5
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $7 \times 8=56$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.