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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

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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

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

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

$open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space minus 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 1 end cell cell space space space space space space space 2 end cell row 3 cell space minus 2 end cell cell space space minus 2 end cell end table close vertical bar space equals space 2 open vertical bar table row cell space 1 end cell cell space space space space 2 end cell row cell negative 2 end cell cell space space minus 2 end cell end table close vertical bar space minus left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar plus left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar space space space space space equals 2 left parenthesis negative 2 plus 4 right parenthesis space plus 1 left parenthesis negative 2 minus 6 right parenthesis minus 1 left parenthesis negative 2 minus 3 right parenthesis space equals space 4 minus 8 plus 5 space equals space 1 space not equal to space 0 therefore space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 1 cell space space space space space 2 end cell row cell negative 2 end cell cell space space minus 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar$
i.e. 2, 8, – 5 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space space space space minus 1 end cell row cell negative 2 end cell cell space space minus 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space minus 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space minus space open vertical bar table row 2 cell space space space space space space minus 1 end cell row 3 cell space space space space minus 2 end cell end table close vertical bar$

i.e. 0, – 1, 1 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row 1 cell space space space space space 2 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space space space space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space minus 1 end cell row 1 cell space space space space space space 1 end cell end table close vertical bar comma space space straight i. straight e. space space space space minus 1 comma space minus 5 comma space space 3 space respectively$
$therefore space space space space adj space space straight A space equals space open square brackets table row 2 cell space space space space 8 end cell cell space space space minus 5 end cell row 0 cell space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space minus 5 end cell cell space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 2 cell space space space space space 0 end cell cell space space minus 1 end cell row 8 cell space minus 1 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space 1 end cell cell space space space space space 3 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 space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space open square brackets table row 2 cell space space space space 0 end cell cell space space space minus 1 end cell row cell space 8 end cell cell space minus 1 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space space 1 end cell cell space space space space space 3 end cell end table close square brackets Now space space space AX space equals space straight B 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 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 open square brackets table row cell space 2 end cell cell space space space space 0 end cell cell space space space space minus 1 end cell row cell space 8 end cell cell space space minus 1 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space 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 1 row 1 row 1 end table close square brackets 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 open square brackets table row cell 2 plus 0 minus 1 end cell row cell 8 minus 1 minus 5 end cell row cell negative 5 plus 1 plus 3 end cell end table close square brackets 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 open square brackets table row cell space space 1 end cell row cell space space space space 2 end cell row cell negative 1 end cell end table close square brackets. therefore space space space space space solution space is space straight x space equals space 1 comma space space space straight y space equals space 2 comma space space space straight z space equals space 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

73 Views

Previous Year Papers

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Long Answer Type

Short Answer Type

Long Answer Type

294. Use matrix method to solve the following system of equations:2x – y – z = 1x + y + 2z = 13x – 2y – 2z = 1

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