Question

If $A=\left[\begin{array}{cc}a& x\\ y& a\end{array}\right]$ and if $xy=1$ , then $\text{det}\left(A{A}^T\right)$ is equal to :

• A. $a^2-1$
• B. $(a^2+1{)}^2$
• C. $1-a^2$
• D. $(a^2-1{)}^2$
• E. $(a-1{)}^2$

Answer 1

The correct option is D $(a^2-1{)}^2$

Given, $A=\left[\begin{array}{cc}a& x\\ y& a\end{array}\right]$

Thus $A^T=\left[\begin{array}{cc}a& y\\ x& a\end{array}\right]$

Therefore, $A{A}^T=\left[\begin{array}{cc}a& x\\ y& a\end{array}\right]\left[\begin{array}{cc}a& y\\ x& a\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+x^2& ay+ax\\ ay+ax& y^2+a^2\end{array}\right]$

$\therefore \text{det}\left(A{A}^T\right)=\left|\begin{array}{cc}{a}^2+x^2& ay+ax\\ ay+ax& y^2+a^2\end{array}\right|$

$=a^2(y^2+a^2)+x^2(y^2+a^2)-(ay+ax{)}^2$

$=a^2{y}^2+a^4+x^2{y}^2+x^2{a}^2-(a^2{y}^2+2a^{2}xy+a^2{x}^2)$

$=a^4+x^2{y}^2-2a^{2}xy$

$=(a^2-xy{)}^2$

$=(a^2-1{)}^2$ .....Given $xy=1$