Browse Questions

# Using properties of determinants prove that $\begin{vmatrix} -bc & b^2+bc & c^2+bc \\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}= (ab+bc+ca)^3$

Toolbox:
• (i)Elementary transformation can be made by interchanging any two rows or two columns
• (ii) By adding or subtracting any two or three rows or columns
• If $A= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
• Then $|A|=a_{11}(a_{22} \times a_{33}-a_{23} \times a_{32})-a_{12}(a_{21} \times a_{33}-a_{23} \times a_{31})+a_{13}(a_{21} \times a_{32}-a_{22} \times a_{31})$
Step 1:
Let $\Delta=\begin{vmatrix} -bc & b^2+bc & c^2+bc \\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}$
Let us multiply $R_1$ by a, $R_2$ by b and $R_3$ by c
$\Delta=\large\frac{1}{abc} \normalsize \begin{vmatrix} -abc & ab^2+abc & ac^2+abc \\ a^2b+abc & -abc & bc^2+abc \\ a^2c+abc & cb^2+abc & -abc \end{vmatrix}$
Let us take a,b and c as the common factor from $C_1,C_2$ and $C_3$ respectively,
$\Delta=\large\frac{abc}{abc} \normalsize \begin{vmatrix} -bc & ab+ac & ac+ab \\ ab+bc & -ac & bc+ab \\ ac+bc & bc+ac & -ab \end{vmatrix}$
Step 2:
Now apply $R_1 \to R_1+R_2+R_3$
$\Delta=\begin{vmatrix} ab+bc+ac & ab+ac+bc & ab+bc+ac \\ ab+bc & -ac & bc+ab \\ ac+bc & bc+ac & -ab \end{vmatrix}$
Now take (ab+bc+ac) as the common factor from $R_1$
$\Delta=(ab+bc+ca)\begin{vmatrix} 1 & 1 & 1 \\ ab+bc & -ac & bc+ab \\ ac+bc & bc+ac & -ab \end{vmatrix}$
Apply $C_1 \to C_1-C_2$ and $C_2 \to C_1-C_3$
$\Delta=(ab+bc+ca)\begin{vmatrix} 0 & 0 & 1 \\ ab+bc+ca & 0 & bc+ab \\ 0 & ab+ac+bc & -ab \end{vmatrix}$
Now expand along $R_1$,
$\Delta=(ab+bc+ca) \bigg[1(ab+bc+ca)^2-0 \bigg]$
$\Delta=(ab+bc+ca)^3$
Solution:Hence proved