|{:(1+x,1,1),(1,1+y,1),(1,1,1+z):}|=xy+y...

`|{:(1+x,1,1),(1,1+y,1),(1,1,1+z):}|=xy+yz+zx+xyz`

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Prove that, [[1+x,1,1],[1,1+y,1],[1,1,1+z]] = xyz(1 + 1/x+1/y+1/z)

If x!=0,y!=0,z!=0 and |[1+x,1,1],[1+y,1+2y,1],[1+z,1+z,1+3z]|=0 , then x^(-1)+y^(-1)+z^(-1) is equal to a.1 b.-1 c.-3 d. none of these

If |{:(x,x^2,x^3+1),(y,y^2,y^3+1),(z,z^2,z^3+1):}|=a and x ney nez , prove that, xyz=-1.

if x+y+z=0 , then show that, |{:(1,1,1),(x,y,z),(x^3,y^3,z^3):}|=0

Prove that |{:(ax,,by,,cz),(x^(2),,y^(2),,z^(2)),(1,,1,,1):}|=|{:(a,,b,,c),(x,,y,,z),(yz,,xz,,xy):}|

Use product {:[( 1,1,2),( 0,2,3),( 3,2,4) ]:} {:[( 2,0,1),(9,2,3),(6,1,2) ]:} to solve the system of equations x-y+2z=1 xy-3z=1 3x-2y+4z =2

Without expanding prove that Delta ={:|( x+y,y+z,z+x) ,( z,x,y),( 1,1,1) |:} =0

|{:(" "1," "z,-y),(-z," "1," "x),(" "y,-x," "1):}|

By using properties of determinants , show that : {:|( x,x^(2) , yz) ,( y,y^(2) , zx ) ,( z , z^(2) , xy ) |:} =( x-y)(y-z) (z-x) (xy+yz+ zx)

Using Cofactors of elements of third column , evaluate , Delta ={:|( 1,x,yz),(1,y,zx),( 1,z,xy) |:}

|{:(a,b,c),(a^2,b^2,c^2),(a^3,b^3,c^3):}|=abc(a-b)(b-c)(c-a)
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|{:(a,b,c),(a^2,b^2,c^2),(b+c,c+a,a+b):}|=(a-b)(b-c)(c-a)(a+b+c)
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|{:(1+x,1,1),(1,1+y,1),(1,1,1+z):}|=xy+yz+zx+xyz
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|{:(x+4,2x,2x),(2x,x+4,2x),(2x,2x,x+4):}|=(5x+4)(x-4)^2
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|{:(1+a,1,1),(1,1+a,1),(1,1,1+a):}|=a^3+3a^2
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|{:(1,b+c,b^2+c^2),(1,c+a,c^2+a^2),(1,a+b,a^2+b^2):}|=(b-c)(c-a)(a-b)
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|{:(a^2,ab,ac),(ab,b^2,bc),(ca,bc,c^2):}|=?
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|{:(-1,b,c),(a,-1,c),(a,b,-1):}|=(a+1)(b+1)(c+1)((a)/(a+1)+(b)/(b+1)+(...
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|{:(x^2+y^2+1,x^2+2y^2+3,x^2+3y^2+4),(y^2+2,2y^2+6,3y^2+8),(y^2+1,2y^2...
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|{:(" "3a," "-a+b,-a+c),(a-b," "3b," "c-b),(a-c," "b-c," "3c):}|...
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|{:(x,a,b),(a,x,b),(a,b,x):}|=(x-a)(x-b)(x+a+b)
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|{:(b+c,a-b,a),(c+a,b-c,b),(a+b,c-a,c):}|=3abc-a^3-b^3-c^3
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Prove |{:(1,a^2+bc,a^3),(1,b^2+ca,b^3),(1,c^2+ab,c^3):}|=-(a-b)(b-c)(c...
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|{:(a^2,a^2-(b-c)^2,bc),(b^2,b^2-(c-a)^2,ca),(c^2,c^2-(a-b)^2,ab):}|=(...
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|{:((a+1)(a+2),a+2,1),((a+2)(a+3),a+3,1),((a+3)(a+4),a+4,1):}|=-2
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Show that the following determinant is a perfact square : |{:(1,a,a^...
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if x+y+z = 0 then prove that |{:(x,y,z),(x^2,y^2,z^2),(y+z,z+x,x+y):}|...
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|{:(1,1,1),(x,y,z),(x^3,y^3,z^3):}|=0
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|{:(x," "4,-2),(4," "x,-2),(4,-2," "x):}|=0
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|{:(x,a,a),(a,x,b),(b,b,x):}|=0
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