<prompt> HebuboldA anzapmatrix 0 1 2 1 0 1 2 1 0 endpmatrix.Kuna viunganishi p, q , na r vile vileboldA3 p boldA2 q boldA r boldI bold0,ambapo boldI na bold0 ni matrix ya utambulisho ya 3 mara 3 na sifuri mtawalia. Weka mara tatu p,q,r.  <response>: Tunaona hilo boldA2 anzapmatrix 0 1 2 1 0 1 2 1 0 mwishopmatrix anzapmatrix 0 1 2 1 0 1 2 1 0 mwishopmatrix anzapmatrix 5 2 1 2 2 2 1 2 5 mwishopmatrix na boldA3 anzapmatrix 0 1 2 1 0 1 2 1 0 mwishopmatrix anzapmatrix 5 2 1 2 2 2 1 2 5 mwishopmatrix anzapmatrix 4 6 12 6 4 6 12 6 4 mwishopmatrix. Kwa hivyo, tunataka p, q, na r ili kukidhianzapmatrix 4 6 12 6 4 6 12 6 4 mwishopmatrix p anzapmatrix 5 2 1 2 2 2 1 2 5 mwishopmatrix q anzapmatrix 0 1 2 1 0 1 2 1 0 mwishopmatrix r anzapmatrix 1 0 0 0 1 0 0 0 1 mwishopmatrix anzapmatrix 0 0 0 0 0 0 0 0 0 endpmatrix.Upande wa kushoto ni sawa naanzapmatrix 5p r 4 2p q 6 p 2q 12 2p q 6 2p r 4 2p q 6 p 2q 12 2p q 6 5p r 4 endpmatrix.Hii inatupa mfumo wa milinganyoanzaalign5p r -4, 2p q -6, p 2q -12, 2p r -4.endalignKutatua mfumo huu, tunapata p,q,r 0,-6,-4.Kumbuka: Nambari ya aina nyingi x3px2qxr ni sifa bainishi ya polinomia mathbf A. 