Probability of getting a non-singular matrix












1












$begingroup$


Out of all Boolean-square matrices of size 2x2 possible, a matrix is choosen at random.The probability that the matrix selected is non-singular is.



My work.



Total matrices possible-> $2^{4}=16$



$begin{bmatrix}
a&b \
c&d
end{bmatrix}$



Determinant of such matrix is given by $ad-bc$



I am calculating number of possible singular boolean matrices I can make.



So, for all such matrices $ad-bc=0$ and hence $ad=bc$



Now, I further divide this into two cases



(1)Product $ad,bc$ both are zero.-Suppose, I choose a and c to be zero, now d and b both have 2 choices-Either 0 or 1->so $2 times 2=4$ ways this way.



Now another way can be when $d=0,b=0$ and $a,c$ can be choosen freely each in 2 ways, so here also total 4 ways.



So, all in all I have 8 ways in which I can have product $ad=bc=0$.But 1 case is common when all $a,b,c,d=0$ and this was counted twice, so total only $7$ ways in which the product $ad=bc=0.$



(2)Product $ad,bc=1$->This has only 1 way when all $a,b,c,d$ are one.



So, from a total of 16 matrices, I can have $7+1=8$ singular matrices.



So, required probability=$displaystyle frac{8}{16}$



Am I correct in both my reasoning and the answer?






probability







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$endgroup$












  • $begingroup$
    What are the operations?
    $endgroup$
    – Wuestenfux
    Jan 15 at 9:26










  • $begingroup$
    Don't consider any operation.Just assume each element can be either 0 or 1.
    $endgroup$
    – user3767495
    Jan 15 at 9:38
















1












$begingroup$


Out of all Boolean-square matrices of size 2x2 possible, a matrix is choosen at random.The probability that the matrix selected is non-singular is.



My work.



Total matrices possible-> $2^{4}=16$



$begin{bmatrix}
a&b \
c&d
end{bmatrix}$



Determinant of such matrix is given by $ad-bc$



I am calculating number of possible singular boolean matrices I can make.



So, for all such matrices $ad-bc=0$ and hence $ad=bc$



Now, I further divide this into two cases



(1)Product $ad,bc$ both are zero.-Suppose, I choose a and c to be zero, now d and b both have 2 choices-Either 0 or 1->so $2 times 2=4$ ways this way.



Now another way can be when $d=0,b=0$ and $a,c$ can be choosen freely each in 2 ways, so here also total 4 ways.



So, all in all I have 8 ways in which I can have product $ad=bc=0$.But 1 case is common when all $a,b,c,d=0$ and this was counted twice, so total only $7$ ways in which the product $ad=bc=0.$



(2)Product $ad,bc=1$->This has only 1 way when all $a,b,c,d$ are one.



So, from a total of 16 matrices, I can have $7+1=8$ singular matrices.



So, required probability=$displaystyle frac{8}{16}$



Am I correct in both my reasoning and the answer?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are the operations?
    $endgroup$
    – Wuestenfux
    Jan 15 at 9:26










  • $begingroup$
    Don't consider any operation.Just assume each element can be either 0 or 1.
    $endgroup$
    – user3767495
    Jan 15 at 9:38














1












1








1





$begingroup$


Out of all Boolean-square matrices of size 2x2 possible, a matrix is choosen at random.The probability that the matrix selected is non-singular is.



My work.



Total matrices possible-> $2^{4}=16$



$begin{bmatrix}
a&b \
c&d
end{bmatrix}$



Determinant of such matrix is given by $ad-bc$



I am calculating number of possible singular boolean matrices I can make.



So, for all such matrices $ad-bc=0$ and hence $ad=bc$



Now, I further divide this into two cases



(1)Product $ad,bc$ both are zero.-Suppose, I choose a and c to be zero, now d and b both have 2 choices-Either 0 or 1->so $2 times 2=4$ ways this way.



Now another way can be when $d=0,b=0$ and $a,c$ can be choosen freely each in 2 ways, so here also total 4 ways.



So, all in all I have 8 ways in which I can have product $ad=bc=0$.But 1 case is common when all $a,b,c,d=0$ and this was counted twice, so total only $7$ ways in which the product $ad=bc=0.$



(2)Product $ad,bc=1$->This has only 1 way when all $a,b,c,d$ are one.



So, from a total of 16 matrices, I can have $7+1=8$ singular matrices.



So, required probability=$displaystyle frac{8}{16}$



Am I correct in both my reasoning and the answer?






probability







share|cite|improve this question











$endgroup$




Out of all Boolean-square matrices of size 2x2 possible, a matrix is choosen at random.The probability that the matrix selected is non-singular is.



My work.



Total matrices possible-> $2^{4}=16$



$begin{bmatrix}
a&b \
c&d
end{bmatrix}$



Determinant of such matrix is given by $ad-bc$



I am calculating number of possible singular boolean matrices I can make.



So, for all such matrices $ad-bc=0$ and hence $ad=bc$



Now, I further divide this into two cases



(1)Product $ad,bc$ both are zero.-Suppose, I choose a and c to be zero, now d and b both have 2 choices-Either 0 or 1->so $2 times 2=4$ ways this way.



Now another way can be when $d=0,b=0$ and $a,c$ can be choosen freely each in 2 ways, so here also total 4 ways.



