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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

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263.
Solve (Use matrix method):
x + y = 0
y + z = 1
z + x = 3

The given equations are
x + y + Oz =0
Ox + y + z = 1
x + 0y + z = 3
These equations can be written as
$open square brackets table row 1 cell space space 1 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 1 end cell row 1 cell space space 0 end cell cell space space 1 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 0 row 1 row 3 end table close square brackets$
or $AX space equals space straight B space where space straight A space equals space open square brackets table row 1 cell space space space 1 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 1 end cell row 1 cell space space 0 end cell cell space space 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space thin space space straight B space equals space open square brackets table row 0 row 1 row 3 end table close square brackets$

$open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 1 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space 1 end cell row 1 cell space space 0 end cell cell space 1 end cell end table close vertical bar space equals space 1 open vertical bar table row 1 cell space space 1 end cell row 0 cell space space 1 end cell end table close vertical bar space minus space 1 open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar plus 0 open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space space 0 end cell end table close vertical bar space space space space space space space equals 1 space left parenthesis 1 minus 0 right parenthesis space minus space 1 left parenthesis 0 minus 1 right parenthesis space plus space 0 left parenthesis 0 minus 1 right parenthesis space equals space 1 left parenthesis 1 right parenthesis space minus space 1 left parenthesis negative 1 right parenthesis space plus space 0 left parenthesis negative 1 right parenthesis space space space space space space space equals 1 plus 1 plus 0 space equals space 2 not equal to 0 space space space space therefore space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 1 cell space space space 1 end cell row 0 cell space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space space space 0 end cell end table close vertical bar$
i.e. 1, 1 1 respectively.
Co-factors of the elements of second row of | A | are
$therefore space space space space space space space adj. space straight A space equals space open square brackets table row cell space space 1 end cell cell space space space 1 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space 1 end cell cell space space space space space space space 1 end cell row cell space space 1 end cell cell space minus 1 end cell cell space space minus 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space space space 1 end cell row 1 cell space space space space 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell space space space 1 end cell cell space space space space space 1 end cell end table close square brackets$
i.e. -1, 1, 1 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row 1 cell space space 0 end cell row 1 1 end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space 0 end cell row 0 cell space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space 1 end cell row 0 cell space space 1 end cell end table close vertical bar$
i.e. 1, -1, 1 respectively.

or $AX space equals space straight B space space space where space straight A space equals open square brackets table row 1 cell space space space 1 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 1 end cell row 1 cell space space 0 end cell cell space space 1 end cell end table close square brackets comma space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 0 row 1 row 3 end table close square brackets$
Co-factors of the elements of first row of | A | are

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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

76 Views

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

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

77 Views

Previous Year Papers

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Long Answer Type

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

263.
Solve (Use matrix method):
x + y = 0
y + z = 1
z + x = 3