Let $A$ and $B$ be two $3 \times 3$ invertible matrices such that $A$ is an idempotent matrix and $$\det(\mathrm{adj}(B))=\det (A)^{12}+\det(A-A^T)^{13}+\det (\mathrm{adj}(A)-I)^{14}.$$ Then find $\det B$.
Note: Here $\det(A)$ denotes determinant of $A$
and $adj(A)$ denotes adjoint of $A$
MY APPROACH:
Since $A$ is idempotent so $A^2=A$ so $\det (A^2)=\det(A)$
$\implies$ $\det(A)=0$ or $1$
But it is given that invertible so $\det (A)=1$.
Now $A^2=A$ $\implies$ $A(A-I)=0$
My doubt: Can i conclude that either $A=0$ or $A=I$ $?$
But I've seen even when both matrices are not zero matrix multiplication can be zero.
Can anybody help me in this question
Best Answer
$\mathrm{adj}(B)B=\det(B)I$, so $\det(\mathrm{adj}(B))\det(B)=\det(B)^3$. Since $\det(\mathrm{adj}(B))=1$ we get $\det(B)=\det(B)^3$, so $\det(B)^2=1$. It follows that $\det(B)=\pm1$.