Question

If $\mathbf{}\mathbf{A}$ and $\mathbf{B}$are invertible $\mathbf{}\mathbf{n}\mathbf{\times }\mathbf{n}$matrices, and if $\mathbf{A}$is similar to$\mathbf{B}$, is$\mathbf{adj}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$ necessarily similar to $\mathbf{}\mathbf{adj}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$?

Short answer

Yes, A is similar to B and $adjA$is similar to$adjB$ .

Step-by-step solution

Matrix Definition. 

Matrix is a set of numbersarranged in rows and columns soas 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 “.$\mathrm{m}\times \mathrm{n}$”

If A is B similar to , isadj(A)  necessarily similar to adj(B)

If A is similar to B, then there exists an invertible$n\times n$matrix T

Such that

$$\begin{gathered} {\mathrm{T}}^{-1}\mathrm{AT}=\mathrm{B} \\ \mathrm{adj}\left({\mathrm{T}}^{-1}\mathrm{AT}\right)=\mathrm{adjB}. \end{gathered}$$

Since is similar to , they have the same determinant.

So,

$$\begin{gathered} \left(\frac{1}{detT}detAdetT\right){\left(T^{-1}AT\right)}^{-1}=\left(detB\right)B^{-1} \\ {\left(T^{-1}\right)}^{-1}(adjA)T^{-1}=adjB. \end{gathered}$$

.

Therefore, $adjA$is similar to $adjB.$