Question

Let $A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right],x\in R$ and $A^4=\left[a_{ij}\right]$. If $a_{11}=109$, then $a_{22}$ is equal to

Answer 1

$A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$

$A^2=A\cdot A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}{x}^2+1& x\\ x& 1\end{array}\right]$

$A^3=A^2\cdot A=\left[\begin{array}{cc}{x}^2+1& x\\ x& 1\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$

$\Rightarrow A^3=\left[\begin{array}{cc}{x}^3+2x& x^2+1\\ x^2+1& x\end{array}\right]$

$A^4=A^3\cdot A=\left[\begin{array}{cc}{x}^3+2x& x^2+1\\ x^2+1& x\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$

$\Rightarrow A^4=\left[\begin{array}{cc}{x}^4+2x^2+x^2+1& x^3+2x\\ x^3+x+x& x^2+1\end{array}\right]$

$\Rightarrow A^4=\left[\begin{array}{cc}{x}^4+3x^2+1& x^3+2x\\ x^3+2x& x^2+1\end{array}\right]$

$a_{11}=109$ (Given)
$\Rightarrow x^4+3x^2+1=109$
$\Rightarrow x^4+3x^2-108=0$
$\Rightarrow (x^2+12)(x^2-9)=0$
$\Rightarrow x=\pm 3$
$\therefore a_{22}=x^2+1=10$