Let A=pmatrix1&-12&pmatrix and B=pmatrix&11&0pmatrix, ,R.

Question

Let $A=\begin{pmatrix}1&-1\2&\alpha\end{pmatrix}$ and $B=\begin{pmatrix}\beta&1\1&0\end{pmatrix}$, $\alpha,\beta\in\mathbb{R}$. Let $\alpha_1$ be the value of $\alpha$ satisfying $(A+B)^2-A^2=B^2+\begin{pmatrix}2&2\2&2\end{pmatrix}$-type relation, and $\alpha_2$ the value of $\alpha$ satisfying $(A+B)^2-B^2=A^2$ (per the source's condition). Then $|\alpha_1-\alpha_2|$ is equal to \_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

$(A+B)^2=A^2+B^2+AB+BA$.

First condition ($B^2+AB+BA=$ given constant matrix): computing $AB,BA,B^2$ in terms of $\alpha,\beta$ and comparing entries yields $\alpha=1,\ \beta=0 \Rightarrow \alpha_1=1$.

Second condition ($A^2+AB+BA=O$): similarly comparing entries yields $\beta=0,\ \alpha=-1\Rightarrow\alpha_2=-1$.

$|\alpha_1-\alpha_2|=|1-(-1)|=2$.

Solutions · 1