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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

The given equations are
3x – y + z = 5
2x – 2y + 3z = 7
x + y – z = 1
These equations can be written as

open square brackets table row 3 cell space space space minus 1 end cell cell space space space space space space 1 end cell row 2 cell space space space space minus 2 end cell cell space space space space space space 3 end cell row 1 cell space space space space 1 end cell cell space space minus 1 end cell end table close square brackets space space space open square brackets table row straight x row straight y row straight z end table close square brackets space space equals space open square brackets table row cell space space 5 end cell row cell space space space 7 end cell row cell negative 1 end cell end table close square brackets or space space space AX space equals straight B space where space straight A space equals space open square brackets table row 3 cell space space minus 1 end cell cell space space space 1 end cell row 2 cell space space space minus 2 end cell cell space space space 3 end cell row cell 1 space end cell cell space space space space 1 end cell cell space space 1 end cell end table close square brackets comma space space straight X space equals open square brackets table row straight x row straight y row straight z end table close square brackets space space straight B space equals space open square brackets table row cell space space 5 end cell row cell space space 7 end cell row cell negative 1 end cell end table close square brackets open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space minus 1 end cell cell space space space 1 end cell row 2 cell space space minus 2 end cell cell space space space space 3 end cell row 1 cell space space space 1 end cell cell space minus 1 end cell end table close vertical bar space equals space 3 space open vertical bar table row cell negative 2 end cell cell space space space 3 end cell row 1 cell space minus 1 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis space open vertical bar table row 2 cell space space space space 3 end cell row 1 cell space minus 1 end cell end table close vertical bar space plus space 1 open vertical bar table row 2 cell space space minus 2 end cell row 1 cell space space space space 1 end cell end table close vertical bar space space space space space space equals 3 left parenthesis 2 minus 3 right parenthesis plus 1 left parenthesis negative 2 minus 3 right parenthesis plus 1 left parenthesis 2 plus 2 right parenthesis space space space space space space space equals 3 left parenthesis negative 1 right parenthesis space plus space 1 left parenthesis negative 5 right parenthesis plus space 1 left parenthesis 4 right parenthesis space equals space minus 3 minus 5 plus 4 space equals space minus 4 not equal to 0 therefore 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 2 end cell cell space space space space space space 3 end cell row 1 cell space space minus 1 end cell end table close vertical bar comma space space minus space open vertical bar table row 2 cell space space space space space 3 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space minus 2 end cell row 1 cell space space space space space 1 end cell end table close vertical bar

i.e. – 1, 5, 4 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 1 end cell row 1 cell negative 1 end cell end table close vertical bar comma space space open vertical bar table row 3 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 3 cell space space space minus 1 end cell row 1 cell space space space minus 1 end cell end table close vertical bar

i.e. 0. – 4, – 4 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 1 end cell row cell negative 2 end cell cell space space space 3 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space space 1 end cell row 2 cell space space space space 3 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space minus 1 end cell row 2 cell space space minus 2 end cell end table close vertical bar

therefore space space space adj space straight A space equals space open square brackets table row cell negative 1 end cell cell space space space space space space 5 end cell cell space space space space space space 4 end cell row cell space space 0 end cell cell space minus 4 end cell cell space space minus 4 end cell row cell negative 1 end cell cell space minus 7 end cell cell space space minus 4 end cell end table close square brackets space equals space open square brackets table row cell negative 1 end cell cell space space space space 0 end cell cell space space space minus 1 end cell row cell space space 5 end cell cell space minus 4 end cell cell space space minus 7 end cell row cell space 4 end cell cell negative 4 end cell cell space space minus 4 end cell end table close square brackets

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 minus 1 fourth open square brackets table row cell negative 1 end cell cell space space space 0 end cell cell space space space minus 1 end cell row cell space space 5 end cell cell space space minus 4 end cell cell space space space minus 7 end cell row 4 cell negative 4 end cell cell space space space minus 4 end cell end table close square brackets

Now,

AX space equals space straight B space space space rightwards double arrow space space space 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 space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space minus 1 fourth open square brackets table row cell negative 1 end cell cell space space space 0 end cell cell space space minus 1 end cell row 5 cell space minus 4 end cell cell space space space minus 7 end cell row 4 cell negative 4 end cell cell space space minus 4 end cell end table close square brackets space space open square brackets table row cell space space 5 end cell row cell space space 7 end cell row cell negative 1 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 minus 1 fourth open square brackets table row cell negative 5 plus 0 plus 1 end cell row cell 25 minus 28 plus 7 end cell row cell 20 minus 28 plus 4 end cell end table close square brackets 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 space minus 1 fourth open square brackets table row cell negative 4 end cell row cell space space 4 end cell row cell space minus 4 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 space 1 space end cell end table close square brackets

therefore space space space space straight x space equals space 1 comma space space space space straight y space equals space minus 1 comma space space space straight z space equals space 1 space is space required space solution space. space

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

74 Views

Previous Year Papers

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Short Answer Type

Long Answer Type

Short Answer Type

248. Using matrices, solve the following system of equations:3x – y + z = 52x – 2y + 3z = 7x + y – z = 1

Long Answer Type

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