<prompt> HebumathbfM anzapmatrix a b c d endpmatrixkuwa matrix yenye maingizo halisi kiasi kwamba mathbfM3 mathbfI . Weka thamani zote zinazowezekana za a d, zikitenganishwa na koma.  <response>: Tunaweza kuhesabu hilomathbfM3 anzapmatrix a3 2abc bcd a2 b abd bd2 b2 c a2 c acd cd2 bc2 abc 2bcd d3 mwishopmatrix. Kwa hivyo, a2 b abd bd2 b2 c ba2 ad d2 bc 0, na a2 c acd cd2 bc2 ca2 ad d2 bc 0.Zaidi ya hayo,det mathbfM3 det mathbfM3 det mathbfI 1,hivyo det mathbfM 1. Kwa maneno mengine, ad - bc 1.Kutoka kwa mlinganyo ba2 ad bd2 bc 0, ama b 0 au a2 ad d2 bc 0. Ikiwa b 0 , basimathbfM3 anzapmatrix a3 0 a2 c acd cd2 d3 endpmatrix.Kwa hiyo, a3 d3 1, hivyo a d 1, na a d 2. Pia, c c c 0, hivyo c 0. Hivyo, mathbf M mathbfI.Vinginevyo, a2 ad d2 bc 0. Tangu ad - bc 1, hii inakuwaa2 ad d2 tangazo - 1 0,ambayo ina maana a d2 1. Ama a d 1 au a d -1 .Kumbuka hiloanzaalignmathbfM2 - a d mathbfM ad - bc mathbfI anzapmatrix a2 bc ab bd ac cd bc d2 mwishopmatrix - a d anzapmatrix a b c d mwishopmatrix tangazo - bc anzapmatrix 1 0 0 1 mwishopmmatrix anzapmatrix 0 0 0 0 endpmatrix mathbf0.endalignIkiwa a d 1, basimathbfM2 - mathbfM mathbfI mathbf0.Kisha mathbfM mathbfImathbfM 2 - mathbfM mathbfI mathbf0. Kupanua, tunapatamathbfM3 - mathbfM2 mathbfM mathbfM2 - mathbfM mathbfI mathbf0, ambayo hurahisisha mathbfM3 -mathbfI. Huu ni mkanganyiko, kwa sababu mathbfM3 mathbfI.Kisha uwezekano pekee uliobaki ni kwamba a d -1. Kumbuka hilomathbfM anzapmatrix 0 -1 1 -1 mwishopmatrixinatosheleza mathbfM3 mathbfI, hivyo - 1 ni thamani inayowezekana ya a d.Kwa hivyo, thamani pekee zinazowezekana za a d ni 2, -1. 