InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Prove that `[[x, x^2 , 1+px^3], [y, y^2, 1+py^3] ,[z, z^2, 1+pz^3]] = (1+pxyz)(x-y)(y-z)(z-x)` |
|
Answer» Let the given determinant be `Delta`. Then, `Delta = |{:(x, x^(2), 1+px^(3)),(y, y^(2), 1+py^(3)),(z, z^(2), 1+pz^(3)):}|` `= |{:(x, x^(2), 1),(y, y^(2), 1),(z, z^(2), 1):}| + |{:(x, x^(2), px^(3)),(y, y^(2), py^(3)),(z, z^(2), pz^(3)):}|" "["say"Delta_(1) + Delta_(2)]` `=(-1)* |{:(x, 1, x^(2)),(y, 1, y^(2)),(z, 1, z^(2)):}| + p*|{:(x, x^(2), x^(3)),(y, y^(2), y^(3)),(z, z^(2), z^(3)):}|["by"C_(2) harr C_(3) "in"Delta_(1) ["taking p common from"C_(3) "in"Delta_(2)]` `=(-1)(-1)* |{:(1, x, x^(2)),(1, y, y^(2)),(1, z, z^(2)):}| + (pxyz)*|{:(1, x, x^(2)),(1, y, y^(2)),(1, z, z^(2)):}|["by"C_(1) harr C_(2)] ["taking x, y, z common from"R_(1), R_(2), R_(3)"resp."]` ` =(1+pxyz)|{:(1, x, x^(2)), (1, y, y^(2)),(1, z, z^(2)):}|` `=(1+pxyz) |{:(1, x, x^(2)),(0, y-x, y^(2)-x^(2)),(0, z-x, z^(2)-x^(2)):}| ["by"R_(2) to R_(2) - R_(1) "and"R_(3) to R_(3)-R_(1) ]` `=(1+pxyz)(y-x)(z-x)* |{:(1, x, x^(2)),(0, 1, y+x),(0, 1, z+x):}| ["taking (y-x) common from"R_(2) "and (z-x) common from"R_(3)]` `=(1+pxyz)(y-x)(z-x)*1*|{:(1, y+x),(1, z+x):}|` `=(1+pxyz)(y-x)(z-x)*{(z+y)-(y-x)}` `= (1+pxyz)(y-x)(z-x)(z-y)` `=(1+pxyz)(x-y)(y-z)(z-x)=RHS. `therefore` LHS = RHS. |
|