prove that determinant of x x2 yz

y y2 zx = (x-y)(y-z)(z-x)(xy+yz+zx)

z z2 xy

We want to evaluate:


First, subtract the second row from the first row. We know that
the determinant will not change . So, we can say that the above
expression (required determinant) equals,



As you see, from the first row, we can factor (x-y):


Again we can subtract the third row from the second row:Again we can factor (y-z) from the second row:


Now, we can subtract the second row from the first. We'll have:


We can now factor (x-z) from the first row:


We can now subtract z times the first column from the second column:


We have to make as many zeros as possible, since they make some parts of the expansion vanish.Now,expand the determinant.= 0 ( ... ) - 1 (xy) + z(-x-y) = -(xy+xz+yz)So all in all, the determinant will be:(x-y)(y-z)(x-z)(-(xy+xz+yz))Let's apply the negative sign in (x-z) and write it as (z-x). We'll have(x - y)(y - z)(z - x)(xy + xz + yz),

 

  • 3

 x    x2    yz

y     y2    zx

z     z2    xy

=(x-y)(y-z)(z-x)(xy+yz+zx)

pls tell me the answer fast ty

  • 1

Its a determinant and i have to prove it and y2 = y2 and z2=z

  • 3

It is a determinant and y2=y2 and z2=z2

  • 4
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