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Determinants

Short Answer Type

241.

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

77 Views

242.

Use matrix method to solve the system of equations:
5x - 7y = 2
7x - 5y = 3

93 Views

243.

Use matrix method to solve the system of equations:
x + 2y = 4
2x + 5y = 9

89 Views

244.

Use matrix method to solve the system of equations:
2x + 3y = 5
3x - y = 2

75 Views

245.

Use matrix method to solve the system of equations:
3x + 5y = 8
2x - y = 1

74 Views

246.

Use matrix method to solve the system of equations:
5x - y = 4
3x + 7y = 10

75 Views

Long Answer Type

247. Using matrices, solve the following system of linear equations:
x + 2y – 3z = – 4
2x + 3y + 2z = 2
3x – 3y – 4z = 11

89 Views

Short Answer Type

248. Using matrices, solve the following system of equations:
3x – y + z = 5
2x – 2y + 3z = 7
x + y – z = 1

73 Views

Long Answer Type

249. Using matrices, solve the following system of equations:
2x – y + z = 0
x + y – z = 6
3x – y – 4 z = 7

74 Views

250.
Using matrices, solve the following system of equations:
x + 2y + z =1
2x – y + z = 5
3x + y – z = 0

The given equations are
x + 2y + z = 1
2x – y + z = 5
3x + y – z = 0
These equations can be written as
$open square brackets table row 1 cell space space space space space space space 2 end cell cell space space space space space space 1 end cell row 2 cell space space space minus 1 end cell cell space space space space space space 1 end cell row 3 cell space space space space space 1 end cell cell space space minus 1 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 5 row 0 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 1 cell space space space space space 2 end cell cell space space space space space 1 end cell row 2 cell space minus 1 end cell cell space space space space space 1 end cell row 3 cell space space space 1 end cell cell space space minus 1 end cell end table close square brackets comma 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 5 row 0 end table close square brackets therefore space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space 2 end cell cell space space 1 end cell row 2 cell space minus 1 end cell cell space 1 end cell row 3 cell space space 1 end cell cell space minus 1 end cell end table close vertical bar space equals space 1 space open vertical bar table row cell negative 1 end cell cell space space space space 1 end cell row 1 cell space minus 1 end cell end table close vertical bar space minus space 2 open vertical bar table row 2 cell space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar plus 1 open vertical bar table row 2 cell space space minus 1 end cell row 3 cell space space space space 1 end cell end table close vertical bar space space space space space space space space space space space space space space space space space equals 1 left parenthesis 1 minus 1 right parenthesis minus 2 left parenthesis negative 2 minus 3 right parenthesis plus 1 left parenthesis 2 plus 3 right parenthesis space space space space space space space space space space space space space space space space space space equals space 1 left parenthesis 0 right parenthesis minus 2 left parenthesis negative 5 right parenthesis plus 1 left parenthesis 5 right parenthesis space equals space 0 plus 10 plus 5 space equals space 15 not equal to 0 therefore space 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
$open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space minus 1 end cell row 3 cell space space space space 1 end cell end table close vertical bar$

i.e. 0, 5, 5 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 2 cell space space space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 2 end cell row 3 cell space space 1 end cell end table close vertical bar$

i.e. 3, – 4, 5 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row 2 cell space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space space space 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar$
i.e. 3, 1, – 5 respectively.
$therefore space space space adj space straight A space equals space open square brackets table row 0 cell space space space space 5 end cell cell space space space space space 5 end cell row 3 cell space space minus 4 end cell cell space space space space space 5 end cell row 3 cell space space 1 end cell cell space minus 5 end cell end table close square brackets to the power of apostrophe space equals space space open square brackets table row 0 cell space space space space space space 3 end cell cell space space space space space space 3 end cell row 5 cell space space minus 4 end cell cell space space space space space space 1 end cell row 5 cell space space space space space space 5 end cell cell space space minus 5 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 over 15 open square brackets table row 0 cell space space space space 3 end cell cell space space space 3 end cell row 5 cell space minus 4 end cell cell space space space space 1 end cell row 5 cell space space space space 5 end cell cell space minus 5 end cell end table close square brackets Now space space space AX space equals space straight B 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 open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 15 open square brackets table row 0 cell space space space 3 end cell cell space space space space 3 end cell row 5 cell space minus 4 end cell cell space space space space 1 end cell row 5 cell space space space space 5 end cell cell space minus 5 end cell end table close square brackets space space space space open square brackets table row 1 row 5 row 0 end table close square brackets rightwards double arrow 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 15 open square brackets table row cell 0 plus 15 plus 0 end cell row cell 5 minus 20 plus 0 end cell row cell 5 plus 25 plus 0 end cell end table close square brackets space space space space 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 15 open square brackets table row cell space space space 15 end cell row cell negative 15 end cell row cell space space space space space 30 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 open square brackets table row cell space space space space 1 end cell row cell negative 1 end cell row cell space space space 2 end cell end table close square brackets therefore space space space space space straight x space equals space 1 comma space space space straight y space equals space minus 1 comma space space space straight z space equals space 2 space is space the space required space solution. space$