CT = -C.Also (AB)T=BTATTaking TRANSPOSE of XTCX:XTCX = (XTCX)T = XTCT(XT)T (using (AB)T=BTAT) = XTCTX = XT(-C)XXTCX = - XTCXXTCX + XTCX = 02XTCX = 0XTCX = 0Therefore, XTCX is a NULL matrix.