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Determinants

Question
CBSEENMA12034730
Wired Faculty App

Use matrix method to solve the system of equations:
3x – 2y = 7
5x + 3y = 1

Solution

 The given equations are
3x – 2y = 7
5x + 3y = 1
The equations can be written as
    open square brackets table row 3 cell space space minus 2 end cell row 5 cell space space space space space 3 end cell end table close square brackets space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 7 row 1 end table close square brackets
or         AX space equals space straight B comma space where space straight A space equals space open square brackets table row 3 cell space space space minus 2 end cell row 5 cell space space space space space 3 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y end table close square brackets comma space space space space straight B space equals space open square brackets table row 7 row 1 end table close square brackets space space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space space minus 2 end cell row 5 cell space space space space 3 end cell end table close vertical bar space equals space 9 plus 10 space equals space 19 space not equal to 0 space space space space space space rightwards double arrow space space space straight A to the power of negative 1 end exponent space exists. space space space space space adj. space straight A space equals space open square brackets table row 3 cell space space space minus 5 end cell row 2 cell space space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space space open vertical bar table row cell space space space 3 end cell cell space space space space 2 end cell row cell negative 5 end cell cell space space space space 3 end cell end table close vertical bar space space space space space space space space space space straight A to the power of negative 1 end exponent space space equals fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 19 open square brackets table row 3 cell space space space space 2 end cell row cell negative 5 end cell cell space space space 3 end cell end table close square brackets Now space space space AX space equals space straight B space space space space space rightwards double arrow space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B rightwards double arrow space space space space space open square brackets table row straight x row straight y end table close square brackets space equals space 1 over 19 open square brackets table row 3 cell space space space space 2 end cell row cell negative 5 end cell cell space space space 3 end cell end table close square brackets space open square brackets table row 7 row 1 end table close square brackets space space space space rightwards double arrow space open square brackets table row straight x row straight y end table close square brackets space equals space 1 over 19 open square brackets blank close square brackets rightwards double arrow space space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row cell space space 23 over 19 end cell row cell negative 32 over 19 end cell end table close square brackets space space space space space space space space space space space space space space rightwards double arrow space space space space space space straight x space equals space 23 over 19 comma space space straight y space equals negative 32 over 19 space is space the space solution.

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