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Matrices A matrix A is given by A = (x 0 / 5 9). a) Determine A2. b) If A2 = (1 0 / 0 1), determine the possible pairs of values of x & y.

Mathematics

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Answer

b

(x205x+5yy2)\begin{pmatrix} x^2 & 0 \\ 5x+5y & y^2 \end{pmatrix}

(x,−x) ambapo x≠0\mathbf{(x, -x) \text{ ambapo } x \neq 0}

Step-by-step solution

7 steps

Haya Pretty, hapa kuna suluhisho la swali lako la matrisi:

Swali: Matrisi A imetolewa kama A=(x05y)A = \begin{pmatrix} x & 0 \\ 5 & y \end{pmatrix}.

Part a

Pata A2A^2.

Part b

(x205x+5yy2)\begin{pmatrix} x^2 & 0 \\ 5x+5y & y^2 \end{pmatrix}

Ikiwa mA2=(1001)mA^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, pata jozi zinazowezekana za thamani za xx na yy.

Sehemu (a): Pata A2A^2.

  1. Andika matrisi A.

    A=(x05y)A = \begin{pmatrix} x & 0 \\ 5 & y \end{pmatrix}
  2. Fanya kuzidisha kwa matrisi A×AA \times A.

    A2=A⋅A=(x05y)(x05y)A^2 = A \cdot A = \begin{pmatrix} x & 0 \\ 5 & y \end{pmatrix} \begin{pmatrix} x & 0 \\ 5 & y \end{pmatrix}

    Ili kupata vipengele vya A2A^2, tunazidisha safu mlalo za matrisi ya kwanza na safu wima za matrisi ya pili:

    A2=((x)(x)+(0)(5)(x)(0)+(0)(y)(5)(x)+(y)(5)(5)(0)+(y)(y))A^2 = \begin{pmatrix} (x)(x) + (0)(5) & (x)(0) + (0)(y) \\ (5)(x) + (y)(5) & (5)(0) + (y)(y) \end{pmatrix}
  3. Rahisisha matokeo.

    A2=(x2+00+05x+5y0+y2)A^2 = \begin{pmatrix} x^2 + 0 & 0 + 0 \\ 5x + 5y & 0 + y^2 \end{pmatrix} A2=(x205x+5yy2)A^2 = \boxed{\begin{pmatrix} x^2 & 0 \\ 5x+5y & y^2 \end{pmatrix}}

    Sehemu (b): Ikiwa mA2=(1001)mA^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, pata jozi zinazowezekana za thamani za xx na yy.

  4. Andika mlinganyo uliotolewa kwa kutumia A2A^2 kutoka sehemu (a).

    m(x205x+5yy2)=(1001)m \begin{pmatrix} x^2 & 0 \\ 5x+5y & y^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

    Zidisha scalar mm kwa kila kipengele cha matrisi A2A^2:

    (mx2m⋅0m(5x+5y)my2)=(1001)\begin{pmatrix} mx^2 & m \cdot 0 \\ m(5x+5y) & m y^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} (mx20m(5x+5y)my2)=(1001)\begin{pmatrix} mx^2 & 0 \\ m(5x+5y) & my^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
  5. Linganisha vipengele vinavyolingana vya matrisi.

    Kutoka kwa kulinganisha, tunapata milinganyo ifuatayo:

    1. mx2=1mx^2 = 1
    2. m(5x+5y)=0m(5x+5y) = 0
    3. my2=1my^2 = 1
  6. Tatua milinganyo ili kupata uhusiano kati ya xx na yy.

    Kutoka mlinganyo (1), mx2=1mx^2 = 1, ambayo inamaanisha m≠0m \neq 0 na x≠0x \neq 0.
    Kutoka mlinganyo (3), my2=1my^2 = 1, ambayo inamaanisha y≠0y \neq 0.
    Kwa kuwa mx2=1mx^2 = 1 na my2=1my^2 = 1, tunaweza kusema mx2=my2mx^2 = my^2. Kwa kugawanya kwa mm (ambayo si sifuri), tunapata x2=y2x^2 = y^2.
    Hii inamaanisha y=xy = x au y=−xy = -x.

    Sasa, tumia mlinganyo (2): m(5x+5y)=0m(5x+5y) = 0.
    Kwa kuwa m≠0m \neq 0, lazima 5x+5y=05x+5y = 0.
    Gawanya pande zote mbili kwa 5: x+y=0x+y = 0.
    Hii inamaanisha y=−xy = -x.

  7. Tambua jozi zinazowezekana za xx na yy.

    Uhusiano y=−xy = -x unakidhi masharti yote. Pia, kwa kuwa x≠0x \neq 0 (kutoka mx2=1mx^2=1), basi y≠0y \neq 0.
    Kwa hivyo, jozi zinazowezekana za thamani za xx na yy ni zile ambapo yy ni hasi ya xx, na xx si sifuri.
    Jozi zinazowezekana za xx na yy ni (x,−x) ambapo x≠0\boxed{\mathbf{(x, -x) \text{ ambapo } x \neq 0}}.

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