Let A=pmatrixa_1_2pmatrix and B=pmatrixb_1_2pmatrix be two 21 matrices with real entries s…

Question

Let $A=\begin{pmatrix}a_1\a_2\end{pmatrix}$ and $B=\begin{pmatrix}b_1\b_2\end{pmatrix}$ be two $2	imes1$ matrices with real entries such that $A=XB$, where $X=\dfrac1{\sqrt3}\begin{pmatrix}1&-1\1&k\end{pmatrix}$, $k\in\mathbb{R}$. If $a_1^2+a_2^2=\dfrac23(b_1^2+b_2^2)$ and $(k^2+1)b_2^2
e-2b_1b_2$, then the value of $k$ is

Answer: 1

Accepted Answer

$\sqrt3a_1=b_1-b_2$ and $\sqrt3a_2=b_1+kb_2$. Squaring and adding:
$$3(a_1^2+a_2^2)=(b_1-b_2)^2+(b_1+kb_2)^2=2b_1^2+b_2^2(1+k^2)+2b_1b_2(k-1).$$

Given $a_1^2+a_2^2=\dfrac23(b_1^2+b_2^2)$: $2(b_1^2+b_2^2)=2b_1^2+b_2^2(1+k^2)+2b_1b_2(k-1)$.

$$\Rightarrow b_2^2(k^2-1)+2b_1b_2(k-1)=0 \Rightarrow (k-1)\big[b_2^2(k+1)+2b_1b_2\big]=0.$$

Since $(k^2+1)b_2^2\ne-2b_1b_2$ excludes the second factor being zero (given as the non-degenerate case), we need $k=1$.

Solutions · 1