Show that the solution of the equation `[(x, y),(z, t)]^(2)=O` is `[(x,y),(z,t)]=[(pm sqrt(alpha beta),-beta),(alpha,pm sqrt(alpha beta))]` where `alpha, beta` are arbitrary.

Explanation

Given equation is `[(x,y),(z,t)]^(2)=[(0,0),(0,0)]`
`implies [(x,y),(z,t)][(x,y),(z,t)]=[(x^(2)+yz,xy+yt),(zx+tz,zy+t^(2))]=[(0,0),(0,0)]`
`implies x^(2)+yz=0` (1)
`y(x+t)=0` (2)
`z(x+t)=0` (3)
`yz+t^(2)=0` (4)
From (1) and (4), we have `x^(2)=t^(2)` or `x= pm t`
Case I : If `x=t`, then from (2) and (3), we get
`y=0, z=0`
then from (1), x=0=t.
Case II : If x = -t, then (2) and (3) are satified for all values of y and z.
If we take `y=-beta, z=alpha`, then from (1), `x= pm sqrt(alpha beta)=-t`
Obviously, case I is included in case II `(alpha=0=beta)`.
Hence, the general solution of the given equation is
`x=-t= pm sqrt(alpha beta), y=-beta, z=alpha`
`implies [(x,y),(z,t)]=[(pm sqrt(alpha beta),-beta),(alpha, pm sqrt(alpha beta))]`, where `alpha, beta` are arbitrary