Describing determinants of order $n$ as determinants of order $n-1$

Asked Sep 05 '21 at 04:10

Active Sep 05 '21 at 04:10

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I'm positive there's a sufficient answer for this here, but I can't seem to find it. I'd appreciate being redirected to the original question if anyone knows of one.

Take the augmented matrix below as an example, $$ \pmatrix{a_1 \ a_2 \ a_3 \\ b_1 \ b_2 \ b_3 \\ c_1\ c_2\ c_3}$$ Why is it that I am allowed to take the first component $a_1$, cover up every other element corresponding to its row and column, and multiply it to the remaining matrix $\pmatrix{b_1 \ b_3 \\ c_2 \ c_3}$, and then repeat this toward the last element on the first row? And why do the signs alternate? Like so: $$a_1\pmatrix{b_2 \ b_3 \\ c_2 \ c_3}-a_2\pmatrix{b_1 \ b_3 \\ c_1 \ c_3}+a_3\pmatrix{b_1 \ b_2 \\ c_1 \ c_2}=$$

asked Sep 05 '21 at 04:10

Lex_i

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Read Introduction to Linear Algebra by Gilbert Strang. There's a chapter for determinant. – sea yellow Sep 05 '21 at 04:17
Maybe because it is defined like that. – Lalit Tolani Sep 05 '21 at 04:20
@LalitTolani It can be defined in multiple ways, and other properties can be derived from it. – sea yellow Sep 05 '21 at 04:30
@seayellow, I agree with you "Laplace expansion " and "Leibniz Formula " are widely used to find determinants of a matrix. Interestingly we can find determinants using "Graph theory" also. – Manish Saini Sep 05 '21 at 05:37
@ManishSaini Great stuff. Roger. – sea yellow Sep 05 '21 at 06:13
Try solving for $\vec{x}$. $$\pmatrix{a_{11} \ a_{12} \ a_{21} \ a_{22} }\vec{x} = \pmatrix{c_1\c_2}$$ Then try it with bigger matrices and vectors. I think you'll see why determinants are defined the way they are. – John Joy Sep 05 '21 at 07:05
@Lex_i You might find this post about the determinant helpful – Ben Grossmann Sep 06 '21 at 21:16

Describing determinants of order $n$ as determinants of order $n-1$

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