<prompt> Acha omega iwe mzizi usio halisi wa x3 1, na uruhusumathbfM anzapmatrix -omega2 - omega 1 0 endpmatrix.Tafuta jumla ya maingizo ya mathbfM mathbfM2 mathbfM3 nukta mathbfM2009.  <response>: Kwa kuwa omega3 1, omega3 - 1 0. Kishaomega - 1omega2 omega 1 0.Tangu omega neq 1, omega2 omega 1 0.Tunakusanya nguvu chache za kwanza za mathbfM:anzaalignmathbfM2 anzapmatrix -omega2 - omega 1 0 mwishopmatrix anzapmatrix -omega2 - omega 1 0 mwishopmatrix anzapmatrix omega4 - omega omega3 -omega2 -omega mwishopmatrix anzapmatrix 0 1 -omega2 -omega mwishopmatrix, mathbfM3 anzapmatrix 0 1 -omega2 -omega mwishopmatrix anzapmatrix -omega2 - omega 1 0 mwishopmatrix anzapmatrix 1 0 omega4 - omega omega3 mwishopmatrix anzapmatrix 1 0 0 1 mwishopmatrix.endalignKwa kuwa mathbfM3 mathbfI, nguvu za mathbfM ni za muda na kipindi cha 3, kwa hivyoanzaalignmathbfM mathbfM2 mathbfM3 vidoti mathbfM2009 670 mathbfM 670 mathbfM 2 669 mathbfM3 670 anzapmatrix -omega2 - omega 1 0 mwishopmatrix 670 anzapmatrix 0 1 -omega2 -omega mwisho pmatrix 669 anzapmatrix 1 0 0 1 mwishopmatrix anzapmatrix -670 omega2 669 -670 omega 670 670 - 670 omega2 -670 omega 669 mwishopmatrix.endalignJumla ya maingizo ni basianzaalign-670 omega2 669 -670 omega 670 670 - 670 omega2 -670 omega 669 -1340 omega2 - 1340 omega 2678 1340 2678 4018.endalignKwa suluhu la haraka zaidi, tunaweza kutambua kuwa jumla ya maingizo katika mathbfM, mathbfM2, na mathbfM3 ni zote ni sawa na 2. Kwa hivyo, jumla tunayotafuta ni 2009 cdot 2 4018. 