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Number of $n$x$n$ matrices that can be constructed with $m$ different matrix elements

According to me it should be $m^{n*n}$. Is it correct?

and the follow up question with this concept in use

All elements of a $2$x$2$ matrix "A" can have values either $0$ or $1$. The probability that any element gets a value $(0$ or $1)$ is $1/2$. If all elements of this matrix are chosen at random, what is the probability that the determinant of this matrix is positive?

My answer to the above problem is → $3/16$

in Calculus | 24 views
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First one seems right.

Second one:

How can you make a 0/1 matrix of order 2 negative ? By ensuring the determinant evaluates as $0 - 1 = -1$

i.e. The leading diagonal has to multiply to 0, and other diagonal has to multiple to 1.
Therefore, the leading diagonal can be (0,0), (0,1),(1,0) and the other diagonal is (1,1).

So I think out of 16 possible combinations, 3 combinations will yield a negative determinant:

$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$, $\begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}$, $\begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix}$.

Probability of a negative determinant $= \frac{3}{16}$

Probability of a positive determinant $= 1- \frac{3}{16} = \frac{13}{16}$

What is the right answer ?

Edit: my definition of positive is 'non-negative', so I've included $0$ too
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0 is neither positive nor negative right?
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Ya that is why u got 13/16

or else it should have been 3/16 i guess

Btw 0 is neither positive nor negative right?
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I knew that'd be a question. That is why with matrices, you'd use the term non-singular instead. Or non-zero positive in case of integers. Or natural numbers.

But yes, you're right.
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Okay..

Thanks.. :)