Question

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Answer 1

write the matrix

Answer 2

familiar wid row operations ?

Answer 3

no?

Answer 4

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

Answer 5

they're easy,
we will be using below row operations mostly :-
1) you can add/subtract a row from another row

Answer 6

idea is to get a upper triangular matrix using row operations

Answer 7

upper triangular matrix looks like below :-

1 x x
0 1 x
0 0 1

Answer 8

all elements below the diagonal will be 0

Answer 9

you wid me still ? :)

Answer 10

confused slightly on how u got the 1 x x upper triang

Answer 11

thats just an example of upper triangular matrix
x means any number

Answer 12

dotn wry about that example if that confuses u

Answer 13

ok

Answer 14

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

Answer 15

Focus on first column.
lets make it :-
1
0
0

Answer 16

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

subtract Row 1 from Row 2
R2 - R1

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
0 & 2 & 0 & -4 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

Answer 17

check if i did the subtraction correctly

Answer 18

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

subtract Row 1 from Row 2
R2 - R1

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
0 & 2 & 0 & -4 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

subtract Row 1 from Row 3
R3 - R1

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
0 & 2 & 0 & -4 \\
0 & 2 & -2 & 4 \\ \end{array} \right]\]

Answer 19

so when u substract you replace the row used first, not second?

Answer 20

You got it !

Answer 21

R2 - R1

messes wid Row 2

Answer 22

we're done wid first column !
look at first column, we have all 0's below the diagonal :-
1
0
0

Answer 23

lets do the same for second column

Answer 24

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

subtract Row 1 from Row 2
R2 - R1

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
0 & 2 & 0 & -4 \\
1 & 1 & -1 & 7 \\ \end{array} \right]\]

subtract Row 1 from Row 3
R3 - R1

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
0 & 2 & 0 & -4 \\
0 & 2 & -2 & 4 \\ \end{array} \right]\]


subtract Row 2 from Row 3
R3 - R2

\[ \left[ \begin{array}{cccc}
1 & -1 & 1 & 3\\
0 & 2 & 0 & -4 \\
0 & 0 & -2 & 8 \\ \end{array} \right]\]

Answer 25

fine so far ?

Answer 26

yes

Answer 27

We're done !

Answer 28

totally confused now, after we got the 1st column to 1
0
0
then what happened?

Answer 29

after that we fixed the second column

Answer 30

ive just subtracted Row 2 from Row 3

Answer 31

see that, after that step, all elements below diagonal are 0

Answer 32

so we have an upper triangular matrix. so we stop. and do back substitute

Answer 33

to find teh values of x, y,z

Answer 34

ok holdon, at the point of the 1st column, i have 1 -1 1 3, 0 2 0 -4, - 2 -2 4 as the 3 rows, then you did r3 - r2.. so r3 would be - 0 0 2 0?

Answer 35

check ur subtraction again. r3-r2 wud be, 0 0 -2 8

Answer 36

btw, ur third is incorrect... check ur work again.
compare it wid my work above

Answer 37

ok so after r3-r2, we do?

Answer 38

we're done wid elimination.
we simply find the values of x, y, z now

Answer 39

First look at last row, wat do we have ?

Answer 40

i thought point was to get 0's in diagonal, no?

Answer 41

no need. just the 0's below the diagonal will do

Answer 42

look at last row,

0 0 -2 8
-2z = 8 -------------(1)

Answer 43

look at second row,

0 2 0 -4
2y = -4 -------------(2)

Answer 44

look at first row,

1 -1 1 3
x -y + z = 3 -------------(3)

Answer 45

from (1), u wil get value of z
from (2), u wil get value of y

put y, z values in (3), you will get value of z

Answer 46

you mean x for 1?

Answer 47

(1) is the first equation ive labelled above

Answer 48

from (1),

-2z = 8

divide -2 both sides

z = -4

Answer 49

so x-5, -4, y = -2?

Answer 50

y = -2

is correct

Answer 51

x = 5

Answer 52

x=5, z -4

Answer 53

good job !

Answer 54

ty

Answer 55

np :)