The Matrix Has You

If $a,b,c,d,e,f,g,h,i$ are in geometric progression with common ratio $r$ , then the second row is $r^3$ times the first row. Thus taking $r^3$ times the first row from the second row we see that $\mathrm{det}\,M \; = \; \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right| \; = \; \left| \begin{array}{ccc} a & b & c \\ 0 & 0 & 0 \\ g & h & i \end{array}\right| \; = \; 0$

If $a,b,c,d,e,f,g,h,i$ are in arithmetic progression with common difference $u$ , first taking the second row from the third and then the first row from the second, and then the second row from the third row again, we obtain $\mathrm{det}\,M = \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right| \; = \; \left|\begin{array}{ccc}a & b & c \\ d & e & f \\ 3u & 3u & 3u \end{array}\right| \; = \; \left|\begin{array}{ccc} a & b & c \\ 3u & 3u &3u \\ 3u & 3u & 3u \end{array}\right| \; = \; \left|\begin{array}{ccc} a & b & c \\ 3u & 3u & 3u \\ 0 & 0 & 0 \end{array}\right| \; = \; 0$

For all arithmetic sequences, let $M = \begin{pmatrix} a - 4d & a - 3d & a - 2d \\ a - d & a & a + d \\ a + 2d & a + 3d & a + 4d \end{pmatrix}$ . Then

$\det(M)$

$\small{= (a - 4d)a(a + 4d) + (a -3d)(a + d)(a + 2d) + (a - 2d)(a - d)(a + 3d) - (a - 2d)a(a + 2d) - (a - 4d)(a + d)(a + 3d) - (a - 3d)(a - d)(a + 4d)}$

$= (a^3 - 16ad^2) + (a^3 - 7 a d^2 - 6 d^3) + (a^3 - 7 a d^2 + 6 d^3) - (a^3 - 4 a d^2) - (a^3 - 13 a d^2 - 12 d^3) - (a^3 - 13 a d^2 + 12 d^3)$

$= 0$

For all geometric sequences, let $M = \begin{pmatrix} ar^{-4} & ar^{-3} & ar^{-2} \\ ar^{-1} & a & ar \\ ar^2 & ar^3 & ar^4 \end{pmatrix}$ . Then

$\det(M)$

$= ar^{-4} \cdot a \cdot ar^4 + ar^{-3} \cdot ar \cdot ar^2 + ar^{-2} \cdot ar^{-1} \cdot ar^3 - ar^{-2} \cdot a \cdot ar^2 - ar^{-4} \cdot ar \cdot ar^3 - ar^{-3} \cdot ar^{-1} \cdot ar^4$

$= a^3 + a^3 + a^3 - a^3 - a^3 - a^3$

$= 0$

Both arithmetic and geometric sequences result in $\det(M) = 0$ , so the statement is true .