Question

Show that $\mathbf{}\mathbf{A}\mathbf{\left(}\mathbf{adjA}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{=}\mathbf{\left(}\mathbf{adjA}\mathbf{\right)}\mathbf{A}\mathbf{}$for all noninvertible $\mathbf{n}\mathbf{\times }\mathbf{n}\mathbf{}$matrices A. See Exercise 42.

Short answer

Therefore,

$\mathrm{A}\left(\mathrm{adjA}\right)=\left(\mathrm{adjA}\right)\mathrm{A}=0$and it is showed.

Step-by-step solution

Matrix Definition. 

Matrix is a set of numbersarranged in rows and columnsso as to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be an “m by n” matrix, written “.$\mathrm{m}\times \mathrm{n}$”

To show that  A(adjA)=0=(adjA)A.

Let,

$B=adjA.$

Then, the entry in the i -th row and j -th column of $\left(A\right(adjA\left)\right)$equals

$\sum _{j=1}^n{a}_{i,j},bji=\sum _{j=1}^n{a}_{i,j}{(-1)}^{i+j}det{A}_{i,j}$

which is, in fact, the Laplace expansion of the determinant along the i -th row, and as such, equals 0 .

Therefore,

$A(adjA)=(adjA)A=0.$