Consider the set of all boolean square matrices of order $3 \times 3$ as shown below where a,b,c,d,e,f can be either 0 or 1.
$\begin{bmatrix}
a&b&c\\
0&d&e\\
0&0&f
\end{bmatrix}$
Out of all possible boolean matrices of above type, a matrix is chosen at random.The probability that the matrix is singular is?
My Work
The above matrix is a triangular matrix and it's determinant can be 0, if either one or more from a,d and f are 0 in which case 0 will be an eigen value of the matrix and hence determinant 0.
Number of ways in which I can set $a,d,f$ to zero are: $\binom{3}{1}+\binom{3}{2}+\binom{3}{3}=7$ ways.
Now, total given boolean matrices possible are
$2^6=64$
So, the required probability must be $\frac{7}{64}$
Is my answer correct?
Best Answer
To be singular, we need $a=0$ or $d=0$ or $f=0$.
To be non-singular, we need $a=d=f=1$.
Hence, the probaility is $$1-\frac{1}{2^3}=\frac{7}{8}.$$
Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 \times 8=56$.