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Find matrix X given that AX - X = [(-4),(-10)] where A = [(2,3),(-2,4)]

Mathematics

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2 parts
a

5 cm\text{5 cm}

b

3852 cm\frac{\sqrt{385}}{2} \text{ cm}

Step-by-step solution

9 steps

Hapa kuna hatua za kutatua swali la 5:

Part a· 3 steps

Kokotoa thamani ya x inayowakilisha urefu wa kodi NK.

  1. Tumia kanuni ya makutano ya nje ya kodi (Power of a Point Theorem).

    Kwa kodi mbili SM na NK zinazokutana nje kwenye P, bidhaa ya sehemu za kodi moja ni sawa na bidhaa ya sehemu za kodi nyingine.
    Hivyo, PM×PS=PK×PNPM \times PS = PK \times PN.

  2. Badilisha thamani ulizopewa

    Tunajua SM=15 cmSM = 15 \text{ cm}, MP=18 cmMP = 18 \text{ cm}, PK=22 cmPK = 22 \text{ cm}, na NK=x cmNK = x \text{ cm}.
    PS=PM+SM=18+15=33 cmPS = PM + SM = 18 + 15 = 33 \text{ cm}.
    PN=PK+NK=22+x cmPN = PK + NK = 22 + x \text{ cm}.
    Badilisha hizi kwenye kanuni:

    18×33=22×(22+x)18 \times 33 = 22 \times (22 + x)
  3. Tatua kwa x

    594=22(22+x)594 = 22(22 + x)

    Gawanya pande zote mbili kwa 22:

    59422=22+x\frac{594}{22} = 22 + x 27=22+x27 = 22 + x

    Toa 22 kutoka pande zote mbili:

    x=27−22x = 27 - 22 x=5 cmx = 5 \text{ cm}

    Thamani ya x ni 5 cm\boxed{\text{5 cm}}.

Part b

3852 cm\frac{\sqrt{385}}{2} \text{ cm}

Kokotoa rediasi ya duara.

  1. Kwa kuwa SOK ni kipenyo cha duara, pembe yoyote inayoundwa na kipenyo kwenye mzingo wa duara ni 90∘90^\circ.

    Hivyo, ∠SNK=90∘\angle SNK = 90^\circ na ∠SMK=90∘\angle SMK = 90^\circ. Hii inamaanisha △SNK\triangle SNK na △SMK\triangle SMK ni pembetatu za pembe mraba.

  2. Tumia Theorem ya Pythagoras kwenye △SNK\triangle SNK na △SMK\triangle SMK.

    Katika △SNK\triangle SNK: SN2+NK2=SK2SN^2 + NK^2 = SK^2.
    Tunajua NK=5 cmNK = 5 \text{ cm} (kutoka sehemu a)) na SK=2rSK = 2r (kipenyo ni mara mbili ya rediasi).

    SN2+52=(2r)2SN^2 + 5^2 = (2r)^2 SN2+25=4r2(Mlinganyo 1)SN^2 + 25 = 4r^2 \quad \text{(Mlinganyo 1)}

    Katika △SMK\triangle SMK: SM2+MK2=SK2SM^2 + MK^2 = SK^2.
    Tunajua SM=15 cmSM = 15 \text{ cm} na SK=2rSK = 2r.

    152+MK2=(2r)215^2 + MK^2 = (2r)^2 225+MK2=4r2(Mlinganyo 2)225 + MK^2 = 4r^2 \quad \text{(Mlinganyo 2)}
  3. Tumia Kanuni ya Kosine (Law of Cosines) kwenye △PSN\triangle PSN na △PMK\triangle PMK ili kueleza SN2SN^2 na MK2MK^2 kwa kutumia cos⁡(∠P)\cos(\angle P).

    Katika △PSN\triangle PSN: PS=33 cmPS = 33 \text{ cm} na PN=22+5=27 cmPN = 22 + 5 = 27 \text{ cm}.

