Using matrix method, solve the following system of equations:2x&n

Subject

Mathematics

Class

CBSE Class 12

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.
Advertisement

 Multiple Choice QuestionsShort Answer Type

11.

Write the intercept cut off by the plane 2x + y – z = 5 on x-axis.


 Multiple Choice QuestionsLong Answer Type

12.

Prove the following:

cot-1   1 + sin x +  1 - sin x 1 + sin x -  1 - sin x  = x2,   x   0, π4 


13.

Find the value of  tan-1  xy  - tan-1  x - yx + y


14.

Using properties of determinants, prove that

  - a2      ab         ac     ba -b2      bc    ca  cb  - c2  = 4 a2b2c2


Advertisement
15.

Find the value of ‘a’ for which the function f defined as

f ( x ) =  a sin π2 ( x + 1 ),       x  0tan x - sin x x3,            x > 0 

is continuous at x = 0.


16.

Differentiate  X x cos x +  x2 + 1x2 - 1  w.r.t. x


17.

If   x = a  θ - sin θ ,   y =  1 + cos θ ,    find d2ydx2


18.

Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of t heradius of the base. How fast is the sand cone increasing when the height is 4 cm?


Advertisement
19.

Find the points on the curve  x2 + y2 – 2x – 3= 0  at  whichthe tangents are parallel to x-axis.


Advertisement

20.

Using matrix method, solve the following system of equations:

2x + 3y + 10z = 4,       4x - 6y + 5z,       6x + 9y - 20z;    x, y, z  0


The given system of equation is  2x + 3y + 10z = 4,    4x - 6y + 5z = 1,   6x  + 9y - 20z = 2

The given system of equation can be written as 

 2       3      104  - 6        56      9  - 20  1 x  1y 1y = 412Or  AX = B,  where  A =  2       3       104  - 6         56      9  - 20 ,    X = 1 x  1y 1y,      and  B = 412 Now,   A  =  2       3       104  - 6         56       9  - 20                      = 2 ( 120 - 45 ) - 3 ( - 80 - 30 ) + 10 ( 36 + 36 )                     = 1200  0

Hence, the unique solution of the system of equation is given by  X = A- 1 B

Now, the cofactors of  A  are computed as:

 

C11 = ( - 1 )2   120 - 45  = 75,             C12 = ( - 1 )3   - 80 - 30  = 110,           C13 = ( - 1 )4  36 + 36   = 72C21 = ( - 1 )3   - 60 - 90  = 150,        C22 = ( - 1 )4  - 60 - 90  = 150,            C23 = ( - 1 )5  18 - 18  = 0C31 = ( - 1 )4  15+ 60   =75,                 C32 = ( - 1 )5   10 - 40  =30,                  C33 = ( - 1 )6   - 12 - 12  = - 24

 

 Adj A =  75     110      72150     -100          075      30 - 24 T =  75     150       75110     -100          3072         0   - 24  A-1 = Adj A A  = 11200  75     150       75110     -100          3072         0   - 24 X =  A-1  B     = 11200  75     150       75110     -100          3072         0   - 24   4 1 2     =11200  3000 + 150 + 150440 - 100 + 60288 + 0 - 48   = 11200   600 400 240 

 

X =  6001200 40012002401200   =    121315      1x1y1z   =   121315  1x = 12,      1y = 13,    and    1z = 15 x = 2,   y = 3,    and    z = 5

Thus, solution of given system of equation is given by  x = 2,   y = 3,   z = 5.


Advertisement
Advertisement