# What is the probability of the matrix being Singular

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 $$\frac{7}{64}$$

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 $$7 \times 8=56$$.

• So, where I went wrong? Commented Jan 22, 2019 at 8:25
• you forgot to multiply by $8$ for the values of $b,c,e$. Commented Jan 22, 2019 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.