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Determinants

Question
CBSEENMA12034873
Wired Faculty App

Use matrix method to solve the following system of equations:
3x + 14y + 7z = 14
2x – y + 3z = 4
x + 2y – 3z = 0

Solution

The given equations are
3x + 14y + 7z = 14
2x – y + 3z = 4
x + 2y – 3z = 0
These equations can be written as
                       open square brackets table row 3 cell space space space space space space 4 end cell cell space space space space 7 end cell row 2 cell space space minus 1 end cell cell space space space space space 3 end cell row 1 cell space space space space space space 2 end cell cell space minus 3 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space space open square brackets table row 14 row 4 row 0 end table close square brackets
or    AX space equals space straight B space space where space straight A space equals space open square brackets table row 3 cell space space space space space space 4 end cell cell space space space space space 7 end cell row 2 cell space space minus 1 end cell cell space space space space space 3 end cell row 1 cell space space space space space 2 end cell cell space minus 3 end cell end table close square brackets comma space space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 14 row 4 row 0 end table close square brackets
open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space space space 4 end cell cell space space space space 7 end cell row 2 cell space minus 1 end cell cell space space space space 3 end cell row 1 cell space space space space 2 end cell cell negative 3 end cell end table close vertical bar space equals space 3 open vertical bar table row cell negative 1 end cell cell space space space space space space space 3 end cell row cell space 2 end cell cell space space minus 3 end cell end table close vertical bar space minus space 4 open vertical bar table row 2 cell space space space space space space 3 end cell row 1 cell space space minus 3 end cell end table close vertical bar space plus space 7 open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space space space space space 2 end cell end table close vertical bar space space space space space equals 3 space left parenthesis 3 minus 6 right parenthesis space minus space 4 left parenthesis negative 6 minus 3 right parenthesis space plus space 7 left parenthesis 4 plus 1 right parenthesis space equals space minus 9 plus 36 plus 35 space equals space 62 not equal to 0 space space space space rightwards double arrow space space straight A to the power of negative 1 end exponent space exists.
Co-factors of the elements of first row of | A | are
open vertical bar table row cell negative 1 end cell cell space space space space space 3 end cell row 2 cell space space minus 3 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space space space space 3 end cell row 1 cell space space minus 3 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space minus 1 end cell row 1 cell space space space space space space 2 end cell end table close vertical bar
i.e.,  – 3,  9, 5 respectively.
Co-factors of the elements of second row of | A | are
negative open vertical bar table row 4 cell space space space space space space 7 end cell row 2 cell space space minus 3 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space space space 7 end cell row 1 cell space space minus 3 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space space 4 end cell row 1 cell space space space 2 end cell end table close vertical bar
i.e.,   26, – 16, – 2 respectively.
Co-factors of the elements of third row of | A | are
open vertical bar table row cell space 4 end cell cell space space space 7 end cell row cell negative 1 end cell cell space space 3 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space space 7 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space space open vertical bar table row 3 cell space space space space space space 4 end cell row 2 cell space space space minus 1 end cell end table close vertical bar
i.e., 19,  5,  – 11 respectively.
therefore space space space space adj. space straight A space equals space open square brackets table row cell negative 3 end cell cell space 9 end cell cell space space space space 5 end cell row 26 cell negative 16 end cell cell space minus 2 end cell row 19 5 cell space minus 11 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 3 end cell cell space space space 26 end cell cell space space space space 19 end cell row cell space 9 end cell cell space minus 16 end cell cell space space space 5 end cell row cell space 5 end cell cell space minus 2 end cell cell space minus 11 end cell end table close square brackets therefore space space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 62 open square brackets table row cell negative 3 end cell cell space space space space 26 end cell cell space space space space 19 end cell row 9 cell space space minus 16 end cell cell space space space space 5 end cell row 5 cell space minus 2 end cell cell space minus 11 end cell end table close square brackets Now comma space space AX space equals space straight B space space space space space space space rightwards double arrow space 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 space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 62 open square brackets table row cell negative 3 end cell cell space space space 26 end cell cell space space space 19 end cell row 9 cell space space minus 16 end cell cell space space space space 5 end cell row 5 cell space space minus 2 end cell cell space minus 11 end cell end table close square brackets space open square brackets table row 14 row 4 row 0 end table close square brackets rightwards double arrow space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 62 open square brackets table row cell negative 42 plus 104 plus 0 end cell row cell 126 minus 64 plus 0 end cell row cell 70 minus 8 plus 0 end cell end table close square brackets space space space space rightwards double arrow space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space space space equals space 1 over 62 open square brackets table row 62 row 62 row 62 end table close square brackets rightwards double arrow space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 1 row 1 row 1 end table close square brackets
⇒ x = 1, y = 1, z = 1 is the required solution.

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