Why is it true that $\mathrm{adj}(A)A = \det(A) \cdot I$?

This is a statement in linear algebra that I can't seem to understand the proof behind.

For a square matrix $A$, why is: $$\mathrm{adj}(A)A = \det(A) \cdot I$$

Any explanation would be greatly appreciated.

There is a nice short explanation. You probably know that the determinant of any matrix can be expanded row or column-wise using the minors:

$$\det A=\sum_{i=1}^n a_{ij}(-1)^{i+j}\det A(i\mid j)$$

or

$$\det A=\sum_{j=1}^n a_{ij}(-1)^{i+j}\det A(i\mid j)$$

for any $j$ (resp. $i$) or your liking where we obtain $A(i\mid j)$ by striking out the $i$-th row and the $j$-th column.

We define the coefficients of the adjoint $\hat A$ by $$(\hat A)_{ij}=(-1)^{{i+j}}\det A(j\mid i)$$

Now, upon matrix multiplication, we have

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

If $k=\ell$, then $$(A\cdot \hat A)_{\ell\ell}=\sum_{i=1}^n(-1)^{i+\ell}a_{\ell i}\det A(\ell \mid i)=\det A$$ since we're expanding the determinant through the $\ell$-th row.

If $k\neq \ell$

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

for it is the expansion of the determinant of the matrix $A^{k\ell}$ defined by $$(A^{k\ell})=\begin{cases} a_{ij} \text{ if }i\neq \ell\\a_{kj}\text{ if }i=\ell\end{cases}$$ which has two equal rows.

• I can't seem to understand, why is this true? "since we're expanding the determinant through the ℓ-th column. If k≠ℓ (A⋅A^)kℓ = 0" – vondip Mar 31 '13 at 10:18
• @vondip That sentence refers to the $\ell =k$ part. See the first two summations at the beginning. – Pedro Tamaroff Mar 31 '13 at 16:32
• Sorry but i too, can't understand why if k≠ℓ (A⋅A^)kℓ = 0. is it possible to make it even simpler? – George Smyridis Feb 10 '16 at 21:05
• @GeorgeSmyridis math.stackexchange.com/questions/1404189 – BCLC Sep 19 '16 at 14:45
• – BCLC Sep 19 '16 at 14:45

Another way of stating Cramer's Rule. See e.g. Wikipedia.

• it seems to be as though Cramer's Rule relays on this. But what is the right way to prove this theorem? – vondip Mar 29 '13 at 9:43

Let $A=[\alpha_{ij}]$ be matrix of order $n$ on field $K$ and $s,r \in \{1,\dots,n\}$, $n>1,$ fixed elements. Let $\hat{A}_{rs}=(-1)^{r+s} \det A_{rs}$ . Then $$\det A = \alpha_{1s}\hat{A}_{1s}+\alpha_{2s}\hat{A}_{2s}+\dots+\alpha_{ns}\hat{A}_{ns}$$ and similary $$\det A = \alpha_{r1}\hat{A}_{r1}+\alpha_{r2}\hat{A}_{r2}+\dots+\alpha_{rn}\hat{A}_{rn}\tag{1}$$ Then we find that $$\alpha_{i1}\hat{A}_{r1}+\dots+\alpha_{in}\hat{A}_{rn}=\begin{cases} 0, &\text{ for } i \neq r,\\ \det A,& \text{ for } i=r.\end{cases}\tag{2}$$

And similary for $\alpha_{1i}\hat{A}_{1s}+\dots+\alpha_{ni}\hat{A}_{ns}=0 \quad (i \neq s).$

Now conclude that for every matrix of order $n$ on field $K$ it belongs one matrix

$$\operatorname{adj}A = \begin{bmatrix} \hat{A}_{11} & \cdots & \hat{A}_{n1} \\ \vdots & \ddots & \vdots \\ \hat{A}_{1n} & \cdots & \hat{A}_{nn} \end{bmatrix}$$ from $M_n(K)$, such that $$A \cdot \operatorname{adj}A = \det A \cdot I$$

Edit (explanation for $(2)$): From $(1)$, left side of $(2)$ is determinant of matrix $B$ which arise from matrix $A$ when we change $r$-th row its $i$-th row. If $i \neq r$ then that matrix have two same rows $(B_{r\to}=A_{i\to}=B_{i\to}$), so its determinant is zero.

• Why is it true that for each i≠r, 0 aiAr is 0 ? – vondip Mar 29 '13 at 11:36
• @vondip I edited the answer. – Cortizol Mar 29 '13 at 19:38

It is well known that the determinant, viewed as function of the columns of a square matrix, is linear in each of the columns. This means that if we fix any matrix $A\in M_n(K)$ and a column index $j$, and denote by $f_{A,j}$ the function $K^n\to K$ defined by $f_{A,j}(x_1,\ldots,x_n)=\det(A')$ where $A'$ is the matrix obtained by replacing column $j$ of $A$ by the entries $x_1,\ldots,x_n$, then $f_{A,j}$ is a linear function, so there exist constants $c_1,\ldots,c_n$ such that $f_{A,j}(x_1,\ldots,x_n)=c_1x_1+\cdots+c_nx_n$. These constants of course depend on $A$ and on $j$ (although the column $j$ of $A$ is irrelevant, as it gets systematically replaced), and indeed the Laplace expansion of $\det(A')$ by column $j$ gives explicit expressions for these coefficients. By definition, the $1\times n$ matrix $(c_1~c_2~\ldots~c_n)$ is precisely row $j$ of the adjugate matrix $\def\adj{\operatorname{adj}}\adj A$.

Now consider the matrix product $(c_1~c_2~\ldots~c_n)\cdot A=(p_1~p_2~\ldots~p_n)$. This can be computed separately by the columns of $A$: one has $p_i=f_{A,j}(A_{1,i},\ldots,A_{n,i})$ (the arguments form column $i$ of $A$), which is $\det(A')$ where $A'$ is the matrix obtained by replacing column $j$ of $A$ by (a copy of) column $i$ of $A$. Obviously one has $A'=A$ if $i=j$, so that $p_i=\det A$, and all other values $p_i$ are zero due to the alternating property of the determinant, since in these cases we are taking the determinant of a matrix $A$ whose columns $i$ and $j$ are identical, with $i\neq j$. In formula $$p_i=\delta_{i,j}\det A \qquad\text{for i\in\{1,\ldots,n\}.}$$ But what we have computed is just row $j$ of the matrix product $\adj A\cdot A$, and what we have said is valid for all $j\in\{1,\ldots,n\}$. Combining all these computations one gets $$\adj A\cdot A=I_n\det A$$ (where the right hand side is just a scalar multiple of the identity matrix $I_n$), which is the identity in the question.

All in all you can see there is not much more to this identity than the multi-linear and alternating properties (by columns) of the determinant, Laplace expansion and the definition of the adjugate matrix. Since it is a formal algebraic identity, one may take any commutative ring in the place of $K$ (but commutativity is essential; multi-linearity depends on it). To prove that $A\cdot\adj A$ also gives the same result, one needs the multi-linear and alternating properties by rows of the determinant.