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Determinants

Long Answer Type

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

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282. Use matrix method to solve the following system of equations:
2x – y - z = 7
3x + y – z = 7
x + y – z = 3

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283. Use matrix method to solve the following system of equations:
x + y + z = 1
x – 2y + 3z = 2
x – 3y + 5z = 3

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

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285. Use matrix method to solve the following system of equations:
x + 2y + z = 7
x + z = 11
2x – 3y = 1

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286. Use matrix method to solve the following system of equations:
x + 2y – 3z = 4
x + 3 + 2z = 2
3x – 3y + 4z = 11

These equations can be written as

open square brackets table row 1 cell space space space space space 2 end cell cell space space minus 3 end cell row 2 cell space space space space space 3 end cell cell space space space space 2 end cell row 3 cell space space minus 3 end cell cell space minus 4 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 negative 4 end cell row cell space space 2 end cell row 11 end table close square brackets

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 space space space 2 end cell cell space space space minus 3 end cell row 2 cell space space space space space 3 end cell cell space space space space space space 2 end cell row 3 cell space minus 3 end cell cell space space minus 4 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 cell negative 4 end cell row cell space space 2 end cell row cell space 11 end cell 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 space space 2 end cell cell space space minus 3 end cell row 2 cell space space space space space 3 end cell cell space space space space space space 2 end cell row 3 cell space minus 3 end cell cell space minus 4 end cell end table close vertical bar space equals space 1 open vertical bar table row cell space space 3 end cell cell space space space space space space space 2 end cell row cell negative 3 end cell cell space space minus 4 end cell end table close vertical bar minus space 2 open vertical bar table row 2 cell space space space space space space 2 end cell row 3 cell space space space minus 4 end cell end table close vertical bar plus left parenthesis negative 3 right parenthesis space open vertical bar table row 2 cell space space space space space 3 end cell row 3 cell space space space minus 3 end cell end table close vertical bar space space space space space space space equals 1 space left parenthesis negative 12 plus 6 right parenthesis space minus space 2 left parenthesis negative 8 minus 6 right parenthesis space minus space 3 left parenthesis negative 6 minus 9 right parenthesis space space space space space space space space equals negative 6 plus 28 plus 45 equals space 67 not equal to 0

Co-factors of the elements of first row of | A | are

open vertical bar table row cell space space 3 end cell cell space space space space space space space 2 end cell row cell negative 3 end cell cell space space space minus 4 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space space space 2 end cell row 3 cell space space space minus 4 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space 3 end cell row 3 cell space minus 3 end cell end table close vertical bar

i.e. – 6, 14, – 15 respectively
Co-factors of the elements of second row of | A | are

negative open vertical bar table row 2 cell space space space minus 3 end cell row cell negative 3 end cell cell space 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 minus 3 end cell row 3 cell space space minus 4 end cell end table close vertical bar space space minus open vertical bar table row 1 cell space space space space space space space 2 end cell row 3 cell space space minus 3 end cell end table close vertical bar

i.e. 17, 5, 9 respectively.
Co-factors of the elements of third row of | A | are

open vertical bar table row 2 cell space space space minus 3 end cell row 3 cell space space 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 space minus 3 end cell row 2 cell space space space space space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space space 3 end cell end table close vertical bar

i.e. 13, – 8, – 1 respectively

therefore space space space adj space straight A space equals space open square brackets table row cell negative 6 end cell cell space space space 14 end cell cell space minus 15 end cell row 17 cell space space space space 5 end cell cell space space space space 9 end cell row 13 cell space minus 8 end cell cell negative 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 6 end cell cell space space 17 end cell cell space space 13 end cell row cell space space 14 end cell cell space space 5 end cell cell space minus 8 end cell row cell negative 15 end cell cell space 9 end cell cell space minus 1 end cell end table close square brackets therefore 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 1 over 67 open square brackets table row cell negative 6 end cell cell space space 17 end cell cell space space space 13 end cell row 14 cell space space space 5 end cell cell space minus 8 end cell row cell negative 15 end cell cell space 9 end cell cell space minus 1 end cell end table close square brackets Now comma space space space AX space equals space straight B rightwards double arrow space space space space straight X space space equals straight A to the power of negative 1 end exponent straight B 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 over 67 open square brackets table row cell negative 6 end cell cell space space 17 end cell cell space 13 end cell row 14 5 cell negative 8 end cell row cell negative 15 end cell 9 cell negative 1 end cell end table close square brackets space space open square brackets table row cell negative 4 end cell row 2 row 11 end table close square brackets space space space rightwards double arrow 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 67 open square brackets table row cell 24 plus 34 plus 143 end cell row cell negative 56 plus 10 minus 88 end cell row cell 60 plus 18 minus 11 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 over 67 open square brackets table row cell space space space space 201 end cell row cell negative 134 end cell row cell space space space 67 end cell end table close square brackets space 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 open square brackets table row cell space space space 3 end cell row cell negative 2 end cell row cell space space space 1 end cell end table close square brackets therefore space space space space space space space space space solution space is space space space straight x space equals space 3 comma space space space straight y space equals space minus 2 comma space space space straight z space equals space 1 space space

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

73 Views

288. Use matrix method to solve the following system of equations:
3x + 14y + 7z = 14
2x – y + 3z = 4
x + 2y – 3z = 0

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289. Use matrix method to solve the following system of equations:
5x + 3y + z = 16
2x + y + 3 z = 19
x + 2y+ 4 z = 25

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290. Use matrix method to solve the following system of equations:
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0