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How many $4 \times 4$ matrices with entries from ${0, 1}$ have odd determinant?

Hint: Use modulo $2$ arithmetic.

  1. $20160$
  2. $32767$
  3. $49152$
  4. $57343$
  5. $65520$
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18 votes
OPTION A is correct

Whenever the $1^{st}$ row is $0$ then its determent is $0$, and similarly if any $2$ or more rows are linearly dependent then its $|det|=0$

In order to find the odd determinant the $1^{st}$ row must be non zero
$\Rightarrow$ totally$(2^{4}-1)$ possibilities $\large\mid\frac{0}{1} \; \frac{0}{1}\;  \frac{0}{1} \;\frac{0}{1}\large\mid$ like totally $=16-1$

$2{nd}$ row must be non zero and not linearly depending on $1^{st}$ row  $\Rightarrow$ totally $(2^{4}-2)$ possibilities
For $3{rd}$ row it must be non-zero as well as not linearly depending on first $2$ rows (not start with $0$) $\Rightarrow$ totally $(2^{4}-4)$

For 4th row $\Rightarrow (2^{4}-8)$

Total possibilities $=(2^{4}-2^{0}) * (2^{4}-2^{1}) * (2^{4}-2^{2}) *(2^{4}-2^{3})={15}*{14}*{12}*{8}=20160.$
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Whenever the 1st  row is 0 then its determent is 0, and similarly if any 2 or more rows are linearly dependent then its |det|=0

1st row should not be all zero- hence #Cases=(2^4-1)

2nd row should be non-zero and not linearly depending on 1st row- So #Cases=(2^4-2)

3rd row, it cannot be either of those rows(i.e  1st or 2nd row) OR the sum (sum of 1st and 2nd row) OR all 0 entry in 3rd row, so #Cases=(2^4-4)

4th row, it cannot be either of those rows( i.e 1st , 2nd or 3rd row)- #Cases=3

OR cannot be the linear combination of any two row of 1st ,2nd and 3rd – #Cases=3C2=3

OR cannot be the linear combination of all row 1st ,2nd and 3rd – #Cases=1

OR cannot be the all entry are zero— #Cases=1

So total number of cases for 4th row = (2^4-8)

Hence total possible matrix = (2^4-1).(2^4-2).(2^4-4).(2^4-8)= 15*14*12*8= 20160

A should be the answer.

 

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