Recursion in matrix calculation

2020 8 25
1 个回答
更新时间：2020 8 25
3 次查看（30 天）

1 个投票

Hi,
How does recursion work in this context, as in how did the function CalDet manage to calculate the determinant of the minor without explicitly writing an equation?
function [determinant] = CalDet(M)
dimensionM = size(M);
if (dimensionM(1) == 1)
determinant = M(1, 1);
else
determinant = 0;
for i = 1:dimensionM(2)
determinant = determinant + (-1)^(i+1) * M(1, i) * CalDet(MMin(M, 1, i));
end
end
end
function [MatrixMinor] = MMin(M, i, j)
dimensionM = size(M);
MatrixMinor = M([1:(i-1) (i+1):dimensionM(1)], [1:(j-1) (j+1):dimensionM(2)]);
end

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回答（1 个）

James Tursa
James Tursa 2020-8-25
编辑：James Tursa 2020-8-25

1 个投票

This uses recursive calls (CalDet calls CalDet with smaller matrices until the size is 1x1). I.e., the recursion continues all the way down until the input is a 1x1 matrix, at which point the result is simply M(1,1) and then the results get passed back up through the stack of calls.
See Laplace's expansion and the adjugate matrix here:
https://en.wikipedia.org/wiki/Determinant#:~:text=This%20procedure%20can%20be%20extended%20to%20give%20a,methods%20of%20solution%20are%20much%20more%20computationally%20efficient.
BTW, this is not a good numerical technique.

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