Question

Let $A_P=\left[\begin{array}{cc}0& p\\ p& 1\end{array}\right]$ and $B_q=\left[\begin{array}{cc}q& 1\\ 0& q\end{array}\right]$ where $p,q\in \{-1,0,1\}$. Then

• A. The sum of all different possible values of $det\left(A_p{B}_q\right)$ is equal to $-1$
• B. The least value of $det(A_p+B_q)-det\left(A_p\right)-det\left(B_q\right)$ is $-2$
• C. The number of different values of $det(A_p+B_q)-det\left(A_p\right)-det\left(B_q\right)$ is $5$.
• D. For $p\ne 0$, the matrix $A_p^{-1}(B_1^T{)}^{2021}{A}_p$ is equal to $\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$

Answer 1

Answer: (A), (B), (C), (D)

For $p\ne 0$, the matrix $A_p^{-1}(B_1^T{)}^{2021}{A}_p$ is equal to $\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$

$det\left(A_p{B}_q\right)=|A_p||B_q|=-p^2{q}^2$
Hence, the only possible value of $det\left(A_p{B}_q\right)$ is $-1$ or $0$

Aliter :
$A_p{B}_q=\left[\begin{array}{cc}0& p\\ p& 1\end{array}\right]\times \left[\begin{array}{cc}q& 1\\ 0& q\end{array}\right]=\left[\begin{array}{cc}0& pq\\ pq& p+q\end{array}\right]$
$\Rightarrow det(A_p\cdot B_q)=-p^2{q}^2$
Hence, the only possible value of $det\left(A_p{B}_q\right)$ is $-1$ or $0$.

$A_p+B_q=\left[\begin{array}{cc}q& p+1\\ p& q+1\end{array}\right]$
$\Rightarrow det(A_p+B_q)=q^2+q-p^2-p$

$\therefore det(A_p+B_q)-det\left(A_p\right)-det\left(A_q\right)$
$=(q^2+q-p^2-p)+p^2-q^2$
$=(q-p)=\{\begin{array}{c}-2\\ -1\\ 0\\ 1\\ 2\end{array}\Rightarrow 5\text{possible values}$

$A_p^{-1}=\frac{1}{|A_p|}adj\left(A_p\right)=\frac{1}{-p^2}\left[\begin{array}{cc}1& -p\\ -p& 0\end{array}\right]$
$\Rightarrow A_p^{-1}=\frac{1}{p^2}\left[\begin{array}{cc}-1& p\\ p& 0\end{array}\right]$

Also, $B_1=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
$\Rightarrow B_1^T=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$
$\therefore (B_1^T{)}^{2021}=\left[\begin{array}{cc}1& 0\\ 2021& 1\end{array}\right]$

Now,
$A_p^{-1}(B_1^T{)}^{2021}{A}_p=\frac{1}{p^2}\left[\begin{array}{cc}-1& p\\ p& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2021& 1\end{array}\right]\left[\begin{array}{cc}0& p\\ p& 1\end{array}\right]$
$=\frac{1}{p^2}\left[\begin{array}{cc}-1& p\\ p& 0\end{array}\right]\left[\begin{array}{cc}0& p\\ p& 2021p+1\end{array}\right]$
$=\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$