Put the famous Pascal triangle into a matrix. It could go intoa lower triangular $L$ or its transpose $L'$ or a symmetric matrix $S$: $$L=matrix{ [ 1 0 0 0 ] cr [ 1 1 1 1 ] cr [ 1 1 0 0 ] cr[ 1 2 1 0 ] cr[ 1 3 3 1 ] cr}quad L' =matrix{ [ 1 1 1 1]cr [ 0 1 2 3 ]cr [ 0 0 1 3 ] cr[ 0 0 0 1 ] cr}quad S = matrix{[ 1 2 3 4]cr [ 1 3 6 10]cr [ 1 4 10 20]cr}$$These binomial numbers come from a recursion, or from the formulafor $i$ choose $j$, or functionally from taking powers of $(1 + x)$. The amazing thing is that $L imes L' = S$. (OK for $4 imes 4$)It follows that $S$ has determinant 1. The matrices have otherunexpected properties too, that give beautiful examples in teachinglinear algebra. The proof of $L L' = S$ comes 3 ways: 1. By induction using the recursion formula for the matrix entries. 2. By an identity for the coefficients $i+j$ choose $j$ in $S$. 3. By applying both sides to the column vector $[ 1 x x^2 x^3 ... ]'$.The third way also gives a proof that $S^3 = -I$ but we doubt that result. The rows of the ``hypercube matrix" $L^2$ count corners and edgesand faces and ... in n dimensional cubes.