Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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

The given equations are
2x – y + z = 0
x + y – z = 6
3x – y – 4z = 7
These equations can be written as
$open square brackets table row 2 cell space space minus 1 end cell cell space space space space space space 1 end cell row 1 cell space space space space space space 1 end cell cell space space minus 1 end cell row 3 cell space space space minus 1 end cell cell space space space minus 4 end cell end table close square brackets space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 0 row 6 row 7 end table close square brackets$

or $space space AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space minus 1 end cell cell space space space 1 end cell row 1 cell space space space space space 1 end cell cell space minus 1 end cell row 3 cell space space minus 1 end cell cell space minus 4 end cell end table close square brackets comma space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 0 row 6 row 7 end table close square brackets$

Now $straight A space equals space open square brackets table row 2 cell space space minus 1 end cell cell space space space 1 end cell row 1 cell space space space space 1 end cell cell space minus 1 end cell row 3 cell space minus 1 end cell cell space minus 4 end cell end table close square brackets$

$therefore space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space minus 1 end cell cell space space space space space 1 end cell row 1 cell space space space space space 1 end cell cell space minus 1 end cell row 3 cell space space minus 1 end cell cell space space minus 4 end cell end table close vertical bar space space equals space 2 space open vertical bar table row 1 cell space space minus 1 end cell row cell negative 1 end cell cell space space space minus 4 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space minus 4 end cell end table close vertical bar plus open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar space space space space space space space space space space space space space space space equals 2 left parenthesis negative 4 minus 1 right parenthesis space plus space 1 left parenthesis negative 4 plus 3 right parenthesis space plus space 1 left parenthesis negative 1 minus 3 right parenthesis space space space space space space space space space space space space space space space space space equals 2 left parenthesis negative 5 right parenthesis space plus space 1 left parenthesis negative 1 right parenthesis space plus space 1 left parenthesis negative 4 right parenthesis space equals space minus 10 minus 1 minus 4 space equals space minus 15 not equal to space 0 therefore space space space space straight A to the power of negative 1 end exponent space space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row cell space space 1 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space minus 4 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space minus 4 end cell end table close vertical bar space space comma space space open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar$
i.e. – 5, 1, – 4 respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 1 end cell cell space space space 1 end cell row cell negative 1 end cell cell space minus 4 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space space space space 1 end cell row 3 cell space space minus 4 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space minus 1 end cell row 3 cell space space minus 1 end cell end table close vertical bar$

i.e. – 5, – 11, – 1 respectively.
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 1 end cell cell space space space space 1 end cell row 1 cell negative 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space space 1 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space minus 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar$
i.e. 0, 3, 3 respectively.
$adj space straight A space equals space open square brackets table row cell negative 5 end cell cell space space space 1 end cell cell space space minus 4 end cell row cell negative 5 end cell cell space space minus 11 end cell cell space space space minus 1 end cell row 0 3 cell space 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 cell negative 5 end cell cell space space space minus 5 end cell cell space space space 0 end cell row 1 cell space minus 11 end cell cell space space 3 end cell row cell negative 4 end cell cell space minus 1 end cell cell space space 3 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 space minus 1 over 15 open square brackets table row cell negative 5 end cell cell space space minus 5 end cell cell space space space 0 end cell row cell space space 1 end cell cell space space minus 11 end cell cell space space 3 end cell row cell negative 4 end cell cell space space minus 1 end cell cell space space 3 end cell end table close square brackets Now space space space space space space space AX space equals space straight B space space space 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 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 minus 1 over 15 open square brackets table row cell negative 5 end cell cell space space space minus 5 end cell cell space space space 0 end cell row 1 cell space space minus 11 end cell cell space space 3 end cell row cell negative 4 end cell cell space minus 1 end cell cell space 3 end cell end table close square brackets space open square brackets table row 0 row 6 row 7 end table close square brackets rightwards double arrow 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 minus 1 over 15 open square brackets table row cell 0 minus 30 plus 0 end cell row cell 0 minus 66 plus 21 end cell row cell 0 minus 6 plus 21 end cell end table close square brackets space space space 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 minus 1 over 15 open square brackets table row cell negative 30 end cell row cell negative 45 end cell row 15 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 open square brackets table row cell space space space 2 end cell row cell space space space 3 end cell row cell negative 1 end cell end table close square brackets therefore space space space space straight x space equals space 2 comma space space space straight y space equals space 3 comma space space space straight z space equals space minus 1 space space is space the space solution.$