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I am going over the proof of $\mathrm{adj}(A)A = \det(A) \cdot I$ as for the first part I understand why the elements on the diagonal are $Det(A)$ but as for the second part.

why if $k\neq l$ we get: $$(A\cdot\hat A)_{k\ell}=\sum_{i=1}^n (-1)^{i+\ell}a_{ki} \det A(\ell\mid i)=0$$

Do we define $\hat A$ differently from the first case? why does it have two identical rows?

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gbox
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1 Answers1

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Comparing the two expressions: $$\sum_{i=1}^n (-1)^{i+\ell}a_{li} \det A(\ell\mid i)\quad (1)$$ and $$\sum_{i=1}^n (-1)^{i+\ell}a_{ki} \det A(\ell\mid i) \quad (2)$$

You said you know why the first one is $\det(A)$. It is the expansion along the $l$th row, where the $l$th row has elements $a_{l1},a_{l2},\dots, a_{ln}$.

Now the only difference of $(2)$ from $(1)$ is, it changes the elements of $l$th row into $a_{k1}, a_{k2},\dots, a_{kn}$. That's why you can consider $(2)$ still as $l$th row expansion, but with a different matrix. This matrix is the original matrix in which the $l$th row is replaced with $k$th row. But this matrix then has two identical rows. So the determinant is zero.

KittyL
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  • And we can change the matrix because in the expansion remove the $l$th row? – gbox Jul 23 '15 at 14:57
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    @gbox: Yes, the $\det A(l| i)$ is the determinant of matrix by removing the $l$th row and $i$th column. It does not change what ever $l$th row is. – KittyL Jul 23 '15 at 15:16