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Matrices

Short Answer Type

21. If

open square brackets table row cell straight x plus 2 straight y 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 negative 3 end cell row 8 2 end table close square brackets

find the values of x and y.

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22. If

open square brackets table row cell straight x minus straight y 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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23. Find the values of.x, y and z from the following equations :

open square brackets table row 4 3 row straight x 5 end table close square brackets equals open square brackets table row y z row 1 5 end table close square brackets

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24. Find the values of.x, y and z from the following equations :

open square brackets table row cell straight x plus straight y end cell 2 row cell 5 plus straight z end cell cell straight x space straight y end cell end table close square brackets equals open square brackets table row 6 2 row 5 8 end table close square brackets

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25. Find the values of.x, y and z from the following equations :

open square brackets table row cell straight x plus straight y plus straight z end cell row cell straight x plus straight z end cell row cell straight y plus straight 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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26. Find the values of a. b, c and rf from the equation :

open square brackets table row cell straight s minus straight b end cell cell 2 straight a plus straight c end cell row cell 2 straight a minus straight b 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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27.

If $open square brackets table row straight x cell 3 straight x minus straight y end cell row cell 2 straight x plus straight z end cell cell 3 straight y minus straight w end cell end table close square brackets equals open square brackets table row 3 2 row 4 7 end table close square brackets$ find x, y, z w.

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Long Answer Type

28. 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

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Short Answer Type

29.

If $open square brackets table row cell straight a plus straight b end cell 2 row 5 cell space straight a space straight b end cell end table close square brackets equals open square brackets table row 6 2 row 5 8 end table close square brackets$ find the values of a and b.

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30. Find the values of x and y if

$open square brackets table row cell straight x plus 10 end cell cell straight y squared plus 2 straight y end cell row 0 cell negative 4 end cell end table close square brackets equals open square brackets table row cell 3 x plus 4 end cell 3 row 0 cell y squared minus 5 y end cell end table close square brackets$

We are given that

$open square brackets table row cell straight x plus 10 end cell cell straight y squared plus 2 straight y end cell row 0 cell negative 4 end cell end table close square brackets equals open square brackets table row cell 3 straight x plus 4 end cell 3 row 0 cell straight y squared minus 5 straight y end cell end table close square brackets$

From the definition of equality of matrices, we have,
x + 10 = 3x + 4, y²+2y=3, -4= y²–5y
Now x + 10 = 3 x + 4 ⇒2x= 6 ⇒ ,x = 3
y² + 2 y – =3 ;⇒ y²+2y – 3 =0 ⇒ (y + 3) ( y –1) = 0 ⇒ y = 1. –3
– 4 = y ² – 5 y ⇒ y² – 5 y + 4 = 0 ⇒ ( y – 1) ( y – 4) = 0 ;⇒ y – 1. 4
Since both the equations y² + 2 y =3 and - 4 ='y² -5 y are to be satisfied, so we have y'= 1
∴ x = 3. y = I.

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