A=XB
[a1a2]=31[11−1k][b1b2]
[3a13a2]=[b1−b2b1+kb2]
b1−b2=3a1…(1)
b1+kb2=3a2…(2)
Given, a12+a22=32(b12+b22)
(1)2+(2)2
(b1+b2)2+(b1+kb2)2=3(a12+a22)
a12+a22=32b12+3(1+k2)b22+32b1b2(k−1)
Given, a12+a22=32b12+32b22
On comparing we get
3k2+1=32⇒k2+1=2
⇒k=±1…(3)
32(k−1)=0⇒k=1…(4)
From both we get k=1