Pertanyaan

Untuk A , B , dan X matriks berordo 2 × 2 . Buktikan bahwa: ( k A ) − 1 = k 1 ⋅ A − 1 , k bilangan bulat positif

Untuk $A$ , $B$ , dan $X$ matriks berordo $2 \times 2$ .

Buktikan bahwa: $(k A)^{- 1} = \frac{1}{k} \cdot A^{- 1}$ , $k$ bilangan bulat positif

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A. Septianingsih

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Pembahasan

Misal space straight A equals open parentheses table row straight a straight b row straight c straight d end table close parentheses

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses kA close parentheses to the power of negative 1 end exponent end cell equals cell 1 over straight k straight A to the power of negative 1 end exponent end cell row cell open square brackets straight k. open parentheses table row straight a straight b row straight c straight d end table close parentheses close square brackets to the power of negative 1 end exponent end cell equals cell 1 over straight k. fraction numerator 1 over denominator ad minus bc end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row cell open square brackets open parentheses table row ak bk row ck dk end table close parentheses close square brackets to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row cell fraction numerator 1 over denominator adk squared minus bck squared end fraction open parentheses table row dk cell negative bk end cell row cell negative ck end cell ak end table close parentheses end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row cell fraction numerator 1 over denominator straight k squared left parenthesis ad minus bc right parenthesis end fraction open parentheses table row dk cell negative bk end cell row cell negative ck end cell ak end table close parentheses end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row cell fraction numerator straight k over denominator straight k squared left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell equals cell fraction numerator 1 over denominator straight k left parenthesis ad minus bc right parenthesis end fraction open parentheses table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close parentheses end cell row blank blank cell left parenthesis terbukti right parenthesis end cell row blank blank blank row blank blank blank end table$