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# The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:

$\begin{array}{1 1} 27 \\ 18 \\ 81 \\ 512 \end{array}$

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A)
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• There are 9 elements in $3\times 3$ matrix and one element can be filled in two ways either 0 or 1.
$\Rightarrow$ Total possible matrices =$2^9$=512.

Ans:(D).

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A)
The number of elements in a $3$ X $3$ matrix is the product $3$ X $3 = 9$.
Given this, the total possible matrices that can be selected is $2^9 = 512$.
Hence, the solution is $(D): 512$.