Options are diagonal: diag(1,1,1), diag(1,1,−1), diag(1,−1,−1), diag(−1,−1,−1)
To find: which A is not B2 for any real 3×3 matrix B
A=B2⇒detA=(detB)2≥0
Plan: use det; then verify the rest really are squares
det(diag(d1,d2,d3))=d1d2d3
(1) det=1;(2) det=−1
(3) det=1;(4) det=−1
det=−1<0⇒(2), (4) cannot be squares
I=I2⇒(1) is a square
(3)=1 0
0 R, R²=-I₂
R=0 -1
1 0
R²=-1 0
0 -1=-I₂
B=1 0 0
0 0 -1
0 1 0 ⇒ B²=diag(1,-1,-1)
(2) det=−1<0⇒not a square
(4) −I3, det=−1<0⇒not a square
Not squares: (2) and (4)