python: function that calculates determinant recursively but for matrix that are tridiagonal. that is is uses det$($A$)=$

j$=0$

$(1)$j det$($B$0,$j$)$A$[0,$j$]$ with Bi$,$j the matrix built from A by removing the i$-$th row and j$-$th column. so for tridiagonal the loop is not for i in range $($n$)$ but in range $[0,1].$ for such algorithm i have to calculate time complexity. the answer is # loop iteration: $1$ add $+1$ power $+2$ mult $=4=$ p # $0\mathrm{.}2$ pts

# recursion:

# n $=1$: $0$ # $0\mathrm{.}1$ pts

# n $=2$: T$(2)=$ p $+$ p $=2$ p # $0\mathrm{.}1$ pts $(2$ loop iterations, no cost for determinant$)$

# n $=3$: T$(3)=2$ x $($p $+$ T$(2))=2$ x $(3$ p$)=6$ p # $0\mathrm{.}1$ pts

# n $=4$: T$(4)=2$ x $($p $+$ T$(3))=2$ x $(7$ p$)=14$ p

# n $=5$: T$(5)=2$ x $($p $+$ T$(4))=2$ x $(15$ p$)=30$ p

# For writing$/$recognizing the geometric sum: # $0\mathrm{.}4$ pts

# n: T$($n$)=2$ x $($p $+$ T$($n$-1))=(2^$ n $-2)$ p # $0\mathrm{.}1$ pts $($solving the geo sum$)$

# $=>$ total: O$($n$)=$ T$($n$)$ but i only understand cases n$=0$ and n$=1.$ could you explain the next cases and from where this pattern comes from?