Question

http://prntscr.com/nxyt02

Answer 1

@Narad

Answer 2

If you have a matrix A of dimensions m x n and another matrix B of dimensions n x p, the

product AB is of dimensions m x p

Answer 3

so C?

Answer 4

Yes, this is the correct option

Answer 5

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Answer 6

This is the same as before
A has dim m x n
B has dim n x p
AB has dim m x p

Answer 7

B?

Answer 8

No, try again

Answer 9

C

Answer 10

Yes, this is the correct option

Answer 11

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Answer 12

Question nº32

If you have 2 simulaneous equations
\[ax+by=c..................(1)\]and
\[dx+ey=f..........................(2)\]
Then, the substitution method is
to get x from the first equation and substitute in the second equation
\[x=\frac{ (c-by) }{ a }\]
\[d*\frac{ (c-by) }{ a }+ey=f\]
The last equation has only one unknown y, which can be solved
Then, you can get x

Answer 13

Im a bit confused loll

Answer 14

This is the general case
The first equation is
\[x-2y=5\]
Can you get x in terms of y?

Answer 15

x=(5-2y)/x ?

Answer 16

no, this is easy

Answer 17

Add 2y on both sides of the equation

Answer 18

y=7y?

Answer 19

No,
\[x-2y+2y=5+2y\]
What can you conclude?

Answer 20

7y since 5+2y

Answer 21

you can't add 5 to 2y

Answer 22

7y is not (5+2y)
7y is (5y+2y)

Answer 23

so just 7?

Answer 24

No,
x=5+2y
That's all

Answer 25

thats the answer?

oh i thought we had to add em together

Answer 26

Not yet the answer, now substitute this value in equation (2)

Answer 27

How?

Answer 28

The second equation is
2x-4y=1
replace x by the value you obtained from equation (1)

Answer 29

this is way too confusing

Answer 30

x=5+2y

Then,
2*(5+2y))-4y=1
Develop this equation!

Answer 31

thats 10+4y

Answer 32

10+4y-4y=1

Answer 33

now what

Answer 34

What simplification can be made?

Answer 35

add 4y to both sides?

Answer 36

you have +4y and -4y, then

Answer 37

You obtain

10=1
Is this possible?

Answer 38

no

Answer 39

So the system is inconsistent, no solutions

Answer 40

x-2y=5

x=2y+5

2x-4y=1

2(2y+5)-4y=1

4y+10-4y=1

10=1

This is not possible, the system is inconsistent

Answer 41

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Answer 42

Express y in terms of x in both equations

Answer 43

x=3x+5

Answer 44

y=3x+5

and!!!!

Answer 45

and? well there isnt numbers for the second equation except the 3

Answer 46

y=3-x

Answer 47

okay now what?

Answer 48

y=3x+5 and y=3-x

What can you observe?

Answer 49

both y=3

Answer 50

no,

3x+5=3-x

Answer 51

Solve for x?

Answer 52

x=-1/2

Answer 53

yes, then y=????

Answer 54

-1/2

Answer 55

y=3-x

x=-1/2

y=???

Answer 56

Y=3-(-1/2+

Answer 57

yes, continue

Answer 58

y=7/2

Answer 59

Yes you got it
The solutions are (-1/2,7/2)

Answer 60

okay can you just put the steps in order please

Answer 61

y-3x=5, y=5+3x

y+x=3, y=3-x

5+3x=3-x

4x=3-5=-2

x=-2/4=-1/2

y=3-x=3-(-1/2)=7/2

Answer 62

okay

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Answer 63

could we go just a little faster because this is timed

Answer 64

Express 2y in terms of x

First equation

3x+2y=7

Answer 65

you mean -1 ?

Answer 66

x+2y=7

Then

2y=7-3x

Answer 67

Express in the same way 2y for the second equation?

Answer 68

5x-2y=1

Therefore,

2y= ?????

Answer 69

We can do this differently, just add the 2 equations

3x+2y=7

5x-2y=1

Answer 70

I mean we have to do it using the elimination method

Answer 71

Yes, that's the question??

Answer 72

Yes

Answer 73

so what do I do

Answer 74

could you just tell me it please because I dont have a lot of time left and I need to finish it before the timer ends

Answer 75

3x+5x+2y-2y=7+1

8x=8

x=1

2y=7-3x=7-3*1=4

y=2

The solution is (1,2)

Answer 76

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Answer 77

The equations are

3x-2y=1

and

4y=6x-2

Divide the last equation by 2

Answer 78

2y?

Answer 79

2y=???????

