Question

If A = $\left[\begin{array}{ccc}i&0\\0&-i\end{array}\right],$ B = $\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]$ and C = $\left[\begin{array}{ccc}0&i\\i&0\end{array}\right]$ and I is the unit matrix of order 2, then show that
i. A² = B² = C² = -I
ii. AB = -BA = -C

Answer 1

i. A² = B² = C² = -I

If
$A=\left[\begin{array}{ccc}i&0\\0&-i\end{array}\right]\\\\for\:\:A^{2}=\left[\begin{array}{ccc}i&0\\0&-i\end{array}\right]\times\left[\begin{array}{ccc}i&0\\0&-i\end{array}\right]\\\\\left[\begin{array}{ccc}i^{2}&0\\0&i^{2}\end{array}\right]$

As i^2= -1

So
$A^{2}=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]$
...eq1

by the same way

$B = \left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]$
$B^{2}=\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]\times\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]\\\\=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]$
...eq2

$C=\left[\begin{array}{ccc}0&i\\i&0\end{array}\right] \\\\C^{2}=\left[\begin{array}{ccc}0&i\\i&0\end{array}\right] \times \left[\begin{array}{ccc}0&i\\i&0\end{array}\right]\\\\=\left[\begin{array}{ccc}i^{2}&0\\0&i^{2}\end{array}\right]$

So
$C^{2}=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]$
...eq3
as we know that
$I=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]\\\\so\\\\-I=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]$
...eq4

from eq1,2,3,4
we can see that A² = B² = C² = -I

ii. AB = -BA = -C

$AB=\left[\begin{array}{ccc}i&0\\0&-i\end{array}\right] \times \left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]\\\\=\left[\begin{array}{ccc}0&-i\\-i&0\end{array}\right]$
...eq5

$-BA= \left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\times \left[\begin{array}{ccc}i&0\\0&-i\end{array}\right]\\\\=\left[\begin{array}{ccc}0&-i\\-i&0\end{array}\right]$
...eq6

from eq5 and eq6
it is clear that AB = -BA = -C