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A is matrix (5 3 / 1 7) and I is the 2 x 2 identity matrix. Determine the values of h and k for which / = hI + kA.

Mathematics

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Answer

0.28%\text{0.28\%}

h=0,k=1h=0, k=1

Step-by-step solution

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Haya Chacha, hapa kuna suluhisho la maswali yako:

15.
Ili kukokotoa asilimia ya hitilafu, tunatumia fomula:

Asilimia ya Hitilafu=∣Thamani Iliyopimwa−Thamani Halisi∣Thamani Halisi×100%\text{Asilimia ya Hitilafu} = \frac{|\text{Thamani Iliyopimwa} - \text{Thamani Halisi}|}{\text{Thamani Halisi}} \times 100\%
  1. Tambua thamani halisi na thamani iliyopimwa.

    Thamani Halisi (kipenyo halisi) =5 mm= 5 \text{ mm}
    Thamani Iliyopimwa (kipenyo kilichorekodiwa) =4.986 mm= 4.986 \text{ mm}

  2. Kokotoa tofauti kamili (absolute error).

    Tofauti kamili=∣4.986 mm−5 mm∣\text{Tofauti kamili} = |4.986 \text{ mm} - 5 \text{ mm}| Tofauti kamili=∣−0.014 mm∣\text{Tofauti kamili} = |-0.014 \text{ mm}| Tofauti kamili=0.014 mm\text{Tofauti kamili} = 0.014 \text{ mm}
  3. Kokotoa asilimia ya hitilafu.

    Asilimia ya Hitilafu=0.014 mm5 mm×100%\text{Asilimia ya Hitilafu} = \frac{0.014 \text{ mm}}{5 \text{ mm}} \times 100\% Asilafu ya Hitilafu=0.0028×100%\text{Asilafu ya Hitilafu} = 0.0028 \times 100\% Asilimia ya Hitilafu=0.28%\text{Asilimia ya Hitilafu} = 0.28\%

    Asilimia ya hitilafu katika kipimo hiki ni \boxed{\text{0.28%}}.

    16.
    Tunapewa matrix A=(5317)A = \begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} na II ni matrix ya utambulisho ya 2×22 \times 2, ambayo ni I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. Tunahitaji kupata thamani za hh na kk kwa mlinganyo A=hI+kAA = hI + kA.

  4. Badilisha matrices AA na II kwenye mlinganyo.

    (5317)=h(1001)+k(5317)\begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} = h \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + k \begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix}
  5. Fanya kuzidisha kwa scalar kwa hIhI na kAkA.

    (5317)=(h⋅1h⋅0h⋅0h⋅1)+(k⋅5k⋅3k⋅1k⋅7)\begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} = \begin{pmatrix} h \cdot 1 & h \cdot 0 \\ h \cdot 0 & h \cdot 1 \end{pmatrix} + \begin{pmatrix} k \cdot 5 & k \cdot 3 \\ k \cdot 1 & k \cdot 7 \end{pmatrix} (5317)=(h00h)+(5k3kk7k)\begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} = \begin{pmatrix} h & 0 \\ 0 & h \end{pmatrix} + \begin{pmatrix} 5k & 3k \\ k & 7k \end{pmatrix}
  6. Ongeza matrices upande wa kulia.

    (5317)=(h+5k0+3k0+kh+7k)\begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} = \begin{pmatrix} h + 5k & 0 + 3k \\ 0 + k & h + 7k \end{pmatrix} (5317)=(h+5k3kkh+7k)\begin{pmatrix} 5 & 3 \\ 1 & 7 \end{pmatrix} = \begin{pmatrix} h + 5k & 3k \\ k & h + 7k \end{pmatrix}
  7. Linganisha vipengele vinavyolingana vya matrices ili kuunda milinganyo.

    Kutoka kwa kipengele cha juu kulia:

    3=3k3 = 3k

    Kutoka kwa kipengele cha chini kushoto:

    1=k1 = k

    Kutoka kwa milinganyo hii miwili, tunaweza kuona kwamba k=1k=1.

  8. Badilisha k=1k=1 kwenye mlinganyo wowote unaohusisha hh na kk.

    Tutatumia kipengele cha juu kushoto:

    5=h+5k5 = h + 5k 5=h+5(1)5 = h + 5(1) 5=h+55 = h + 5 h=5−5h = 5 - 5 h=0h = 0
  9. Thibitisha kwa kutumia kipengele cha chini kulia

    7=h+7k7 = h + 7k 7=0+7(1)7 = 0 + 7(1) 7=77 = 7

    Hii inathibitisha thamani za hh na kk.

    Thamani za hh na kk ni h=0,k=1\boxed{h=0, k=1}.

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