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Determinants

Long Answer Type

271.

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

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272.

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

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273.

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

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274. Use matrix method to solve the following system of equations:
y + 2z = 4
2z + x = 5
x + 2y = 7

The given equations are
y + 2z = 4 or 0x + y + 2 = z = 4
2z + x = 5 or x + 0y + 2z = 5
x + 2y = 7 or x + 2y + 0z = 7
These equations can be written as
$open square brackets table row 0 cell space space 1 end cell cell space space space 2 end cell row 1 cell space 0 end cell cell space space space 2 end cell row cell space 1 end cell cell space space 2 end cell cell space space 0 end cell end table close square brackets space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 4 row 5 row 7 end table close square brackets$
or $AX space equals space straight B space where space straight A space equals space open square brackets table row 0 cell space space 1 end cell cell space space 2 end cell row 1 cell space 0 end cell cell space space 2 end cell row 1 cell space 2 end cell cell space space 0 end cell end table close square brackets comma space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals open square brackets table row 4 row 5 row 7 end table close square brackets$
$open vertical bar straight A close vertical bar equals space open vertical bar table row 0 cell space space 1 end cell cell space space 2 end cell row 1 cell space 0 end cell cell space 2 end cell row 1 2 cell space 0 end cell end table close vertical bar space equals space 0 open vertical bar table row 0 cell space space 2 end cell row 2 cell space space 0 end cell end table close vertical bar minus 1 open vertical bar table row 1 cell space space 2 end cell row 1 cell space space 0 end cell end table close vertical bar plus 2 open vertical bar table row 1 cell space space space 0 end cell row 1 cell space space space 2 end cell end table close vertical bar space space space space space space equals 0 left parenthesis 0 minus 4 right parenthesis space minus 1 space left parenthesis 0 minus 2 right parenthesis space plus space 2 space left parenthesis 2 minus 0 right parenthesis space space space space space space space equals 0 plus 2 plus 4 space equals space 6$
Co-factors of elements of first row of | A | are
$open vertical bar table row 0 cell space space space 2 end cell row 2 cell space space 0 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 2 end cell row 1 cell space space space 0 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space 0 end cell row 1 cell space space 2 end cell end table close vertical bar$

i.e., – 4, 2, 2 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 1 cell space space 2 end cell row 2 cell space space 0 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space space 2 end cell row 1 cell space space space 0 end cell end table close vertical bar comma space space minus open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space space 2 end cell end table close vertical bar$
i.e., 4, – 2, 1 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row 1 cell space space space 2 end cell row 0 cell space space space 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 0 cell space space space 2 end cell row 1 cell space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 0 cell space space space 1 end cell row 1 cell space space space 0 end cell end table close vertical bar$
i.e., 2, 2, – 1 respectively
$therefore space space space adj. space straight A space equals space open square brackets table row cell negative 4 end cell cell space space space space 2 end cell cell space space space 2 end cell row 4 cell negative 2 end cell cell space space 1 end cell row 2 cell space space 2 end cell cell space space 1 end cell end table close square brackets to the power of apostrophe space equals space space open square brackets table row cell negative 4 end cell cell space space space space space 4 end cell cell space space space space space space 2 end cell row 2 cell space space minus 2 end cell cell space space space space space space 2 end cell row 2 cell space space space space 1 end cell cell space space minus 1 end cell end table close square brackets space space space space space space space 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 6 open square brackets table row cell negative 4 end cell cell space space space space 4 end cell cell space space space space 2 end cell row 2 cell space minus 2 end cell cell space space space 2 end cell row 2 cell space space space 1 end cell cell space minus 1 end cell end table close square brackets Now space space space space AX space equals space straight B 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 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 to the power of apostrophe space equals space 1 over 6 open square brackets table row cell negative 4 end cell cell space space space 4 end cell cell space space space space space 2 end cell row 2 cell negative 2 end cell cell space space space space space space 2 end cell row 2 1 cell space minus 1 end cell end table close square brackets space space space open square brackets table row 4 row 5 row 7 end table close square brackets 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 6 open square brackets table row cell negative 16 plus 20 plus 14 end cell row cell 8 minus 10 plus 14 end cell row cell 8 plus 5 minus 7 end cell end table close square brackets space space space 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 1 over 6 open square brackets table row 8 row 12 row 6 end table close square brackets 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 3 row 2 row 1 end table close square brackets space space space rightwards double arrow space space space space straight x space equals space 3 comma space space space straight y space equals space 2 comma space space space straight z space equals space 1$

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275. Use matrix method to solve the following system of equations:
3x – 4y + 2z = – 1
2x + 3y + 5z = 7
x + z = 2

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

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277. Use matrix method to solve the following system of equations:
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6

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

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

78 Views

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

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Previous Year Papers

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Long Answer Type

274. Use matrix method to solve the following system of equations:y + 2z = 42z + x = 5x + 2y = 7

274. Use matrix method to solve the following system of equations:
y + 2z = 4
2z + x = 5
x + 2y = 7