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Diketahui matriks A= (−2   31   −1​)., dan B=(5     134    10​). Jika matriks C= A+B, Invers matriks C adalah ...

Pertanyaan

Diketahui matriks A= open parentheses table row cell negative 2 space space space 3 end cell row cell 1 space space space minus 1 end cell end table close parentheses., dan B=open parentheses table row cell 5 space space space space space 13 end cell row cell 4 space space space space 10 end cell end table close parentheses. Jika matriks C= A+B, Invers matriks C adalah ...

  1. negative 1 over 53 open parentheses table row cell 9 space space space space space minus 16 end cell row cell negative 5 space space space space 3 end cell end table close parentheses space

  2. negative 1 over 53 open parentheses table row cell 9 space space space space space 16 end cell row cell negative 5 space space space space 3 end cell end table close parentheses

  3. negative 1 over 53 open parentheses table row cell 9 space space space space space minus 16 end cell row cell 5 space space space space 3 end cell end table close parentheses

  4. negative 1 over 53 open parentheses table row cell 3 space space space space 16 end cell row cell 5 space space space space 9 end cell end table close parentheses

  5. negative 1 over 53 open parentheses table row cell negative 3 space space space space space 16 end cell row cell 5 space space space space minus 9 end cell end table close parentheses

A. Acfreelance

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Pembahasan

J i k a space C equals open parentheses table row cell a space space space b end cell row cell c space space space space d end cell end table close parentheses comma space m a k a  C to the power of negative 1 end exponent equals fraction numerator 1 over denominator open vertical bar C close vertical bar end fraction space A d j space left parenthesis C right parenthesis  A d j left parenthesis C right parenthesis equals space open parentheses table row cell d space space space space minus b end cell row cell negative c space space space space space space a end cell end table close parentheses  open vertical bar C close vertical bar equals space d e t left parenthesis C right parenthesis equals space a d minus b c  M a k a  C equals space A plus space B  C equals open parentheses table row cell negative 2 space space space space space space space 3 end cell row cell 1 space space space space minus 1 end cell end table close parentheses plus open parentheses table row cell 5 space space space space 13 end cell row cell 4 space space space space space space 10 end cell end table close parentheses equals space open parentheses table row cell 3 space space space space space 16 end cell row cell 5 space space space space space space 9 end cell end table close parentheses  A d j left parenthesis C right parenthesis equals open parentheses table row cell d space space space space minus b end cell row cell negative c space space space space a end cell end table close parentheses equals open parentheses table row cell 9 space space space space space space minus 16 end cell row cell negative 5 space space space space space 3 end cell end table close parentheses  open vertical bar C close vertical bar equals space d e t space left parenthesis C right parenthesis space equals a d space minus space b c space equals space 3.9 minus 16 space. space 5 space equals 27 space minus 80  S e h i n g g a space  C to the power of negative 1 end exponent equals fraction numerator 1 over denominator 27 minus 80 end fraction open parentheses table row cell 9 space space space space minus 16 end cell row cell negative 5 space space space space space 3 end cell end table close parentheses equals negative 1 over 53 open parentheses table row cell 9 space space space space space minus 16 end cell row cell negative 5 space space space space space space space space 3 end cell end table close parentheses

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Nilai x yang memenuhi persamaan matriks (14​23​)(−12​3−5​)=(2x−2​3yz​)+(y4​2z−4​) adalah ....

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