# 1. Using properties of determinants, prove the following:| x y z x2 y2 z2 x3 y3 z3 | = xyz(x - y)(y - z)(z - x) .2. Using properties of determinants, prove the following :| x x2 1+px3 y y2 1+py3 z z2 1+pz3 | = (1+ pxyz)(x - y)(y - z)(z - x) .

1. 2.

To prove:  Proof:

Let,  Expanding along R3, we have: • 10

1. xyz|1   1     1|

|x   y     z|

|x2 y2 z2|.    Now C1->C1-C2 and C3->C3-C2 then expand. Hence Proved

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2. R1-> R1-R2 and. R3-> R3-R2.....Take (x-y) & (z-y) common and then  we shall have

(x-y)(z-y)|1  x+y   x^2+y^2+z^2 |.                     Now, put R1->R1-R3 and take (x-z) common then put C2->C2+C1 then in the next step

y  y^2   1+py^3.                                 put R1->R1-R2 and then expand.....You shall have your answer..

1  z+y.  z^2+y^2+zy|

• -1

1) Take out (xyz)

|  1  1    1

x   y    z

x2  y 2 z2  |  .Now,C1 -->C1-C2 ,C2 -->C2 -C3

(xyz)  |0  0  1

x-y   y-z  z

x2 -y2   y2- z2  z2|

Taking out (x-y)(y-Z) ,

(xyz)(x-y)(y-z)  |0 0  1

0  0   z

x+y  y+z  z2 |  .

On expanding ,

(xyz)(x-y)(y-z)[(y+z)-(x+y)]=  (xyz)(x-y)(y-z)(z-x)

Hope you understood.

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