# then the probability that the matrix $\begin{pmatrix} X_1 & X_2\\ X_3 & X_4 \end{pmatrix}$ is nonsingular?

Let $X_1, X_2, X_3, X_4$ be i.i.d random variables each assuming the values $1$ and $-1$ with probabilities $1/2$. then the probability that the matrix

\begin{pmatrix} X_1 & X_2\\ X_3 & X_4 \end{pmatrix}

is nonsingular?

We have to find out probability of $X_1X_4-X_2X_3 \neq 0$, but how to do?

• Worst comes to worst, there are $16$ cases to check. – user228113 Jan 25 '18 at 6:47
• What could be easier than checking the 16 cases? Draw a small picture that shows that 4 possibilities that columns can take. The answer is easy to see from there. – copper.hat Jan 25 '18 at 6:55

Either $X_1X_4=X_2X_3=1$ ($4$ ways) or $X_1X_4=X_2X_3=-1$ ($4$ ways), so $\frac 8{2^4}=\frac 12$ probability...
It is perhaps (geometrically) easier to check if ${X_1 \over X_2} = {X_3 \over X_4}$. Either side takes values $\pm 1$ with equal probability.