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Determinants

Long Answer Type

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

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282. Use matrix method to solve the following system of equations:
2x – y - z = 7
3x + y – z = 7
x + y – z = 3

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283. Use matrix method to solve the following system of equations:
x + y + z = 1
x – 2y + 3z = 2
x – 3y + 5z = 3

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284. Use matrix method to solve the following system of equations:
2x + y + 2z = 3
x + y + 2z = 2
2x + 3y – z = –2

The given equations are
2x + y + 2z = 3
x + y + 2z = 2
2x + 3y – z = – 2
These equations can be written as
$open square brackets table row 2 cell space space space space 1 end cell cell space space space space space 2 end cell row 1 cell space space space space 1 end cell cell space space space space space 2 end cell row 2 cell space space space 3 end cell cell space space minus 1 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 open square brackets table row cell space space space 3 end cell row cell space space space space 2 end cell row cell negative 2 end cell end table close square brackets or space space space AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space space space space 1 end cell cell space space space space space 2 end cell row 1 cell space space space space space 1 end cell cell space space space space space 2 end cell row 2 cell space space space space space 3 end cell cell space space minus 1 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 straight B space equals space open square brackets table row cell space space space space 3 end cell row cell space space space space 2 end cell row cell negative 2 end cell end table close square brackets open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space space space 1 end cell cell space space space space space 2 end cell row 1 cell space space space space space 1 end cell cell space space space space space 2 end cell row 2 cell space space space space 3 end cell cell space space minus 1 end cell end table close vertical bar space equals space 2 space open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 1 end cell end table close vertical bar space minus space 1 open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar space plus space 2 open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 3 end cell end table close vertical bar space space space space equals space 2 left parenthesis negative 1 minus 6 right parenthesis space minus space 1 left parenthesis negative 1 minus 4 right parenthesis space plus space 2 left parenthesis 3 minus 2 right parenthesis space equals space minus 14 plus 5 plus 2 space equals space minus 7 space not equal to space 0 therefore space space 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
$open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 1 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space space space space space 3 end cell end table close vertical bar$
i.e. -7, 5, 1 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 1 end cell end table close vertical bar comma space space space$ $open vertical bar table row 2 cell space space space space space space space 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space 1 end cell row 2 cell space space space 3 end cell end table close vertical bar$
i.e. 7, – 6, – 4 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row 1 cell space space space space 1 end cell row 2 cell space space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space 2 end cell row 1 cell space space space 2 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space space 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar$
i.e. 0. – 2, 1 respectively
$therefore space space space space adj space straight A space equals space open square brackets table row cell negative 7 end cell cell space space space space 5 end cell cell space space space space space 1 end cell row 7 cell space minus 6 end cell cell space minus 4 end cell row 0 cell space minus 2 end cell cell space space space space 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 7 end cell cell space space space space 7 end cell cell space space space space space 0 end cell row 5 cell space minus 6 end cell cell space space minus 2 end cell row 1 cell space minus 4 end cell cell space space space space space 1 end cell end table close square brackets$

$straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space minus 1 over 7 open square brackets table row cell negative 7 end cell cell space space space space space space 7 end cell cell space space space space space space 0 end cell row 5 cell space minus 6 end cell cell space space space minus 2 end cell row 1 cell space minus 4 end cell cell space space space space space 1 end cell end table close square brackets$
Now $AX space equals space straight B$
$rightwards double arrow space space space space space straight X equals space straight A to the power of negative 1 end exponent straight B$
$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 minus 1 over 7 open square brackets table row cell negative 7 end cell cell space space space space space 7 end cell cell space space space 0 end cell row 5 cell space space minus 6 end cell cell space minus 2 end cell row 1 cell space space space minus 4 end cell cell space space space space 1 end cell end table close square brackets space open square brackets table row cell space space 3 end cell row cell space space 2 end cell row cell negative 2 end cell end table close square brackets space space rightwards double arrow space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space minus 1 over 7 open square brackets table row cell negative 21 plus 14 plus 0 end cell row cell 15 minus 12 plus 4 end cell row cell 3 minus 8 minus 2 end cell 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 minus 1 over 7 open square brackets table row cell negative 7 end cell row cell space space space 7 end cell row cell negative 7 end cell end table close square brackets space space space space space rightwards double arrow space 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 open square brackets table row cell space space space space 1 end cell row cell negative 1 space end cell row cell space space space 1 end cell end table close square brackets therefore space space space space solution space is space straight x space equals space 1 comma space space space space straight y space equals space minus 1 comma space space space straight z space equals space 1$

75 Views

285. Use matrix method to solve the following system of equations:
x + 2y + z = 7
x + z = 11
2x – 3y = 1

77 Views

286. Use matrix method to solve the following system of equations:
x + 2y – 3z = 4
x + 3 + 2z = 2
3x – 3y + 4z = 11

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287. Use matrix method to solve the following system of equations:
x + y + z = 6
2 x – y + z = 3
x – 2y + 3z = 6

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

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289. Use matrix method to solve the following system of equations:
5x + 3y + z = 16
2x + y + 3 z = 19
x + 2y+ 4 z = 25

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290. Use matrix method to solve the following system of equations:
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0