Find ,x, y , a and b if from Mathematics Relations and Function

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 Multiple Choice QuestionsShort Answer Type

321. Construct a 3 x 4 matrix, whose elements are given by :

straight a subscript straight i space straight j end subscript equals 1 half space vertical line space minus 3 space straight i space plus space straight j right square bracket
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322. Construct a 3 x 4 matrix, whose elements are given by :

straight a subscript straight i space straight j end subscript equals 1 half space vertical line space minus 3 space straight i space plus space straight j right square bracket
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323. If space open square brackets table row cell straight x plus 2 straight y space space end cell cell negative straight y end cell row cell 3 straight x end cell 4 end table close square brackets equals open square brackets table row cell negative 4 end cell 3 row 6 4 end table close square brackets find the values of .x and y.
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324. If space open square brackets table row cell straight x plus 2 straight y space space end cell cell 3 straight y end cell row cell 4 straight x end cell 2 end table close square brackets equals open square brackets table row 0 cell space minus 3 end cell row 8 cell space space space 2 end cell end table close square brackets find the values of x and y
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325. If space open square brackets table row cell straight x minus straight x end cell cell 2 straight x plus straight z end cell row cell 2 straight x minus straight y end cell cell 3 straight z plus straight w end cell end table close square brackets equals open square brackets table row cell negative 1 end cell 5 row 0 13 end table close square brackets find x. y. z. w.
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326. Find the values of.x, y and z from the following equations :  open square brackets table row cell straight x plus straight y space space space space end cell 2 row cell 5 plus straight z space space space space end cell cell straight x space straight y end cell end table close square brackets equals space open square brackets table row 6 2 row 5 8 end table close square brackets
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327. Find the values of.x, y and z from the following equations :  open square brackets table row cell x plus y plus z end cell row cell x plus z end cell row cell y plus z end cell end table close square brackets equals open square brackets table row 9 row 5 row 7 end table close square brackets
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328. Find the values of a. b, c and rf from the equation :

open square brackets table row cell straight a minus straight b space space space space space space space space space space end cell cell 2 straight a plus straight c end cell row cell 2 straight a minus straight b space space space space space space space space space end cell cell 3 straight c plus straight d end cell end table close square brackets equals open square brackets table row cell negative 1 end cell 5 row 0 13 end table close square brackets
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329. If space open square brackets table row straight x cell 3 straight x minus straight y end cell row cell 2 straight x plus straight z end cell cell 3 straight y minus straight w end cell end table close square brackets equals open square brackets table row 3 2 row 4 7 end table close square brackets find x, y, x, w'
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330. Find ,x, y , a and b if

open square brackets table row cell 2 straight x minus 3 straight y end cell cell straight a minus straight b end cell 3 row 1 cell straight x plus 4 straight y end cell cell 3 straight a plus 4 straight b end cell end table close square brackets equals open square brackets table row 1 cell negative 2 end cell 3 row 1 6 29 end table close square brackets


We are given that 

open square brackets table row cell 2 straight x minus 3 straight y end cell cell straight a minus straight b end cell 3 row 1 cell straight x plus 4 straight y end cell cell 3 straight a plus 4 straight b end cell end table close square brackets equals open square brackets table row 1 cell negative 2 end cell 3 row 1 6 29 end table close square brackets

From the definition of equality of matrices, we have,
2    x – 3 v = 1    ...(I)
x + 4 y = 6 ...(2)
a – b= – 2 ...(3)
3a + 4 b = 29    ...(4)
(1) and (2) can be writta/i«s
2 x – 3 y – I = 0
x + 4 y – 6 = 0

therefore space space space space space space space space fraction numerator straight x over denominator 18 plus 4 end fraction equals fraction numerator straight y over denominator negative 1 plus 12 end fraction equals fraction numerator 1 over denominator 8 plus 3 end fraction
therefore space space space space space space space space space straight x over 22 equals straight y over 11 equals 1 over 11
therefore space space space space space space space space straight x equals 22 over 11 equals 2 comma space space space straight y space equals 11 over 11 equals 1

(3) and (4) can be written as
a–b+2-0
3 a + 4 b – 29 = 0

therefore space space space space space fraction numerator straight a over denominator 29 minus 8 end fraction equals fraction numerator straight b over denominator 6 plus 29 end fraction equals fraction numerator 1 over denominator 4 plus 3 end fraction space space rightwards double arrow space straight a over 21 equals straight b over 35 equals 1 over 7
therefore space space space space space straight a equals 21 over 7 equals 3 comma space space space space straight b equals 35 over 7 equals 5
∴ solution is x = 2, y= 1, a = 3, b = 5.

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