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Determinants

Long Answer Type

261.

Solve the following system of equations by matrix method:
2x – 3y + 5z =11
3 x + 2y – 4 z = – 5
x + y – 2 z = –3

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262.
Use matrix method to solve the equations:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2

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

$or space space space space AX space equals space straight B space space space space where space space 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 negative 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 comma 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 3 row 2 row 2 end table close square brackets space space space space space 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 1 end cell row 2 cell space 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 vertical bar space equals space 1 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 3 end cell end table close vertical bar minus 1 open vertical bar table row 2 cell space space space space 1 end cell row 1 cell space space space 3 end cell end table close vertical bar plus 1 open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space space minus 2 end cell end table close vertical bar space space space space space space space space space space space space equals 1 left parenthesis negative 3 plus 2 right parenthesis space minus space 1 left parenthesis 6 minus 1 right parenthesis plus 1 left parenthesis negative 4 plus 1 right parenthesis space equals space minus 1 minus 5 minus 3 space equals space minus 9 space not equal to space 0 therefore space space space space space space straight A to the power of negative 1 end exponent space exists. space space space space space space space space space space space space straight A subscript 11 space equals 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 equals space minus 3 plus 2 space equals space minus 1 space space space space space space space space space space space straight A subscript 12 space equals space minus open vertical bar table row 2 cell space space 1 end cell row 1 cell space space space 3 end cell end table close vertical bar space equals space minus left parenthesis 6 minus 1 right parenthesis space equals space minus 5 space space space space space space space space space space space straight A subscript 13 space equals space 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 equals space minus 4 plus 1 space equals space minus 3$

$straight A subscript 21 space equals space minus open vertical bar table row 1 cell space space space 1 end cell row cell negative 2 end cell cell space space space 3 end cell end table close vertical bar space equals space minus left parenthesis 3 plus 2 right parenthesis space equals space minus 5 straight A subscript 22 space equals space open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space space space 3 end cell end table close vertical bar space equals space 3 minus 1 space equals space 2 straight A subscript 23 space equals space minus open vertical bar table row 1 cell space space space space space space 1 end cell row 1 cell space space minus 2 end cell end table close vertical bar space equals space minus left parenthesis 2 minus 1 right parenthesis space equals space 3 straight A subscript 31 space equals space open vertical bar table row 1 cell space space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar space equals space 1 plus 1 space equals space 2 straight A subscript 32 space equals space minus open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell end table close vertical bar space equals space minus left parenthesis 1 minus 2 right parenthesis space equals space 1 straight A subscript 33 space equals space open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar space equals space minus 1 minus 2 space equals space minus 3$

$therefore space space space adj. space straight A space equals space open square brackets table row cell negative 1 end cell cell space 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 space 3 end cell row cell space 2 end cell cell space space space space 1 end cell cell space space 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 space space space 2 end cell row cell negative 5 end cell cell space 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 space 3 end cell cell space space minus 3 end cell end table close square brackets space space 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 9 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 1 end cell row cell negative 3 end cell cell space space 3 end cell cell negative 3 end cell end table close square brackets Now comma space space space AX space equals space straight B 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 space straight X space equals space straight A to the power of negative 1 end exponent straight B 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 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 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 space space open square brackets table row 3 row 2 row 2 end table close square brackets space space 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 9 open square brackets table row cell negative 3 minus 10 plus 4 end cell row cell negative 15 plus 4 plus 2 end cell row cell negative 9 plus 6 minus 6 end cell end table close square brackets space space space space space space 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 9 open square brackets table row cell negative 9 end cell row cell negative 9 end cell row cell negative 9 end cell end table close square brackets rightwards double arrow space 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 1 row 1 row 1 end table close square brackets therefore space space space space space space straight x space equals space 1 comma space space space space straight y space equals space 1 comma space space space space straight z space equals space 1.$

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263.

Solve (Use matrix method):
x + y = 0
y + z = 1
z + x = 3

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264. Using matrices, solve the following system of linear equations:
2x + y + z = 7
x – y – z = – 4
3x + 2y + z = 10

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265.

Using matrices, following system of linear equations:
x – y + 2 z = 1
2 y – 3z = 1
3x – 2y + 4z = 2

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266.

Use matrix method to solve the following system of equations:
x – y + 2z = 7
3x + 4 y – 5 z = – 5
2x – y + 3z = 12

81 Views

267.

Use matrix method to solve the following system of equations:
x – y + 2z = 7
3x + 4 y – 5 z = – 5
2x – y + 3z = 12

110 Views

268.

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

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269.

Use matrix method to solve the following system of equations:
4x + 2y + 3z = 2
x + y + z = 1
3x + y – 2z = 5

75 Views

270.

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

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Previous Year Papers

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Test Yourself

Long Answer Type

262. Use matrix method to solve the equations:x + y + z = 32x – y + z = 2x – 2y + 3z = 2

262.
Use matrix method to solve the equations:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2