Suppose A is a singular square matrix.

It is known that the $(i,j)$-entry of $A[adj(A)] = a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$ $$ \begin{cases} det(A) & \text{if $i = j$} \\[2ex] 0, & \text{if $i ≠ j$} \end{cases} $$

So $A[adj(A)] = det(A)I$

Since $A$ is singular, $det(A) = 0$

and hence $A[adj(A)] = 0I = 0$

While i understand that if $A$ is singular, $det(A) = 0$. I do not understand how does the above chain of $A[adj(A)] = a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$ leads to the conclusion that $A[adj(A)] = det(A)I$.

Please explain. Thanks :)


The $(i,j)$ entry of the identity matrix $I$ is $1$ if $i=j$ and $0$ if $i \neq j$. It follows that the $(i,j)$-entry of $\det(A) I$ is $$ \begin{cases} \det(A) & i = j\\ 0 & i \neq j \end{cases} $$ Because they have the same $(i,j)$ entries, we can conclude that $A\operatorname{adj}(A)$ and $\det(A)I$ are the same matrix.

  • $\begingroup$ Thanks. However, how does it has any relation with $a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$? $\endgroup$ – idolo Sep 27 '17 at 14:13
  • 2
    $\begingroup$ $a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$ is just the expansion of the sum which gives you the $i,j$ entry, using the definition of matrix multiplication. Note that when $i=j$, $a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$ is just a Laplace expansion of the determinant. $\endgroup$ – Omnomnomnom Sep 27 '17 at 15:08

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