Matematika

Pertanyaan

Jika [tex]\left(\begin{array}{ccc}a-b& &-b\\ & & \\0& &1\end{array}\right) ^{-1} = \left(\begin{array}{ccc}a& &1\\ & & \\-a+2b& &1\end{array}\right) [/tex],nilai ab = ...

a.2
b.1
c.-3/5
d.-1
e.-2

1 Jawaban

  • [tex] \frac{1}{a-b} \left[\begin{array}{ccc}1&b\\0&a-b\end{array}\right] = \left[\begin{array}{ccc}a&1\\-a+2b&1\end{array}\right] \\ \left[\begin{array}{ccc}\frac{1}{a-b}&\frac{b}{a-b}\\0&1\end{array}\right] = \left[\begin{array}{ccc}a&1\\-a+2b&1\end{array}\right] [/tex]

    *
    0 = -a + 2b
    -2b = -a
    a = 2b

    *
    1/(a-b) = a
    1/(2b-b) = 2b
    1/b = 2b
    1 = 2b.b
    1 = 2b²
    b² = 1/2
    b = √1/2

    *
    0 = -a + 2b
    0 = -a + 2√1/2
    -2√1/2 = -a
    a = 2√1/2

    *
    ab
    = 2√1/2 . √1/2
    = 2. 1/2
    = 1

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