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Determinants

Short Answer Type

241.

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

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242.

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

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243.
Use matrix method to solve the system of equations:
x + 2y = 4
2x + 5y = 9

The given equations are
x + 2y = 4
2x + 5y = 9
These equations can be written as
$open square brackets table row 1 cell space space space 2 end cell row 2 cell space space 5 end cell end table close square brackets space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 4 row 9 end table close square brackets$

or $straight A space straight X space equals space straight B$
where $straight A space equals space open square brackets table row 1 cell space space 2 end cell row 2 cell space space 5 end cell end table close square brackets space straight X space space equals space open square brackets table row straight x row straight y end table close square brackets comma space space space straight B space equals open square brackets table row 4 row 9 end table close square brackets$

$open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space space 5 end cell end table close vertical bar space equals space 5 minus 4 space equals space 1 space not equal to 0 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 straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space open vertical bar straight A close vertical bar over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 1 open square brackets table row 5 cell space space minus 2 end cell row cell negative 2 end cell cell space space space space 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell space space space 5 end cell cell space space space minus 2 end cell row cell negative 2 end cell cell space space space space 1 end cell end table close square brackets Now space space space space AX space equals space straight B space space space rightwards double arrow space space space straight X space equals space straight A to the power of negative 1 end exponent straight B space space space space space space space space space space space space space rightwards double arrow space space 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 space 5 end cell cell space space space space minus 2 end cell row cell negative 2 end cell cell space space space space space space space 1 end cell end table close square brackets space open square brackets table row 4 row 9 end table close square brackets rightwards double arrow space space space space space space 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 2 row 1 end table close square brackets therefore space space space space straight x space equals space 2 comma space space space space straight y space equals space 1 space is space the space required space solution.$