Mathematics - Determinant Question with Solution | TestHub

Apply column operations $C_2\to C_2-C_1$ and $C_3\to C_3-C_2$. This yields common factors $(y-x)$ from $C_2$ and $(z-y)$ from $C_3$.$$\Delta =(y-x)(z-y)\left|\begin{array}{ccc}(a+x{)}^2& 2a+x+y& 2a+y+z\\ (b+x{)}^2& 2b+x+y& 2b+y+z\\ (c+x{)}^2& 2c+x+y& 2c+y+z\end{array}\right|$$ Next, apply$C_3\to C_3-C_2$. This gives a common factor $(z-x)$ from the new $C_3$.$$\Delta =(y-x)(z-y)(z-x)\left|\begin{array}{ccc}(a+x{)}^2& 2a+x+y& 1\\ (b+x{)}^2& 2b+x+y& 1\\ (c+x{)}^2& 2c+x+y& 1\end{array}\right|$$

Let the remaining$3\times 3$ determinant be ${\Delta }^{\prime }$. Apply row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$. Expanding along the third column gives a $2\times 2$ determinant:$${\Delta }^{\prime }=\left|\begin{array}{cc}(b-a)(2x+a+b)& 2(b-a)\\ (c-a)(2x+a+c)& 2(c-a)\end{array}\right|=(b-a)(c-a)\left|\begin{array}{cc}2x+a+b& 2\\ 2x+a+c& 2\end{array}\right|$$

Evaluating the$2\times 2$ determinant: $(b-a)(c-a)[2(2x+a+b)-2(2x+a+c)]=(b-a)(c-a)[2(b-c)]=2(b-a)(c-a)(b-c)$. Substituting back,$\Delta =(y-x)(z-y)(z-x)\cdot 2(b-a)(c-a)(b-c)$. Rearranging factors:$(y-x)(z-y)(z-x)=-(x-y)\cdot -(y-z)\cdot -(x-z)=-(x-y)(y-z)(x-z)$. And$2(b-a)(c-a)(b-c)=2(-(a-b))(-(a-c))(-(c-b))=-2(a-b)(a-c)(c-b)$. So,$\Delta =[-(x-y)(y-z)(x-z)]\cdot [-2(a-b)(a-c)(c-b)]=2(x-y)(y-z)(x-z)(a-b)(a-c)(c-b)$. Since$(a-c)(c-b)=(c-a)(b-c)$, the expression is $2(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$.