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probability that a matrix is singular
- Thread starter jacks
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Re: probability
To that end, note that there are $2^3$ possible row vectors. Each row vector $v$ has a negative $-v$ in the set of possible row vectors. Since a basis cannot include both a vector and its negative, let's restrict our attention to one vector $v$ from each of the pairs $v$ and $-v$. This leaves us with $(1/2) \; 2^3 = 4$ row vectors to consider. For example, we might choose the set $E = \{(1,1,1), (1,1,-1), (1,-1,1), (-1,1,1)\}$. Since the order of the row vectors does not affect the (non)singularity of a matrix, let's consider just the $\binom{4}{3} = 4$ subsets of size 3 taken from $E$. It's easy to check that each of the 4 3 by 3 matrices thus produced is nonsingular.
Taking into account the possible orderings of the three row vectors in a matrix, we must multiply by $3!$; and taking into account that each vector could be replaced by its negative, we must multiply by $2^3$. So all together, there are
$\binom{4}{3} \; 3! \; 2^3$
nonsingular matrices whose elements are -1 or 1.
So the probability that such a matrix is singular is
$$1 - \frac{\binom{4}{3} \; 3! \; 2^3}{2^9}$$.
There are $2^9$ matrices in all, which we are assume are equally likely. We would like to count those which are nonsingular.A $3 \times 3$ matrices are formed using the the elements of $\left\{-1,1\right\}$. Then the probability that it is Singular, is
To that end, note that there are $2^3$ possible row vectors. Each row vector $v$ has a negative $-v$ in the set of possible row vectors. Since a basis cannot include both a vector and its negative, let's restrict our attention to one vector $v$ from each of the pairs $v$ and $-v$. This leaves us with $(1/2) \; 2^3 = 4$ row vectors to consider. For example, we might choose the set $E = \{(1,1,1), (1,1,-1), (1,-1,1), (-1,1,1)\}$. Since the order of the row vectors does not affect the (non)singularity of a matrix, let's consider just the $\binom{4}{3} = 4$ subsets of size 3 taken from $E$. It's easy to check that each of the 4 3 by 3 matrices thus produced is nonsingular.
Taking into account the possible orderings of the three row vectors in a matrix, we must multiply by $3!$; and taking into account that each vector could be replaced by its negative, we must multiply by $2^3$. So all together, there are
$\binom{4}{3} \; 3! \; 2^3$
nonsingular matrices whose elements are -1 or 1.
So the probability that such a matrix is singular is
$$1 - \frac{\binom{4}{3} \; 3! \; 2^3}{2^9}$$.