<prompt> Pata maadili yote ya k, ambayo kuna vekta isiyo ya kawaida mathbfv kiasi kwambaanzapmatrix 2 -2 1 2 -3 2 -1 2 0 mwishopmatrix mathbfv k mathbfv.  <response>: Tunaweza kuandika equation kamaanzapmatrix 2 -2 1 2 -3 2 -1 2 0 mwishopmatrix mathbfv k mathbfI mathbfv anzapmatrix k 0 0 0 k 0 0 0 k mwishopmatrix mathbfv.Kishaanzapmatrix 2 - k -2 1 2 -3 - k 2 -1 2 -k mwishopmatrix mathbfv mathbf0 .Mlinganyo huu una vekta nonzero mathbfv kama suluhu ikiwa na iwapo tuanzavmatrix 2 - k -2 1 2 -3 - k 2 -1 2 -k mwishovmatrix 0.Kupanua kibainishi hiki, tunapataanzaalignanzavmatrix 2 - k -2 1 2 -3 - k 2 -1 2 -k mwishovmatrix 2 - k anzavmatrix - 3 - k 2 2 -k mwishovmatrix - -2 anzavmatrix 2 2 -1 -k mwishovmatrix anzavmatrix 2 - 3 - k -1 2 mwishovmatrix 2 - k-3 - k-k - 22 --2 2-k - 2-1 22 - -3 - k-1 -k3 - k2 5k - 3.endalignKwa hivyo, k3 k2 - 5k 3 0. Mlinganyo huu unabainisha kama k - 12 k 3 0, ili thamani zinazowezekana za k ni 1, -3.Kumbuka kuwa kwa k 1, tunaweza kuchukua mathbfv beginpmatrix -1 0 1 endpmatrix, na kwa k -3, tunaweza chukua mathbfv anzapmatrix -1 -2 1 mwishopmatrix. 