# Matrix expression of the scalar product between two vectors

My question is in regarding the adjugate matrix in the expression my teacher gave me of the scalar product: $$\vec{x}\cdot \vec{y}=x^+ G y$$ being $$x^+=\overline{x}^T$$ the traspose of the conjugate, in the complex plane. I know that in the real plane, $$x^+=x^T$$ so the expression becomes $$x^T Gy$$ as everyone is familiar with. But the thing is that if this is true, why when finding the inverse of a matrix $$A$$ (let's assume now all the elements are real of this matrix) with cofactors the expression is: $$A^{-1}=\frac{1}{\text{det} A} (A^+)^T$$ and until now, in that expression the adjugate was used finding the cofactors as in with each element of it being $$A_{ij}=(-1)^{i+j} \alpha_{ij}$$ and then transposing it, while due to the logic before, since all terms are real it should just be $$x^+=x^T$$? . Have I been conflicting two different terms or how are both things the same?

• Can you please edit your question to include a precise definition of $A^{+}$? Commented Apr 7, 2023 at 10:40
• @ancientmathematician isn't there already a precise definition? I wrote "being $x^+$ the adjugate, or the transpose of the conjugate", also known as $\text{adj }A$ or $A^*$, but my teachers writes it like that Commented Apr 7, 2023 at 10:44
• If all that you mean by $A^{+}$ is the adjugate of $A$ that's nothing at all to do with $x^{+}=\bar{x}^T$. Commented Apr 7, 2023 at 10:51
• @Aley20 Typically (in my experience), the conjugate-transpose of $A$ is called the adjoint of $A$. In contrast, the "adjugate" usually refers to the transpose of the cofactor matrix. The conjugate-transpose of $A$ is typically denoted as $A^*$ if you're a mathematician and as $A^\dagger$ if you're a physicist, which is what I suspect you had in mind when you wrote $A^+$. Commented Apr 7, 2023 at 22:52
• @Aley20 Note that $\operatorname{adj}(A)$ always refers to the matrix that I call the adjugate and $A^*$ always refers to the conjugate-transpose, which is completely unrelated to $\operatorname{adj}(A)$. (Actually, that last statement is a bit of a lie: physicists often use $A^*$ to refer to the conjugate without any transpose, i.e. instead of $\bar A$). Commented Apr 7, 2023 at 22:55