<prompt> Tafuta matrix mathbfM, yenye maingizo halisi, kama hayomathbfM3 anzapmatrix 19 30 -45 -71 mwishopmatrix.  <response>: HebumathbfM anzapmatrix a b c d mwishopmatrix.KishaanzaalignmathbfM3 anzapmatrix a b c d mwishopmatrix anzapmatrix a b c d mwishopmatrix anzapmatrix a b c d mwishopmatrix anzapmatrix a2 bc ab bd ac cd bc d2 mwishopmatrix anzapmatrix a b c d mwishopmatrix anzapmatrix a3 2abc bcd a2 b abd bd2 bcd a2 c acd c2 bcd abc 2bcd d3 mwishopmmatrix.endalignTunapata kwa kulinganisha maingizoanzaaligna3 2abc bcd 19, ba2 tangazo d2 bc 30, ca2 tangazo d2 bc -45, abc 2bcd d3 -71.endalignKutoka milinganyo ya pili na ya tatu, fracbc -frac3045 -frac23. Let b 2t na c -3t kwa nambari fulani halisi t.Kuondoa equations ya kwanza na ya nne, tunapataa3 - d3 abc - bcd 90, ambayo vipengele kama a - da2 ad d2 bc 90. Ikilinganisha na mlinganyo b a2 ad d2 bc 30, tunapatafraca - db 3,hivyo a - d 3b 6t.Tunajua det mathbfM3 det mathbfM3 ad - bc3. Lakinidet mathbfM3 det anzapmatrix 19 30 -45 -71 mwishopmatrix 19-71 - 30- 45 1,hivyo ad - bc 1. Kisha ad bc 1 -6t2 1.Kuweka usawa wa a - d 6t, tunapataa2 - 2ad d2 36t2.Kisha a2 ad d2 bc 36t2 3ad bc 36t2 3-6t2 1 -6t2 12t2 3. Kuchomeka kila kitu kwenye mlinganyo ba2 ad d2 bc 30, tunapata2t 12t2 3 30.Kisha t4t2 1 5, hivyo 4t3 t - 5 0. Hii inabainisha kama t - 1 4t2 4t 5 0. Sababu ya quadratic haina mizizi halisi, hivyo t 1, ambayo inaongoza kwa b 2 na c -3.Kisha a - d 6 na ad -5. Kutoka a - d 6, a d 6, hivyo d 6d -5. Kishad2 6d 5 d 1d 5 0,hivyo d -1 au d -5. Ikiwa d -1, basi a 5, lakini thamani hizi hazikidhi a3 2abc bcd 19. Ikiwa d -5, basi a 1. Tunaweza kuangalia kwamba ikiwamathbfM anzapmatrix 1 2 -3 -5 mwishopmatrix,kisha mathbfM3 anzapmatrix 19 30 -45 -71 mwishopmatrix. 