Let A=pmatrix0&1&2&0&31&c&0pmatrix, where a,cR.

Question

Let $A=\begin{pmatrix}0&1&2\a&0&3\1&c&0\end{pmatrix}$, where $a,c\in\mathbb{R}$. If $A^3=A$ and the positive value of $a$ belongs to the interval $(n-1,n]$, $n\in\mathbb{N}$, then $n$ is equal to \_\_\_\_\_\_.

Answer: 2

Accepted Answer

Computing $A^3$ and equating to $A$ gives (among other conditions) $a+1+3c=0$ and $2ac+3=0$, i.e. $c=-\dfrac{a+1}3$ and $ac=-\dfrac32$.

Substituting: $a\left(-\dfrac{a+1}3\right)=-\dfrac32 \Rightarrow a^2+a-\dfrac92=0 \Rightarrow 2a^2+2a-9=0$.

Solving: $a=\dfrac{-2\pm\sqrt{4+72}}4=\dfrac{-1\pm\sqrt{19}}2$. The positive root is $a=\dfrac{-1+\sqrt{19}}2\approx1.68$, which lies in $(1,2]$.

So $n=2$.

Solutions · 1