Question

Even if an $\mathbf{n}\mathbf{\times }\mathbf{n}$ matrix A fails to be invertible, we can define the adjoint $\mathbf{adj}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$as in Theorem 6.3.9. The $\mathbf{}\mathbf{ij}$thentry of $\mathbf{}\mathbf{adj}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$is ${\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{i}\mathbf{+}\mathbf{j}}\mathbf{det}\mathbf{\left(}{\mathbf{A}}_{\mathbf{ji}}\mathbf{\right)}$. For which $\mathbf{n}\mathbf{\times }\mathbf{n}$matrices A is$\mathbf{adj}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{0}$ ? Give your answer in terms of the rank of$\mathbf{}\mathbf{A}$. See Exercise 41.

Short answer

Therefore,

$adjA=0$are zero if and only if$rankA\le n-2.$.

Step-by-step solution

Matrix Definition. 

Matrix is a set of numbersarranged in rows and columns so 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 “$m\times n$.”

To define the adj(A).

We know that,

A matrix A has at least one non-zero minor if and only if$\mathrm{rankA}\ge \mathrm{n}-1.$.

Therefore,

$\mathrm{adjA}=0\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{rankA}\le \mathrm{n}-2.$