# Using the factor theorem it is found that

Question:

Using the factor theorem it is found that $a+b, b+c$ and $c+a$ are three factors of the determinant

$\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & c+b & -2 c\end{array}\right|$

The other factor in the value of the determinant is

(a) 4

(b) 2

(c) $a+b+c$

(d) none of these

Solution:

(a) 4

$\Delta=\mid-2 a \quad a+b \quad a+c$

$b+a-2 b \quad b+c$

$c+a \quad c+b \quad-2 c \mid$

Let $a+b=2 C, b+c=2 A$ and $c+a=2 B$.

$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{b}+\mathrm{c}+\mathrm{c}+\mathrm{a}=2 \mathrm{~A}+2 \mathrm{~B}+2 \mathrm{C}$

$\Rightarrow 2(\mathrm{a}+\mathrm{b}+\mathrm{c})=2(\mathrm{~A}+\mathrm{B}+\mathrm{C})$

$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{A}+\mathrm{B}+\mathrm{C}$

Also,

$a=(a+b+c)-(b+c)=(A+B+C)-2 A=B+C-A$

Similarly,

$b=C+A-B, \quad c=A+B-C$

$\Delta=|2 \mathrm{~A}-2 \mathrm{~B}-2 \mathrm{C} \quad 2 \mathrm{C} \quad 2 \mathrm{~B} \quad 2 \mathrm{C} \quad 2 \mathrm{~B}-2 \mathrm{C}-2 \mathrm{~A} \quad 2 \mathrm{~A} \quad 2 \mathrm{~B} \quad 2 \mathrm{~A} \quad 2 \mathrm{C}-2 \mathrm{~A}-2 \mathrm{~B}|=8$

$\times|\mathrm{A}-\mathrm{B}-\mathrm{C} \quad \mathrm{C} \quad \mathrm{B} \quad \mathrm{C} \quad \mathrm{B}-\mathrm{C}-\mathrm{A} \quad \mathrm{A} \quad \mathrm{B} \quad \mathrm{A} \quad \mathrm{C}-\mathrm{A}-\mathrm{B}|$

[taking out 2 common from $\mathrm{R}_{1} \mathrm{R}_{2} \mathrm{R}_{3}$ ]

$=8 \times \mid A-B \quad C+B \quad B$

$B-A \quad B-C \quad A$

$\begin{array}{llll}\mathrm{B}+\mathrm{A} & \mathrm{C}-\mathrm{B} & \mathrm{C}-\mathrm{A}-\mathrm{B} \mid & {\left[\text { Applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}, \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}\right]}\end{array}$

$=8 \times \mid(\mathrm{A}-\mathrm{B}) \quad \mathrm{C}+\mathrm{B} \quad \mathrm{B}$

$\begin{array}{lll}0 & 2 \mathrm{~B} & \mathrm{~A}+\mathrm{B}\end{array}$

$\begin{array}{lll}2 B & 0 & C-B \mid\end{array}$            [Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3}$ ]

$=8 \times\{(\mathrm{A}-\mathrm{B})|2 \mathrm{~B} \quad \mathrm{~A}+\mathrm{B} 0 \quad \mathrm{C}-\mathrm{B}|+(2 \mathrm{~B}) \times|\mathrm{C}+\mathrm{B} \quad \mathrm{B} \quad 2 \mathrm{~B} \quad \mathrm{~A}+\mathrm{B}|\} \quad\left[\right.$ Expanding along $\left.C_{1}\right]$

$=16 \mathrm{~B}\left\{(\mathrm{~A}-\mathrm{B})(C-B)+(C+B)(A+B)-2 B^{2}\right\}$

$=32 \mathrm{ABC}$

$=32\left(\frac{\mathrm{b}+\mathrm{c}}{2}\right)\left(\frac{\mathrm{c}+\mathrm{a}}{2}\right)\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)$

$=4(a+b)(b+c)(c+a)$

Hence, 4 is the other factor of the determinant.