Let A=pmatrixx&11&0pmatrix, xR, and A^4=[a_ij].

Question

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

Answer: 10

Accepted Answer

$A^2=\begin{pmatrix}x^2+1&x\x&1\end{pmatrix}$, and
$$A^4=(A^2)^2=\begin{pmatrix}(x^2+1)^2+x^2 & x(x^2+1)+x\ x(x^2+1)+x & x^2+1\end{pmatrix}.$$

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

$a_{22}=x^2+1=9+1=10$.

Solutions · 1