# Why in the proof of $A\cdot Adj(A)=Det(A)\cdot I_n$ entires not on the diagonal are zero?

I have read the proof. I do not get why the entires that are not on the diagonal are equal to zero, or why the following part of the proof is true:

If $k\neq \ell$

$$(A\cdot\hat A)_{k\ell}=\sum_{i=1}^n (-1)^{i+\ell}a_{ki} \det A(\ell\mid i)=0$$

• It's the determinant of the matrix you get when in $A$ you replace the $\ell$-th row with the $k$-th. The resulting matrix has two identical rows, hence the determinant is $0$. – Daniel Fischer Aug 20 '15 at 18:43
• @DanielFischer so there is an action of replacing a row? It do not derive from the formula? – gbox Aug 20 '15 at 18:45
• I'm not sure what the question is. Interpret the formula bearing Laplace expansion in mind. – Daniel Fischer Aug 20 '15 at 18:53
• Laplace expansion by the $\ell$-th row gives you $$\det B = \sum_{i = 1}^n (-1)^{\ell + i} b_{\ell i}\det B(\ell \mid i).$$ Now, if $B$ is the matrix you obtain from $A$ by replacing the $\ell$-th row with the $k$-th, then $B(\ell \mid i) = A(\ell \mid i)$, and also $b_{\ell i} = a_{k i}$. – Daniel Fischer Aug 20 '15 at 19:23
• Hypothetically. As a Gedankenexperiment. We don't actually do that. But that way we see that, and why, we have $A\cdot \operatorname{Adj} A = (\det A)\cdot I$. – Daniel Fischer Aug 20 '15 at 19:32

Consider the $3\times 3$ case $$\underbrace{\left[\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ \color{blue}{a_{21}} & \color{blue}{a_{22}} & \color{blue}{a_{23}}\\ a_{31} & a_{32} & a_{33}}\right]}_{A}\cdot \underbrace{\left[\matrix{A_{11} & A_{21} & A_{31}\\ A_{12} & A_{22} & A_{32}\\ A_{13} & A_{23} & A_{33}}\right]}_{\text{adj}(A)}.$$ If you multiply the first row with the first column you get exactly the determinant expansion along the first row $$\color{red}{a_{11}}A_{11}+\color{red}{a_{12}}A_{12}+\color{red}{a_{13}}A_{13}=\left|\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}}\right|=\det(A).$$ Now if you multiply the second row with the first column you get $$\color{blue}{a_{21}}A_{11}+\color{blue}{a_{22}}A_{12}+\color{blue}{a_{23}}A_{13}$$ which in the same way can be interpreted as the determinant expansion along the first row, but for what matrix? Well, the cofactors are built of the elements of the second and third rows of $A$, so those rows remain unchanged, and the first row must be made of the blue elements, since it is them who replace the red elements in the formula above, hence $$\color{blue}{a_{21}}A_{11}+\color{blue}{a_{22}}A_{12}+\color{blue}{a_{23}}A_{13}= \left|\matrix{\color{blue}{a_{21}} & \color{blue}{a_{22}} & \color{blue}{a_{23}}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}}\right|.$$ But this matrix has first two rows being equal, which means that the determinant is zero.
P.S. For other rows and columns the idea is exactly the same. The second row times the second column and the third row times the third column will be the determinant expansion along the second and the third rows, respectively, and give $\det(A)$. All other combinations give zero since there are two equal rows in the corresponding determinant.
• this is what we get: $$\color{blue}{a_{21}} \left|\matrix\\ a_{22} & a_{23}\\ a_{32} & a_{33}}\right|+\color{blue}{a_{22}}\left|\matrix{\color{blue}\\ a_{21} & & a_{23}\\ a_{31} & & a_{33}}\right|+\color{blue}{a_{23}}\left|\matrix{\color{blue}\\ a_{21} & & a_{22}\\ a_{31} & & a_{32}}\right|$$ – gbox Aug 20 '15 at 19:58
• Right. Now the last step is to see the same principle for the $n\times n$ case. (This is what @DanielFischer was saying in the comments). – A.Γ. Aug 20 '15 at 20:17