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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

The given equations are
3x + 4y + 2z = 8
2y – 3z = 3
x – 2y + 6z = –2
There equations can be written as

$open square brackets table row 3 cell space space space space space 4 end cell cell space space space space space space 2 end cell row 0 cell space space space space space space 2 end cell cell space space minus 3 end cell row 1 cell space space space minus 2 end cell cell space space space space 6 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 cell space space 8 end cell row cell space space 3 end cell row cell negative 2 end cell end table close square brackets or space space space AX space equals space straight B space where space straight A space equals space open square brackets table row 3 cell space space space space 4 end cell cell space space space space 2 end cell row 0 cell space space space space space 2 end cell cell space minus 3 end cell row 1 cell space minus 2 end cell cell space space space space 6 end cell end table close square brackets comma space 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 cell space space space space 8 end cell row cell space space space 3 end cell row cell negative 2 end cell end table close square brackets 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 3 cell space space space 4 end cell cell space space space space 2 end cell row 0 cell space space space 2 end cell cell space minus 3 end cell row 1 cell space minus 2 end cell cell space space space space space 6 end cell end table close vertical bar space equals space 3 space open vertical bar table row 2 cell space space minus 3 end cell row cell negative 2 end cell cell space space space space 6 end cell end table close vertical bar minus space 4 open vertical bar table row 0 cell space space minus 3 end cell row 1 cell space space 6 end cell end table close vertical bar plus 2 space open vertical bar table row 0 cell space space space space space 2 end cell row 1 cell space minus 2 end cell end table close vertical bar space space space space space space space space space space space space space space space space equals space 3 space left parenthesis 12 minus 6 right parenthesis space minus space 4 left parenthesis 0 plus 3 right parenthesis space plus space 2 left parenthesis 0 minus 2 right parenthesis space space space space space space space space space space space space space space space space space equals 3 left parenthesis 6 right parenthesis minus 4 left parenthesis 3 right parenthesis space plus 2 space left parenthesis negative 2 right parenthesis space equals space 18 minus 12 minus 4 space equals space 2 space not equal to 0 therefore 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
$open vertical bar table row cell space space 2 end cell cell space space space minus 3 end cell row cell negative 2 end cell cell space space space space space 6 end cell end table close vertical bar comma space space space minus open vertical bar table row 0 cell space space space minus 3 end cell row 1 cell space space space space space space 6 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space space space 2 end cell row 1 cell space minus 2 end cell end table close vertical bar$
i.e. 6, –3, –2 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell space 4 end cell cell space space space 2 end cell row cell negative 2 end cell cell space space space 6 end cell end table close vertical bar comma space space space open vertical bar table row 3 cell space space space 2 end cell row 1 cell space space space 6 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space space 4 end cell row 1 cell space minus 2 end cell end table close vertical bar$
i.e. – 28, 16, 10 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row 4 cell space space space space space 2 end cell row 2 cell space space minus 3 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space space space 2 end cell row 0 cell space minus 3 end cell end table close vertical bar comma space space space open vertical bar table row 3 cell space space 4 end cell row 0 cell space space 2 end cell end table close vertical bar$
i.e. – 16, 9, 6 respectively.
$therefore space space space space space adj. space straight A space equals space open square brackets table row 6 cell space space space minus 3 end cell cell space space minus 2 end cell row cell negative 28 end cell cell space space space space 16 end cell cell space space space 10 end cell row cell negative 16 end cell cell space space space space 9 end cell cell space space 6 end cell end table close square brackets to the power of 1 space equals space open square brackets table row cell space space 6 end cell cell space space minus 28 end cell cell space space minus 16 end cell row cell negative 3 end cell cell space space space space 16 end cell cell space space space space 9 end cell row cell negative 2 end cell cell space space space space 10 end cell cell space space space 6 end cell end table close square brackets space 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 half open square brackets table row cell space space space space 6 end cell cell space space space minus 28 end cell cell space space minus 16 end cell row cell negative 3 end cell cell space space space space space 16 end cell cell space space space 9 end cell row cell negative 2 end cell cell space space space space 10 end cell cell space space space 6 end cell end table close square brackets Now space space space space space space space space space space space space AX space equals space straight B 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 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 half open square brackets table row cell space space 6 end cell cell space space minus 28 end cell cell space space minus 16 end cell row cell negative 3 end cell cell space space space space 16 end cell cell space space space space 9 end cell row cell negative 2 end cell cell space space space space 10 end cell cell space space space 6 end cell end table close square brackets space space space open square brackets table row cell space space 8 end cell row cell space space space 3 end cell row cell negative 2 end cell end table close square brackets space space space rightwards double arrow space space space space space open square brackets table row straight x row straight y row 2 end table close square brackets space equals space 1 half open square brackets table row cell 48 minus 84 plus 32 end cell row cell negative 24 space plus space 48 minus 18 end cell row cell negative 16 plus 30 minus 12 end cell end table close square brackets rightwards double arrow 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 half open square brackets table row cell negative 4 end cell row cell space space 6 end cell row cell space 2 end cell end table close square brackets 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 open square brackets table row cell negative 2 end cell row cell space space 3 end cell row cell space 1 end cell end table close square brackets therefore space space space space space space straight x space equals space minus 2 comma space space space space straight y space equals space 3 comma space space space straight z space equals space 1 space is space the space required space solution. space$

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

81 Views

254.

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

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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

89 Views

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

86 Views

260.

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