Answer 80

6x because we cancel out the 2

Answer 81

You must divide all the terms of the equation by 2

Answer 82

(4y)/2=(6x)/2-2/2

Answer 83

2y=3x?

Answer 84

2y=3x-1

as

2/2=1

Answer 85

Ohh okay now what

Answer 86

Compare this equation to the first one
3x-2y=1

Answer 87

its basically flipped

Answer 88

Are they identical !!!!

Answer 89

There in a different order

Answer 90

They are identical but the order is different, so, what can we say?

Answer 91

1 solution

Answer 92

no, the system is consistent and dependent , there are infinetely many solutions

Answer 93

ohh okay

can you put the steps together pls

Answer 94

3x-2y=1, 2y=3x-1

4y=6x-2, 2y=3x-1

The equations are the same, the system is consistent and dependent and there are an infinity of solutions

Answer 95

okay

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Answer 96

If you have the equations
\[ax+by=c\] and
\[dx+ey=f\]
The augmented matrix is
\[A=\left[\begin{matrix}a & b&c \\ d & e&f\end{matrix}\right]\]
Can you write the augmented matrix for the system of equations?

Answer 97

can you please plug it in for me its timed and the timer is almost finished

Answer 98

\[A=\left[\begin{matrix}1 & 1&5\\ 2 & -1&1\end{matrix}\right]\]

Answer 99

Perform row operations on this matrix

R2 <- R2-R1

\[A=\left[\begin{matrix}1 & 1&5 \\ 0 & -3&-9\end{matrix}\right]\]

R2<- R2/-3

\[=\left[\begin{matrix}1 & 1&5 \\ 0 & 1&3\end{matrix}\right]\]

R1<- R1-R2

\[=\left[\begin{matrix}1 & 0&5 \\ 0 &1&3\end{matrix}\right]\]

Answer 100

Correction
\[=\left[\begin{matrix}1 & 0&2 \\ 0 & 1&3\end{matrix}\right]\]

Answer 101

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Answer 102

The solutions are (2,3)

Answer 103

You have to proceed the same way as the last exercise

Answer 104

The augmented matrix is ???

Answer 105

how do we know what a b c d and so on...

Answer 106

The coefficients of x and y and the constant term

Answer 107

\[A=\left[\begin{matrix}3 & 5&-4 \\ 1 & 2&-2\end{matrix}\right]\]

Answer 108

okay and then

Answer 109

Perform the row operations
R2<- 3R2- R1

Answer 110

\[=\left[\begin{matrix}3 & 5&-4 \\ 0 & 1&-2\end{matrix}\right]\]
R1<- R1-5R2

Answer 111

\[= \left[\begin{matrix}3 & 0&6 \\ 0 & 1&-2\end{matrix}\right]\]

Answer 112

R1<-( R1)/2

\[=\left[\begin{matrix}1 & 0&2 \\ 0& 1&-2\end{matrix}\right]\]

The solutiion is (2,-2)

Answer 113

okay
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Answer 114

@Narad

Answer 115

Do you know how to find the equations of the lines

Dotted line and continuous line

Answer 116

not sure

Answer 117

The dotted line passes through the origin
The equation is y=mx
where m is the slope

Answer 118

im thinking a or c?

Answer 119

The dotted line on the graph, what's the slope

Answer 120

im not sure

Answer 121

Find the equation of the dotted line
y=mx
and
The equation of the continuous line is
\[\frac{ x }{ a} + \frac{ y }{ b } =1\]

Answer 122

B?

Answer 123

I only have 10 mins left and I have 4 questions please @Narad what is it

Answer 124

\[y \ > 2x\]
and
\[x+2y \le4\]

Answer 125

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Answer 126

The dotted line is y=-3

The continuous line is y=4x+1ç

The shaded part is

\[y < -3\]
\[y \ge 4x+1\]

Answer 127

okay

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Answer 128

The augmented matrix associated with the equations is
\[\left[\begin{matrix}1 & -3&1&-9\\ 1 & 1&-3&-5\\3&-1&1&11\end{matrix}\right]\]
Perform the Jordan Gauss elimination

The solution is (4,6,5)

Answer 129

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Answer 130

Multiply all the coordinates by 3

The center of dilation is the origin

The first matrix is
\[A=\left[\begin{matrix}-2 & 1&4 \\ 0 & 5&-8\end{matrix}\right]\]

Therefore,

\[3A=\left[\begin{matrix}-6 & 3&12 \\ 0 & 15&-24\end{matrix}\right]\]

Answer 131

okay can I make a new post?

Answer 132

ok