Mathematics - Determinant Question with Solution | TestHub

Apply column operations $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$. This transforms the determinant to:$\Delta =\left|\begin{array}{ccc}(a+1{)}^2& 2a+3& 4a+8\\ (b+1{)}^2& 2b+3& 4b+8\\ (c+1{)}^2& 2c+3& 4c+8\end{array}\right|$ Next, apply the column operation $C_3\to C_3-2C_2$. The third column simplifies to $\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)$.$\Delta =\left|\begin{array}{ccc}(a+1{)}^2& 2a+3& 2\\ (b+1{)}^2& 2b+3& 2\\ (c+1{)}^2& 2c+3& 2\end{array}\right|=2\left|\begin{array}{ccc}(a+1{)}^2& 2a+3& 1\\ (b+1{)}^2& 2b+3& 1\\ (c+1{)}^2& 2c+3& 1\end{array}\right|$ Now, apply row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$. This yields:$\Delta =2\left|\begin{array}{ccc}(a+1{)}^2& 2a+3& 1\\ (b-a)(a+b+2)& 2(b-a)& 0\\ (c-a)(a+c+2)& 2(c-a)& 0\end{array}\right|$ Expanding along the third column, we get:$\Delta =2\times 1\times \left[(b-a)(a+b+2)\cdot 2(c-a)-2(b-a)\cdot (c-a)(a+c+2)\right]$ $\Delta =4(b-a)(c-a)\left[(a+b+2)-(a+c+2)\right]=4(b-a)(c-a)(b-c)$ Rearranging the terms to the standard form $(a-b)(b-c)(c-a)$, we get $4(a-b)(b-c)(c-a)$.