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Determinants

Long Answer Type

291. Use matrix method to solve the following system of equations:
x + y + z = 4
2 x – y + z = - 1
2 x + y – 3 z = – 9

87 Views

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

75 Views

Short Answer Type

293. Use matrix method to solve the following system of equations:
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3

81 Views

Long Answer Type

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

75 Views

295. Use matrix method to solve the following system of equations:
2x + 6y = 2
3x – z = 8
2x - y + z = – 3

82 Views

296. Use matrix method to solve the following system of equations:
x + y + z = 6
x – y + z = 2
2x + y – z = 1

88 Views

297. Use matrix method to solve the following system of equations:
9x – 5y - 11z = 12
x – 3y + z = 1
2x + 3y – 7z = 2

74 Views

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

74 Views

299. Use matrix method to solve the following system of equations:
6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8

77 Views

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

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

$or space space AX space equals space straight B space space space space where space space straight A space equals space open square brackets table row cell space space space 2 end cell cell space space space minus 1 end cell cell space space space space space space space 1 end cell row cell negative 1 end cell cell space space space space space space 2 end cell cell space space minus 1 end cell row cell space space 1 end cell cell space minus 1 end cell cell space space space space 2 end cell end table close square brackets comma space space 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 space open square brackets table row cell space space space space 3 end cell row cell negative 4 end cell row cell space space space 1 end cell 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 cell space space space 2 end cell cell space space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell cell space minus 1 end cell row cell space space 1 end cell cell space space space minus 1 end cell cell space space space space space 2 end cell end table close vertical bar space equals space 2 space open vertical bar table row 2 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space space 2 end cell end table close vertical bar space minus left parenthesis negative 1 right parenthesis space open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row cell space 1 end cell cell space space space space space space 2 end cell end table close vertical bar space plus space 1 space open vertical bar table row cell negative 1 end cell cell space space space space space space space 2 end cell row 1 cell space space space space minus 1 end cell end table close vertical bar space space space space space space space space space space space space equals space 2 left parenthesis 4 minus 1 right parenthesis plus 1 left parenthesis negative 2 plus 1 right parenthesis plus 1 left parenthesis 1 minus 2 right parenthesis space equals space 2 left parenthesis 3 right parenthesis plus 1 left parenthesis negative 1 right parenthesis plus 1 left parenthesis negative 1 right parenthesis space space space space space space space space space space space space equals 6 minus 1 minus 1 space equals space 4 space not equal to space 0 therefore space space space space space space space straight A to the power of negative 1 end exponent space space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row cell space space 2 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close vertical bar comma space space space space minus open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row 1 cell space space space space space 2 end cell end table close vertical bar comma space space space open vertical bar table row cell negative 1 end cell cell space space space space space space space 2 end cell row 1 cell space space space minus 1 end cell end table close vertical bar space space space or space space 4 minus 1 comma space space minus left parenthesis negative 2 plus 1 right parenthesis comma space space 1 minus 2$

i.e. 3, 1, – 1 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space 2 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space 1 end cell row 1 cell space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space space minus 1 end cell end table close vertical bar space space or space space space minus left parenthesis negative 2 plus 1 right parenthesis comma space space 4 minus 1 comma space space space minus left parenthesis negative 2 plus 1 right parenthesis$
i.e. 1, 3, 1 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row cell space space space 2 end cell cell space space space space space space space 1 end cell row cell negative 1 end cell cell space space minus 1 end cell end table close vertical bar comma space space space space space open vertical bar table row cell space space 2 end cell cell space space space space minus 1 end cell row cell negative 1 end cell cell space space space space space space 2 end cell end table close vertical bar space space space space or space space space 1 minus 2 comma space space space minus left parenthesis negative 2 plus 1 right parenthesis comma space space 4 minus 1 straight i. straight e. comma space space space minus 1 comma space space 1 comma space space 3 space respectively.$
$therefore space space space space space space space adj space straight A space equals space open square brackets table row 3 cell space space space space 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 3 end cell cell space space space space space space 1 end cell row cell negative 1 end cell cell space space space space 1 end cell cell space space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space space open square brackets table row 3 cell space space space 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell cell space space 3 end cell end table close square brackets therefore 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 fourth open square brackets table row 3 cell space space space 1 end cell cell space space space minus 1 end cell row 1 cell space space space 3 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell cell space space space space space 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 rightwards double arrow space space space space space straight X 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 fourth open square brackets table row 3 cell space space space 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 3 end cell cell space space space space space space 1 end cell row cell negative 1 end cell cell space space space space 1 end cell cell space space space space space space 3 end cell end table close square brackets space open square brackets table row cell space space space space 3 end cell row cell negative 4 end cell row cell space space 1 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 straight z end table close square brackets space equals 1 fourth open square brackets table row cell 9 minus 4 minus 1 end cell row cell 3 minus 12 plus 1 end cell row cell negative 3 minus 4 plus 3 end cell 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 1 fourth open square brackets table row cell space space space 4 end cell row cell negative 8 end cell row cell negative 4 end cell end table close square brackets space space space space space space rightwards double arrow space space 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 space space space 1 end cell row cell negative 2 end cell row cell negative 1 end cell end table close square brackets therefore space space space space space straight x space equals space 1 comma space space space space straight y space equals space minus 2 comma space space space straight z space equals space minus 1$