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Determinants

Long Answer Type

291. Use matrix method to solve the following system of equations:
x + y + z = 4
2 x – y + z = - 1
2 x + y – 3 z = – 9

87 Views

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

75 Views

Short Answer Type

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

81 Views

Long Answer Type

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

75 Views

295. Use matrix method to solve the following system of equations:
2x + 6y = 2
3x – z = 8
2x - y + z = – 3

82 Views

296. Use matrix method to solve the following system of equations:
x + y + z = 6
x – y + z = 2
2x + y – z = 1

The given equations are
x + y + z = 6
x – y + z = 2
2x + y – z = 1
These equations can be written as
   $open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell cell space minus 1 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 6 row 2 row 1 end table close square brackets$
or    $AX space space equals space straight B space where space straight A space equals space open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space space 1 end cell row 2 cell space space space 1 end cell cell space space minus 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 6 row 2 row 1 end table close square brackets$

$open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space space 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space 1 end cell row 2 cell space space space 1 end cell cell space minus 1 end cell end table close vertical bar space equals space 1 open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar space minus 1 open vertical bar table row 1 cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar plus 1 open vertical bar table row 1 cell space space space space minus 1 end cell row 2 cell space space space space space space space 1 end cell end table close vertical bar space space space space space space equals 1 left parenthesis 1 minus 1 right parenthesis minus 1 left parenthesis negative 1 minus 2 right parenthesis plus 1 left parenthesis 1 plus 2 right parenthesis space equals space 1 left parenthesis 0 right parenthesis minus 1 left parenthesis negative 3 right parenthesis plus 1 left parenthesis 3 right parenthesis space space space space space space space equals 0 plus 3 plus 3 space equals space 6 space not equal to space 0 therefore space space space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of  | A | are
$open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row cell space space 1 end cell cell space minus 1 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space minus 1 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar space space or space space space space space 1 comma space space space minus 1 space comma space space minus left parenthesis negative 1 minus 2 right parenthesis comma space space space space 1 plus 2 straight i. straight e.. space space space 0 comma space 3 comma space 3 space respectively.$
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 1 cell space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell end table close vertical bar space space space or space space minus left parenthesis negative 1 minus 1 right parenthesis comma space space minus 1 minus 2 comma space space minus left parenthesis 1 minus 2 right parenthesis$
i.e. 2, – 3, 1 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell space space 1 end cell cell space space 1 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar space space or space space space 1 plus 1 comma space space minus left parenthesis 1 minus 1 right parenthesis comma space space space space minus 1 minus 1$
i.e. 2, 0, – 2 respectively.
$therefore space space adj space straight A space equals space open square brackets table row 0 cell space space space space space 3 end cell cell space space space space 3 end cell row 2 cell space minus 3 end cell cell space space space space 1 end cell row 2 cell space space space space 0 end cell cell space minus 2 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 0 cell space space space space space space 2 end cell cell space space space space space 2 end cell row 3 cell space space minus 3 end cell cell space space space space space 0 end cell row 3 cell space space space space space space 1 end cell cell space minus 2 end cell end table close square brackets therefore 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 6 open square brackets table row 0 cell space space space space space 2 end cell cell space space space space space 2 end cell row 3 cell space minus 3 end cell cell space space space space space 0 end cell row 3 cell space space space space 1 end cell cell space space minus 2 end cell end table close square brackets Now space space space AX space equals space straight B space space space 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 therefore 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 6 open square brackets table row 0 cell space space space space space 2 end cell cell space space space space space space 2 end cell row 3 cell space minus 3 end cell cell space space space space space 0 end cell row 3 cell space space space space 1 end cell cell space space space minus 2 end cell end table close square brackets space open square brackets table row 6 row 2 row 1 end table close square brackets space space 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 space equals space 1 over 6 open square brackets table row cell 0 plus 4 plus 2 end cell row cell 18 minus 6 plus 0 end cell row cell 18 plus 2 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 1 over 6 open square brackets table row 6 row 12 row 18 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 open square brackets table row 1 row 2 row 3 end table close square brackets therefore space space space space space straight x space equals space 1 comma space space space straight y space equals space 2 comma space space space straight z equals space 3.$

88 Views

297. Use matrix method to solve the following system of equations:
9x – 5y - 11z = 12
x – 3y + z = 1
2x + 3y – 7z = 2

74 Views

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

74 Views

299. Use matrix method to solve the following system of equations:
6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8

77 Views

300. Use matrix method to solve the following system of equations:
2x – y + z = 3
– x + 2y – z = – 4
x – y + 2z = 1

73 Views

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Long Answer Type

Short Answer Type

Long Answer Type

296. Use matrix method to solve the following system of equations:x + y + z = 6x – y + z = 22x + y – z = 1

296. Use matrix method to solve the following system of equations:
x + y + z = 6
x – y + z = 2
2x + y – z = 1