Question

# If $$x+y+z=0$$, then $$\begin{vmatrix} ax & by & cz \\ cy & az & bx \\ bz & cx & ay\end{vmatrix}=xyz\begin{vmatrix} a & b & c\\ c & a & b \\ b & c & a\end{vmatrix}$$.

A
True
B
False

Solution

## The correct option is A TrueGiven,$$x+y+z=0~~~~~~~~~~~~-(1)$$Taking LHS$$\left|\begin{matrix} ax &by&cz\\ cy&az&bx\\ bz&cx&ay \\\end{matrix}\right|$$$$=xa(yza^2-x^2bc)-yb(y^2ac-b^2xz)+zc(c^2xy-z^2ab)\\=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)\\=xyz(a^3+b^3+c^3)-3abc(xyz)$$   .......[since $$x+y+z=0, t$$ therefore,]$$=xyz(a^3+b^3+c^3-3abc)\\=xyz(a+b+c)(a^2+b^2+c^2-ab-ac-bc)~~~~~-(1)$$Taking RHS$$=xyz\left|\matrix{a&b&c\\c&a&b\\b&c&a}\right|$$Applying $$R_1\to R_1+R_2+R_3$$$$=xyz\left|\matrix{a+b+c&a+b+c&a+b+c\\c&a&b\\b&c&a}\right|$$$$=xyz(a+b+c)\left|\matrix{1&1&1\\c&a&b\\b&c&a}\right|$$Applying $$C_2\to C_2-C_1, C_3\to C_3-C_1$$$$=xyz(a+b+c)\left|\matrix{1&0&0\\c&a-c&b-c\\b&c-b&a-b}\right|$$$$=xyz(a+b+c)[(a-c)(a-b)-(c-b)(b-c)]\\=xyz(a+b+c)(a^2+b^2+c^2-ab-bc-ac)~~~~~~-(2)$$from eqs.(1) and (2)LHS=RHSHence,$$\left|\begin{matrix} ax &by&cz\\ cy&az&bx\\ bz&cx&ay \\\end{matrix}\right|=xyz\left|\matrix{a&b&c\\c&a&b\\b&c&a}\right|$$ Mathematics

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