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


So, the required probability must be $\frac{7}{64}$

Is my answer correct?


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 $7 \times 8=56$.

  • $\begingroup$ So, where I went wrong? $\endgroup$ – user3767495 Jan 22 at 8:25
  • 1
    $\begingroup$ you forgot to multiply by $8$ for the values of $b,c,e$. $\endgroup$ – Siong Thye Goh Jan 22 at 8:26

It looks basically correct but I'd word it (fully) as follows:

Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff

$$\;adf=0\iff a=0\,\vee\,d=0\;\vee\, f=0$$

and etc.


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