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

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

87 Views

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

2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3
These equations can be written as
$open square brackets table row 2 cell space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space 2 end cell cell negative 4 end cell row 1 cell space 1 end cell cell negative 2 end cell end table close square brackets space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space space 16 end cell row cell negative 4 end cell row cell negative 3 end cell end table close square brackets$

or $AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space space 2 end cell cell space space minus 4 end cell row 1 cell space space space space 1 end cell cell space minus 2 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space space straight B space equals space open square brackets table row cell space space 16 end cell row cell negative 4 end cell row cell negative 3 end cell end table close square brackets$

Now AX = B $rightwards double arrow space space space space space straight X space equals space straight A to the power of negative 1 end exponent space straight B$

$therefore space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals open square brackets table row 0 cell space space space 1 end cell cell space space space space minus 2 end cell row cell negative 2 end cell cell space space space 9 end cell cell space space minus 23 end cell row cell negative 1 end cell cell space space space 5 end cell cell space minus 13 end cell end table close square brackets space open square brackets table row cell space space space 16 end cell row cell negative 4 end cell row cell negative 3 end cell end table close square brackets space space space space rightwards double arrow space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell 0 minus 4 plus 6 end cell row cell negative 32 minus 36 plus 69 end cell row cell negative 16 minus 20 plus 39 end cell end table close square brackets space rightwards double arrow space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 2 row 1 row 3 end table close square brackets therefore space space space space space space space space space space straight x space equals space 2 comma space space space space straight y space equals space 1 comma space space space straight z space equals space 3 comma space space space which space is space required space solution. space$

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

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

Test Series

Test Yourself

Long Answer Type

Short Answer Type

293. Use matrix method to solve the following system of equations:2x – 3y + 5z = 163x + 2y – 4z = – 4x + y – 2z = – 3

Long 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