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Determinants

Long Answer Type

251.

Using matrices, solve the following system of equations:
2x + y – 3r = 13
x + y – z = 6
2x – y + 4z = – 12

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252.

Using matrices, solve the following system of linear equations.
3x + 4y + 2z = 8
2y –3z = 3
x – 2y + 6z = –2

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253. Using matrices, solve the following system of linear equations:
x – y + z = 1
2x – yz = 2
x – 2y – z

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254.
Using matrices, solve the following system of linear equations:
x – y = 3
2x + 3y + 4z = 17
y + 2 z = 7

The given equations are
x – y = 3
2x + 3y + 4z = 17
y + 2z = 7
These equations can be written as
$open square brackets table row 1 cell negative 1 end cell cell space space 0 end cell row 2 cell space space 3 end cell cell space 4 end cell row 0 1 cell space 2 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 3 row 17 row 7 end table close square brackets$

or $AX equals space straight B space where space straight A space equals space open square brackets table row 1 cell space minus 1 end cell cell space space 0 end cell row 2 cell space space 3 end cell 4 row 0 1 cell space 2 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space straight B space equals space open square brackets table row 3 row 17 row 7 end table close square brackets$

$space Now space open vertical bar straight A close vertical bar space equals open vertical bar table row 1 cell space space minus 1 end cell cell space space 0 end cell row 2 cell space space space space 3 end cell cell space 4 end cell row 0 cell space space 1 end cell cell space 2 end cell end table close vertical bar space equals space 1 open vertical bar table row 3 cell space space 4 end cell row 1 cell space 2 end cell end table close vertical bar space minus space left parenthesis negative 1 right parenthesis space open vertical bar table row 2 cell space space 4 end cell row 0 cell space space 2 end cell end table close vertical bar space plus space 0 space open vertical bar table row 2 cell space space 3 end cell row 0 cell space space 1 end cell end table close vertical bar space space space space space space space equals 1 left parenthesis 6 minus 4 right parenthesis space plus space 1 thin space left parenthesis 4 minus 0 right parenthesis space plus space 0 space equals space 2 plus 4 plus 0 space equals space 6 space not equal to 0 therefore space space space straight A to the power of negative 1 end exponent space exists.$

Co-factors of elements of first row of | A | are
$open vertical bar table row 3 cell space space 4 end cell row 1 cell space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space 4 end cell row 0 cell space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space 3 end cell row 0 cell space space space 1 end cell end table close vertical bar$
i.e. 2, – 4, 2 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space 0 end cell row 1 cell space space 2 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space 0 end cell row 0 cell space space space 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space minus 1 end cell row 0 cell space space space space space 1 end cell end table close vertical bar$
i.e. 2, 2, – 1 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space 0 end cell row cell space space 3 end cell cell space space 4 end cell end table close vertical bar comma space space space space space space minus open vertical bar table row 1 cell space space space space 0 end cell row 2 cell space space space space 4 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space minus 1 end cell row 2 cell space space space space space 3 end cell end table close vertical bar$
i.e. – 4, – 4, 5 respectively.
$therefore space space space space adj space straight A space equals space open square brackets table row cell space 2 end cell cell space space space minus 4 end cell cell space space space space space 2 end cell row cell space 2 end cell cell space space space space space space space 2 end cell cell space minus 1 end cell row cell negative 4 end cell cell space space space minus 4 end cell cell space space space space 5 end cell end table close square brackets to the power of apostrophe space space open square brackets table row 2 cell space space space space space 2 end cell cell space space minus 4 end cell row cell negative 4 end cell cell space space space space space 2 end cell cell space space minus 4 end cell row 2 cell space space minus 1 end cell cell space space space space space 5 end cell end table close square brackets Now space AX space equals space straight B 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 therefore 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 1 over 6 open square brackets table row cell space space 2 end cell cell space space space space space space 2 end cell cell space space minus 4 end cell row cell negative 4 end cell cell space space space space space space 2 end cell cell space space minus 4 end cell row cell space space space 2 end cell cell space space minus 1 end cell cell space space space space space space space 5 end cell end table close square brackets space space open square brackets table row 3 row 17 row 7 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 1 over 6 open square brackets table row cell space space 6 plus 34 minus 28 end cell row cell negative 12 plus 34 minus 28 end cell row cell 6 minus 17 plus 35 end cell end table close square brackets space space space space space 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 1 over 6 open square brackets table row cell space 12 end cell row cell negative 6 end cell row 24 end table close square brackets therefore 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 space open square brackets table row cell space space space space 2 end cell row cell negative 1 end cell row cell space space space 4 end cell end table close square brackets therefore space space space space space straight x space equals space 2 comma space space space straight y space equals space minus 1 comma space space space straight z space equals space 4 space is space the space required space solution.$

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255.

Solve by matrix method:
y + 2z = – 8
x + 2y + 3z = – 14
3x + y + z = – 8

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256.

Solve the following system of equations by matrix method:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4

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257.

Solve the following equation by matrix method:
$2 straight x plus straight y plus straight z space equals space 1 straight x minus 2 straight y minus straight z space equals 3 over 2 space space 3 straight y minus 5 straight z space space equals 9$

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258. Solve system of linear equations, using matrix method:
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2

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259.

Solve the following system of equations by matrix method:
x – y + z = 2
2x – y = 0
2y – z = 1

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260.

Solve the following system of equations by matrix method:
2x – y + z = – 3
3 x – z = – 8
2x + 6y = 2