So, all in all I have 8 ways in which I can have product $ad=bc=0$.But 1 case is common when all $a,b,c,d=0$ and this was counted twice, so total only $7$ ways in which the product $ad=bc=0.$



(2)Product $ad,bc=1$->This has only 1 way when all $a,b,c,d$ are one.



So, from a total of 16 matrices, I can have $7+1=8$ singular matrices.



So, required probability=$displaystyle frac{8}{16}$



Am I correct in both my reasoning and the answer?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 9:19









El borito

664216




664216










asked Jan 15 at 9:11









user3767495user3767495

40218




40218












  • $begingroup$
    What are the operations?
    $endgroup$
    – Wuestenfux
    Jan 15 at 9:26










  • $begingroup$
    Don't consider any operation.Just assume each element can be either 0 or 1.
    $endgroup$
    – user3767495
    Jan 15 at 9:38


















  • $begingroup$
    What are the operations?
    $endgroup$
    – Wuestenfux
    Jan 15 at 9:26










  • $begingroup$
    Don't consider any operation.Just assume each element can be either 0 or 1.
    $endgroup$
    – user3767495
    Jan 15 at 9:38
















$begingroup$
What are the operations?
$endgroup$
– Wuestenfux
Jan 15 at 9:26




$begingroup$
What are the operations?
$endgroup$
– Wuestenfux
Jan 15 at 9:26












$begingroup$
Don't consider any operation.Just assume each element can be either 0 or 1.
$endgroup$
– user3767495
Jan 15 at 9:38




$begingroup$
Don't consider any operation.Just assume each element can be either 0 or 1.
$endgroup$
– user3767495
Jan 15 at 9:38










1 Answer
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$begingroup$

You miss two cases for $ad=cb=0$, if $a=b=0$ and $c=d=1$ and the reverse. The correct probability is $frac{6}{16}$ and can also be obtain like that :



The only matrices that are non singular are permutations (row or columns) of the two following matrices
begin{align*}
begin{bmatrix}
1&0\
0&1
end{bmatrix}&&begin{bmatrix}
1&1\
0&1
end{bmatrix}
end{align*}
The first one have exactly two permutations that are non singular which correspond to exchanging the two rows (or columns, it doesn't matter). The second have four distinct non singular permutation where you exchange rows or columns or both. This also gets you the $frac{6}{16}$ probability of being singular.






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    $begingroup$

    You miss two cases for $ad=cb=0$, if $a=b=0$ and $c=d=1$ and the reverse. The correct probability is $frac{6}{16}$ and can also be obtain like that :



    The only matrices that are non singular are permutations (row or columns) of the two following matrices
    begin{align*}
    begin{bmatrix}
    1&0\
    0&1
    end{bmatrix}&&begin{bmatrix}
    1&1\
    0&1
    end{bmatrix}
    end{align*}
    The first one have exactly two permutations that are non singular which correspond to exchanging the two rows (or columns, it doesn't matter). The second have four distinct non singular permutation where you exchange rows or columns or both. This also gets you the $frac{6}{16}$ probability of being singular.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You miss two cases for $ad=cb=0$, if $a=b=0$ and $c=d=1$ and the reverse. The correct probability is $frac{6}{16}$ and can also be obtain like that :



      The only matrices that are non singular are permutations (row or columns) of the two following matrices
      begin{align*}
      begin{bmatrix}
      1&0\
      0&1
      end{bmatrix}&&begin{bmatrix}
      1&1\
      0&1
      end{bmatrix}
      end{align*}
      The first one have exactly two permutations that are non singular which correspond to exchanging the two rows (or columns, it doesn't matter). The second have four distinct non singular permutation where you exchange rows or columns or both. This also gets you the $frac{6}{16}$ probability of being singular.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You miss two cases for $ad=cb=0$, if $a=b=0$ and $c=d=1$ and the reverse. The correct probability is $frac{6}{16}$ and can also be obtain like that :



        The only matrices that are non singular are permutations (row or columns) of the two following matrices
        begin{align*}
        begin{bmatrix}
        1&0\
        0&1
        end{bmatrix}&&begin{bmatrix}
        1&1\
        0&1
        end{bmatrix}
        end{align*}
        The first one have exactly two permutations that are non singular which correspond to exchanging the two rows (or columns, it doesn't matter). The second have four distinct non singular permutation where you exchange rows or columns or both. This also gets you the $frac{6}{16}$ probability of being singular.






        share|cite|improve this answer









        $endgroup$



        You miss two cases for $ad=cb=0$, if $a=b=0$ and $c=d=1$ and the reverse. The correct probability is $frac{6}{16}$ and can also be obtain like that :



        The only matrices that are non singular are permutations (row or columns) of the two following matrices
        begin{align*}
        begin{bmatrix}
        1&0\
        0&1
        end{bmatrix}&&begin{bmatrix}
        1&1\
        0&1
        end{bmatrix}
        end{align*}
        The first one have exactly two permutations that are non singular which correspond to exchanging the two rows (or columns, it doesn't matter). The second have four distinct non singular permutation where you exchange rows or columns or both. This also gets you the $frac{6}{16}$ probability of being singular.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 9:48









        P. QuintonP. Quinton

        2,0001214




        2,0001214






























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