Question

the number of 3x3 non singular matrices with four entries as 1 and all other entries as 0 is

Answer 1

@ganeshie8 ?

Answer 2

@dan815 ?

Answer 3

@mathslover ?

Answer 4

@vishweshshrimali5 ?

Answer 5

@Miracrown ?

Answer 6

you cannot have a row or a column with all 0's

Answer 7

yes

Answer 8

total ways of filling all elements =9!/4!5!

Answer 9

yes pick 4 elements from 9 :
9 choose 4

Answer 10

ways in which 0 occupies rows and colums=3C1*2

Answer 11

may be we can try this :

(9 choose 4) - ( total number of ways of having all 0's in a line)

Answer 12

ways in which 0 occupies rows and colums=3C1*2?

Answer 13

nope

Answer 14

for example :
if u fill first row with "0 0 0" :
there will be (6 choose 4) different combinations for selecting 1's on remaining two row elements

Answer 15

there will be (6 choose 4) different combinations for selecting 1's on remaining two row elements??

Answer 16

@ganeshie8 ?

Answer 17

?

Answer 18

@ganeshie8 Its not necessary that only matrices having only 0s in rows or columns will have zero determinant.

Like

Answer 19

Ahh true^ more cases to consider then !

Answer 20

this may work : "no two rows or columns should be same"

Answer 21

yeah

Answer 22

*add that constraint to previous

Answer 23

It would be better if we first find out the singular matrices and then subtract from all possible matrices.

Answer 24

1) no zeroes on a row/column
2) no two rows/columns should be same

Answer 25

yes im trying the same above^

Answer 26

understood the constraints but how to find out them?

Answer 27

0 0 0
x x x
x x x

Answer 28

you got four 1's to fill in those 6 x spots :
(6 choose 4) ways

Answer 29

repeat for next two rows :
3*(6 choose 4) = 3(15) = 45 ways

Answer 30

similarly, working the columns give you another 45 ways

but there will be few overlaps which you need to subtract

Answer 31

I think the overlap is 9; so the total number of ways a line(row/col) can have all zeroes is :

45 + 45 - 9 = 81 ways

vishy check :)

Answer 32

One sec.

Answer 33

How did you obtain the overlap ?

Answer 34

0 0 0
1 1 0
1 1 0

thats a overlap : total permutations = 3 per each row
=> 9 in total

Answer 35

we got @mukushla here :)

Answer 36

Okay:

0 0 0
1 1 0
1 1 0

0 0 0
0 1 1
0 1 1

0 1 1
0 1 1
0 0 0

1 1 0
1 1 0
0 0 0

1 0 1
0 0 0
1 0 1

1 0 1
1 0 1
0 0 0

0 0 0
1 0 1
1 0 1

Any else I forgot ?

Answer 37

Okay

1 1 0
0 0 0
1 1 0

0 1 1
0 0 0
0 1 1

Answer 38

There we have 9

Answer 39

yes !

Answer 40

Yeah you are right till this step

Answer 41

good so we nailed one constraint, lets look at second constraint :)

Answer 42

@cody_123 are you wid us ?

Answer 43

?

Answer 44

@vishweshshrimali5 ?

Answer 45

ohh sorry I was busy on another tab.

Answer 46

So did you understand the first constraint

Answer 47

yes

Answer 48

Sorry cody I would be unable to respond you for atleast half and hour. I have some serious work to attend.

Answer 49

Sorry

Answer 50

some hint

Answer 51

@ParthKohli ?

Answer 52

@dan815 ?

Answer 53

@Miracrown

Answer 54

if we have 2 columns same number of matrices=3*3=9

Answer 55

@ganeshie8 ?

Answer 56

hey

Answer 57

if we have 2 columns same number of matrices=3*3=9?

Answer 58

how did y get the columns?

Answer 59

if we fix one of the column with only 1's then there are 9 matrices

Answer 60

Yes !

Answer 61

ok

Answer 62

then?

Answer 63

w8 a second how can u arrange all 1's in one column that would violate the condition of 4 1's

Answer 64

we made a mistake - lets work this thing again :)

Answer 65

0 0 1
0 0 1 = 3 matrices
1 1 0


0 0 1
1 1 0 = 3 matrices
0 0 1


1 1 0
0 0 1 =3 matrices
0 0 1

Answer 66

Wow ! looks perfect !!

Answer 67

will there be any more matrices with same rows ?

Answer 68

no all matrices will be over counted then

Answer 69

i think you're right ! do u have answer for this ? :)

Answer 70

answer should be 126-(81+9)=36?

Answer 71

I think so it should be \(\le 36\)

Need to confirm we did not miss counting any

Answer 72

please see i dont think so there is any

Answer 73

@ganeshie8 ?

Answer 74

@dan815 got some time ?

Answer 75

@hartnn ?

Answer 76

I think we have to fill the diagonal first, if we don't want a singular matrix.

Answer 77

i think those matrices are counted

Answer 78

So there is only one "1" left for the other 6 places.

Answer 79

@cody_123
What do you mean by "counted"? :)

Answer 80

i mean matrices where all the diagonal elements are same are counted in the above cases
like
0 1 1
0 0 1
1 1 0

Answer 81

If they are identical, they should be counted as one.

But...
When the matrix has a zero on the diagonal, it will be singular.
Since the matrix has to be non-singular, all the diagonal terms must be non-zero. This will therefore take care of three "1"s. You can then place the remaining "1" wherever you want.

Answer 82

so it is another constraint?

Answer 83

Yes, it is a constraint that helps you. Because you know where three of the '1"s go, and leaves you with only one left in the other six places.