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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

74 Views

263.

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

74 Views

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

84 Views

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

72 Views

269.
Use matrix method to solve the following system of equations:
4x + 2y + 3z = 2
x + y + z = 1
3x + y – 2z = 5

The given equations are
4x + 2y + 3z = 2
x + y + 2 = 1
3x + y – 2z = 5
These equations can be written as
$open square brackets table row 4 cell space space space space 2 end cell cell space space space space space space 3 end cell row 1 cell space space space space 1 end cell cell space space space space space space 1 end cell row 3 cell space space space 1 end cell cell space space minus 2 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 2 row 1 row 5 end table close square brackets$

$or space space space AX space equals space straight B space space space where space straight A space equals space open square brackets table row 4 cell space space space 2 end cell cell space space space space 3 end cell row 1 cell space space 1 end cell cell space space space 1 end cell row 3 cell space space 1 end cell cell space minus 2 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 2 row 1 row 5 end table close square brackets space space space space space space space space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 4 cell space space space space 2 end cell cell space space space space 3 end cell row 1 cell space space space 1 end cell cell space space space space 1 end cell row 3 cell space space space 1 end cell cell space minus 2 end cell end table close vertical bar space equals space 4 left parenthesis negative 2 minus 1 right parenthesis space minus space 2 left parenthesis negative 2 minus 3 right parenthesis space plus space 3 left parenthesis 1 minus 3 right parenthesis space space space space space space space space space space space space space space space space space space space space equals negative 12 plus 10 minus 6 space equals space minus 8 space not equal to 0 space space space space rightwards double arrow space space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 1 cell space space space space space space 1 end cell row 1 cell space space minus 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space 1 end cell row 3 cell space space space space 1 end cell end table close vertical bar$
or – 3, 5, – 2 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 2 cell space space space space space 3 end cell row 1 cell space minus 3 end cell end table close vertical bar comma space space space space open vertical bar table row 4 cell space space space space space space 3 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 4 cell space space space space 2 end cell row 3 cell space space space 1 end cell end table close vertical bar$
or 7, – 17, 2 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row 2 cell space space space 3 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 4 cell space space space 3 end cell row 1 cell space space 1 end cell end table close vertical bar comma space space open vertical bar table row 4 cell space space space 2 end cell row 1 cell space space space 1 end cell end table close vertical bar$
or – 1, – 1, 2 respectively.
$adj space straight A space equals space open square brackets table row cell negative 3 end cell cell space space space space space 5 end cell cell space space space minus 2 end cell row cell space space space 7 end cell cell space space minus 17 end cell cell space space space space space 2 end cell row cell negative 1 end cell cell space space minus 1 end cell cell space space space space space 2 end cell end table close square brackets to the power of apostrophe space equals space space open square brackets table row cell negative 3 end cell cell space space space space space space 7 end cell cell space space space space minus 1 end cell row cell space space space 5 end cell cell space space minus 17 end cell cell space space minus 1 end cell row cell negative 2 end cell cell space space space space 2 end cell cell space space space space space 2 end cell end table close square brackets 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 8 open square brackets table row cell negative 3 end cell cell space space space 7 end cell cell space space minus 1 end cell row cell space space 5 end cell cell space minus 17 end cell cell space space minus 1 end cell row cell negative 2 end cell cell space space 2 end cell cell space space space space space space 2 end cell end table close square brackets Now comma space space space AX space equals space straight B 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 space space space space 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 over 8 open square brackets table row cell negative 3 end cell cell space space space 7 end cell cell space space minus 1 end cell row cell space space 5 end cell cell negative 17 end cell cell space minus 1 end cell row cell negative 2 end cell cell space space 2 end cell cell space space 2 end cell end table close square brackets space open square brackets table row 2 row 1 row 5 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 8 open square brackets table row cell negative 6 plus 7 minus 5 end cell row cell 10 minus 17 minus 5 end cell row cell negative 4 plus 2 plus 10 end cell end table close square brackets 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 open square brackets table row cell 1 half end cell row cell 3 over 2 end cell row cell negative 1 end cell end table close square brackets therefore space space space space space space space straight x space equals space 1 half comma space space space space space straight y space equals space 3 over 2 comma space space space straight z space equals space minus 1$