

right now, you have just multiplied the last (3rd) row by -3, so that when you add it to the second row, the first number in second row becomes a zero.

Answer 30

ok... so
[0 -5 -4 -3]

Answer 31

yes

Answer 32

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & -5 & -4 &~~~~~~ -3 \\ 1 & 3 & 2 &~~~~~~1\end{matrix}\right] }\)

Answer 33

now, we will multiply the last row times -2, and leave it (for now) like this.

Answer 34

then i do the sam eto the first
[2 -3 -3 -8]

Answer 35

what did you do?

Answer 36

I did the second row to the first is that wrong?

Answer 37

wrong direction, it is

Answer 38

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & -5 & -4 &~~~~~~ -3 \\ 1 & 3 & 2 &~~~~~~1\end{matrix}\right] }\)

multiply the second row times -2, what do you get?

Answer 39

I mean 3rd

Answer 40

3rd row times -2

Answer 41

[-2 -6 -4 -2]

Answer 42

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & -5 & -4 &~~~~~~ -3 \\ -2 & -6 & -4 &~~~~~~-2\end{matrix}\right] }\)
yes.

Answer 43

now add, the 1st row to the 3rd row. What will your third row be?

Answer 44

[0 -4 -3 -7]

Answer 45

yes

Answer 46

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & -5 & -4 &~~~~~~ -3 \\ 0 & -4 & -3 &~~~~~~-7\end{matrix}\right] }\)

Answer 47

multiply the 2nd row times -1.

Answer 48

Yeah i think i've got it. Could you help me with more questions?

Answer 49

I'll open a new thingy though

Answer 50

lets finish this, please. I don't want you to leave with an incorrect answer.

Answer 51

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & -5 & -4 &~~~~~~ -3 \\ 0 & -4 & -3 &~~~~~~-7\end{matrix}\right] }\)

2nd row times -1.

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & 5 & 4 &~~~~~~ 3 \\ 0 & -4 & -3 &~~~~~~-7\end{matrix}\right] }\)

Answer 52

add the 3rd row to the 2nd.

Answer 53

\(\large\color{black}{ \left[\begin{matrix}2 & 2 & 1 &~~~~~~ -5\\ 0 & 1 & 1 &~~~~~~ -4 \\ 0 & -4 & -3 &~~~~~~-7\end{matrix}\right] }\)

Answer 54

i cant post something for some reason

Answer 55

this will be quick, trust me a couple of minutes.

Answer 56

ok, it isn't letting me post numbers

Answer 57

refresh...

Answer 58

I did lets see if this will work.

Answer 59

aha ok so what do I need to be dong right now

Answer 60

multiply the second row times -2, and add it to the 1st row.
[ 0 - 2 -2 8]
+
[ 2 2 1 -5]

Answer 61

what will your first row be?

Answer 62

uh
what is the stuff abover saying to multiply the second row

Answer 63

I multiplied the second row times -2, and got [0 -2 -2 8]
and then I am adding it to the 1st row, to make the first row better.

Answer 64

[ 0 - 2 -2 8]
+
[ 2 2 1 -5]
-------------------

???

Answer 65

oh ok

so it will be 2 0 -1 3

Answer 66

yes

Answer 67

so we get: \(\large\color{black}{ \left[\begin{matrix}2 & 0 & -1 &~~~~~~ 3\\ 0 & 1 & 1 &~~~~~~ -4 \\ 0 & -4 & -3 &~~~~~~-7\end{matrix}\right] }\)

Answer 68

ok

Answer 69

see how I am broaching the needed setup?

Answer 70

yea?

Answer 71

multiply the 2nd row times 4.
add it to the last row.

what will your last row be?

Answer 72

4 or -4

Answer 73

this time, only times positive 4

Answer 74

0 4 4 -16

Answer 75

yes, add this to the last row.
what will your last row be?

Answer 76

0 0 1 -23

Answer 77

yes.

Answer 78

\(\large\color{black}{ \left[\begin{matrix}2 & 0 & -1 &~~~~~~ 3\\ 0 & 1 & 1 &~~~~~~ -4 \\ 0 & 0 & 1 &~~~~~~-23\end{matrix}\right] }\)

Answer 79

well, yes we have that z=-23. but we need the x and y too

Answer 80

yeah

Answer 81

add 3rd row to the 1st row.

Answer 82

your 1st row becomes?

Answer 83

2 0 0 -20

Answer 84

yes.

Answer 85

k

Answer 86

\(\large\color{black}{ \left[\begin{matrix}2 & 0 & 0 &~~~~~~ -20\\ 0 & 1 & 1 &~~~~~~ -4 \\ 0 & 0 & 1 &~~~~~~-23\end{matrix}\right] }\)

Answer 87

divide first row by 2, you get?

Answer 88

1 0 0 -10

Answer 89

\(\large\color{black}{ \left[\begin{matrix}1 & 0 & 0 &~~~~~~ -10\\ 0 & 1 & 1 &~~~~~~ -4 \\ 0 & 0 & 1 &~~~~~~-23\end{matrix}\right] }\)

Answer 90

multiply last row times -1, and add it to the 2nd row.

you second row becomes?

Answer 91

0 0 -1 23

Answer 92

yes you multiplied times -1 correctly.

Answer 93

now add this to the 2nd row

Answer 94

0 1 0 19

Answer 95

yes, so we get:

\(\large\color{black}{ \left[\begin{matrix}1 & 0 & 0 &~~~~~~ -10\\ 0 & 1 & 0 &~~~~~~ 19\\ 0 & 0 & 1 &~~~~~~-23\end{matrix}\right] }\)

Answer 96

clarifications?

Answer 97

Cool thank you so much!!

Nope! I'll start a new question in just a minute!

Answer 98

I have to go light my Hannukah candels, see you:)