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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

80 Views

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

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

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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

The given equations are
x + y + z = 6
2x – y + z = 3
x – 2y + 3z = 6
These equations can be written as
$open square brackets table row 1 cell space space space space 1 end cell cell space space space 1 end cell row 2 cell space minus 1 end cell cell space space 1 end cell row 1 cell space space minus 2 end cell cell space space 3 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 6 row 3 row 6 end table close square brackets$

where $straight A space equals open square brackets table row 1 cell space space space space 1 end cell cell space space space 1 end cell row 2 cell space minus 1 end cell cell space space space 1 end cell row 1 cell negative 2 end cell cell space space space 3 end cell end table close square brackets comma space space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 6 row 3 row 6 end table close square brackets$

$space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space space space 1 end cell cell space space space space 1 end cell row 2 cell space space minus 1 end cell cell space space space space 1 end cell row 1 cell space space space space minus 2 end cell cell space space space space 3 end cell end table close vertical bar space equals space 1 space open vertical bar table row cell negative 1 end cell cell space space space 1 end cell row cell negative 2 end cell cell space space space 3 end cell end table close vertical bar space minus 1 open vertical bar table row 2 cell space space space space space 1 end cell row 1 cell space space space space space 3 end cell end table close vertical bar space plus space 1 open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space minus 2 end cell end table close vertical bar space space space space space space space equals 1 space left parenthesis negative 3 minus 2 right parenthesis space minus space 1 left parenthesis 6 minus 1 right parenthesis space plus space 1 left parenthesis negative 4 plus 1 right parenthesis space equals space minus 1 minus 5 minus 3 space equals space minus 9 not equal to 0 therefore space space space space straight A to the power of negative 1 end exponent space exists$

Co-factors of the elements of first row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space 1 end cell row cell negative 2 end cell cell space space space 3 end cell end table close vertical bar space space minus space open vertical bar table row 2 cell space space space space 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space minus 1 end cell row 1 cell space space minus 2 end cell end table close vertical bar$
i.e., – 1, – 5, – 3 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 1 cell space space space space 1 end cell row cell negative 2 end cell cell space space 3 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space space space space 1 end cell row 1 cell space space minus 2 end cell end table close vertical bar straight i. straight e. space space space minus 5 comma space 2 comma space 3 space respectively$
Co-factors of the elements of third row of | A | are
$open vertical bar table row 1 cell space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space 1 end cell row 2 cell space space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar$
i.e., 2, 1, -3 respectively
$therefore space space adj space straight A space equals space open square brackets table row cell negative 1 end cell cell space space minus 5 end cell cell space space space minus 3 end cell row cell negative 5 end cell cell space space space space space 2 end cell cell space space space space space space 3 end cell row 2 cell space space space 1 end cell cell space minus 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 1 end cell cell space space minus 5 end cell cell space space space 2 end cell row cell negative 5 end cell cell space space space space 2 end cell cell space space space 1 end cell row cell negative 3 end cell cell space space space space 3 end cell cell negative 3 end cell end table close square brackets therefore 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 minus 1 over 9 open square brackets table row cell negative 1 end cell cell space space minus 5 end cell cell space space space space space space space 2 end cell row cell negative 5 end cell cell space space space space 2 end cell cell space space space space space space 1 end cell row cell negative 3 end cell cell space space space 3 end cell cell space space minus 3 end cell end table close square brackets Now comma space space AX space equals space straight B 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 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 equals space minus 1 over 19 open square brackets table row cell negative 1 end cell cell space space minus 5 end cell cell space space space space space 2 end cell row cell negative 5 end cell cell space space space space space space 2 end cell cell space space space space space 1 end cell row cell negative 3 end cell cell space space space space 3 end cell cell space minus 3 end cell end table close square brackets space open square brackets table row 6 row 3 row 6 end table close square brackets 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 19 open square brackets table row cell negative 6 minus 15 plus 12 end cell row cell negative 30 plus 6 plus 6 end cell row cell negative 18 plus 9 minus 18 end cell end table close square brackets 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 19 open square brackets table row cell negative 9 end cell row cell negative 18 end cell row cell negative 27 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 open square brackets table row 1 row 2 row 3 end table close square brackets space space space space rightwards double arrow space space space space space straight x space equals space 1 comma space space space space straight y space equals space 2 comma space space space straight z space equals space 3$

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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

76 Views

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

137 Views

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