If $A$ is a square matrix such that $A^2=A$,then $(I-A)^3$ is

For a $3 \times 3$ matrix $M$,let $\text{trace}(M)$ denote the sum of all the diagonal elements of $M$. Let $A$ be a $3 \times 3$ matrix such that $|A|=\frac{1}{2}$ and $\text{trace}(A)=3$. If $B=\operatorname{adj}(\operatorname{adj}(2A))$,then the value of $|B|+\text{trace}(B)$ equals:

If $A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and $\alpha, \beta, \gamma$ are the roots of the characteristic equation $|A - xI| = 0$,then $\alpha^2 + \beta^2 + \gamma^2 = $

If ${a^2} + {b^2} + {c^2} = -2$ and $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$,then $f(x)$ is a polynomial of degree:

Let $A$ and $B$ be two $3 \times 3$ non-singular matrices such that $\operatorname{det}(A^T B A) = 27$ and $\operatorname{det}(A B^{-1}) = 8$. Then $\operatorname{det}(B^T A^{-1} B) = $

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2}-B^{2})$ is an invertible matrix. If $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$,then the value of the determinant of the matrix $A^{3}+B^{3}$ is equal to: