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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

74 Views

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

73 Views

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

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

82 Views

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

73 Views

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

The given equations are
x + y + z = 9
2 x + 5 y + 7 z = 52
2 x + y – z = 0
These equations can be written as
$open square brackets table row 1 cell space space space 1 end cell cell space space space space space 1 end cell row 2 cell space space 5 end cell cell space space space space 7 end cell row 2 cell space space 1 end cell cell space minus 1 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 9 row 52 row 0 end table close square brackets$
or $AX space equals space straight B space space space where space straight A space equals space open square brackets table row 1 cell space space space 1 end cell cell space space 1 end cell row 2 cell space space space 5 end cell cell space space space 7 end cell row 2 cell space space 1 end cell cell negative 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 9 row 52 row 0 end table close square brackets$

$open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 1 end cell cell space space space 1 end cell row 2 cell space space 5 end cell cell space space space 7 end cell row 2 cell space 1 end cell cell negative 1 end cell end table close vertical bar space equals space 1 open vertical bar table row 5 cell space space space space 7 end cell row 1 cell space minus 1 end cell end table close vertical bar space minus space 1 open vertical bar table row 2 cell space space space space 7 end cell row 2 cell space minus 1 end cell end table close vertical bar plus 1 open vertical bar table row 2 cell space space space 5 end cell row 2 cell space space 1 end cell end table close vertical bar space space space space space space space space equals 1 left parenthesis negative 5 minus 7 right parenthesis minus 1 left parenthesis negative 2 minus 14 right parenthesis space plus space 1 left parenthesis 2 minus 19 right parenthesis space equals space minus 12 plus 16 minus 8 space equals space minus 4 not equal to 0 therefore space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 5 cell space space space space space 7 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space space space 7 end cell row 2 cell space minus 1 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space 5 end cell row 2 cell space space space 1 end cell end table close vertical bar straight i. straight e. space space space space minus 12 comma space space 16 comma space space minus 8 space respectively.$
Cofactors of-the elements of 2nd row of | A | are
$negative open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space 1 end cell row 2 cell space minus 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space space 1 end cell row 2 cell space space space space 1 end cell end table close vertical bar$
i.e., 2, – 3, 1 respectively.
Co-factors of the elements of 3rd row of | A | are
$open vertical bar table row 1 cell space space space space 1 end cell row 5 cell space space space 7 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 7 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 5 end cell end table close vertical bar$
i.e., 2, – 5, 3 respectively.
$therefore space space space space adj space straight A space equals space open square brackets table row cell negative 12 end cell cell space space space 16 end cell cell space space space 8 end cell row cell space space 2 end cell cell negative 3 end cell cell space space 1 end cell row cell space space 2 end cell cell negative 5 end cell cell space space 3 end cell end table close square brackets space equals space open square brackets table row cell negative 12 end cell cell space space space 2 end cell cell space space space space 2 end cell row cell space space 16 end cell cell space minus 3 end cell cell space space minus 5 end cell row cell space space minus 8 end cell cell space space 1 end cell cell space space space space 3 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 minus 1 fourth open square brackets table row cell negative 12 end cell cell space space space space space space 2 end cell cell space space space space space space space 2 end cell row cell space space 16 end cell cell space space minus 3 end cell cell space space space minus 5 end cell row cell negative 8 end cell cell space space space space space 1 end cell cell space space space space 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 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 open square brackets table row straight x row straight y row straight z end table close square brackets space equals space minus 1 fourth open square brackets table row cell negative 12 end cell cell space space space space space 2 end cell cell space space space space 2 end cell row 16 cell space minus 3 end cell cell space minus 5 end cell row cell negative 8 end cell cell space space space space 1 end cell cell space space space space 3 end cell end table close square brackets space open square brackets table row 9 row 52 row 0 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 fourth open square brackets table row cell negative 108 plus 104 plus 0 end cell row cell 148 minus 156 plus 0 end cell row cell negative 72 plus 52 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 equals space minus 1 fourth open square brackets table row cell negative 4 end cell row cell negative 12 end cell row cell negative 20 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 3 row 5 end table close square brackets space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight x space equals space 1 comma space space space space space straight y space equals space 3 comma space space space space straight z space equals space 5.$