    SN2=PS2+PN2−2⋅PS⋅PN⋅cos⁡(∠P)SN^2 = PS^2 + PN^2 - 2 \cdot PS \cdot PN \cdot \cos(\angle P) SN2=332+272−2⋅33⋅27⋅cos⁡(∠P)SN^2 = 33^2 + 27^2 - 2 \cdot 33 \cdot 27 \cdot \cos(\angle P) SN2=1089+729−1782cos⁡(∠P)SN^2 = 1089 + 729 - 1782 \cos(\angle P) SN2=1818−1782cos⁡(∠P)(Mlinganyo 3)SN^2 = 1818 - 1782 \cos(\angle P) \quad \text{(Mlinganyo 3)}

    Katika △PMK\triangle PMK: PM=18 cmPM = 18 \text{ cm} na PK=22 cmPK = 22 \text{ cm}.

    MK2=PM2+PK2−2⋅PM⋅PK⋅cos⁡(∠P)MK^2 = PM^2 + PK^2 - 2 \cdot PM \cdot PK \cdot \cos(\angle P) MK2=182+222−2⋅18⋅22⋅cos⁡(∠P)MK^2 = 18^2 + 22^2 - 2 \cdot 18 \cdot 22 \cdot \cos(\angle P) MK2=324+484−792cos⁡(∠P)MK^2 = 324 + 484 - 792 \cos(\angle P) MK2=808−792cos⁡(∠P)(Mlinganyo 4)MK^2 = 808 - 792 \cos(\angle P) \quad \text{(Mlinganyo 4)}
  4. Linganisha Mlinganyo 1 na 3, na Mlinganyo 2 na 4.

    Kutoka Mlinganyo 1 na 3:

    4r2−25=1818−1782cos⁡(∠P)4r^2 - 25 = 1818 - 1782 \cos(\angle P) 4r2=1843−1782cos⁡(∠P)(Mlinganyo 5)4r^2 = 1843 - 1782 \cos(\angle P) \quad \text{(Mlinganyo 5)}

    Kutoka Mlinganyo 2 na 4:

    4r2−225=808−792cos⁡(∠P)4r^2 - 225 = 808 - 792 \cos(\angle P) 4r2=1033−792cos⁡(∠P)(Mlinganyo 6)4r^2 = 1033 - 792 \cos(\angle P) \quad \text{(Mlinganyo 6)}
  5. Tatua mfumo wa milinganyo 5 na 6.

    Linganisha Mlinganyo 5 na 6:

    1843−1782cos⁡(∠P)=1033−792cos⁡(∠P)1843 - 1782 \cos(\angle P) = 1033 - 792 \cos(\angle P)

    Kusanya maneno ya cos⁡(∠P)\cos(\angle P) upande mmoja na namba upande mwingine:

    1843−1033=1782cos⁡(∠P)−792cos⁡(∠P)1843 - 1033 = 1782 \cos(\angle P) - 792 \cos(\angle P) 810=990cos⁡(∠P)810 = 990 \cos(\angle P) cos⁡(∠P)=810990=8199=911\cos(\angle P) = \frac{810}{990} = \frac{81}{99} = \frac{9}{11}

    Sasa badilisha thamani ya cos⁡(∠P)\cos(\angle P) kwenye Mlinganyo 6:

    4r2=1033−792(911)4r^2 = 1033 - 792 \left(\frac{9}{11}\right) 4r2=1033−(72×9)4r^2 = 1033 - (72 \times 9) 4r2=1033−6484r^2 = 1033 - 648 4r2=3854r^2 = 385
  6. Kokotoa r.

    r2=3854r^2 = \frac{385}{4} r=3854r = \sqrt{\frac{385}{4}} r=3852 cmr = \frac{\sqrt{385}}{2} \text{ cm}

    Rediasi ya duara ni 3852 cm\boxed{\frac{\sqrt{385}}{2} \text{ cm}}.

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