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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using properties of determinants, prove that:
$\begin{vmatrix}3\text{a}&-\text{a+b}&-\text{a+c}\\-\text{b+a}&3\text{b}&-\text{b+c}\\-\text{c+a}&\text{c+b}&3\text{c}\end{vmatrix}=3 (\text{a + b + c}) (\text{ab + bc + ca})$
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Answer

$\triangle=\begin{vmatrix}3\text{a}&-\text{a + b}&-\text{a + c}\\-\text{b + a}&3\text{b}&-\text{b + c}\\-\text{c + a}&\text{c + b}&3\text{c}\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_2 + C_3$, we have:
$\triangle=\begin{vmatrix}\text{a + b + c}&-\text{a + b}&-\text{a + c}\\\text{a + b + c}&3\text{b}&-\text{b + c}\\\text{a + b + c}&-\text{c + b}&3\text{c}\end{vmatrix}$
$= (\text{a + b + c})\begin{vmatrix}1&-\text{a + b}&-\text{a + c}\\1&3\text{b}&-\text{b + c}\\1&-\text{c + b}&3\text{c}\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1​​​​​​​$ and $R_3 \rightarrow R_3 - R_1​​​​​​​$​​​​​​​, we have:
$\triangle=(\text{a + b + c})\begin{vmatrix}1&-\text{a + b}&-\text{a + c}\\0&2\text{b + a}&\text{a}-\text{b}\\0&\text{a}-\text{c}&2\text{c + a}\end{vmatrix}$
Expanding along C, we have:
$\triangle$ $= (a + b + c) [(2b + a) (2c + a) - (a - b) (a - c)]$
$= (a + b + c) [4bc + 2ab + 2ac + a^2 - a^2 + ac + ba - bc]$
$= (a + b + c) (3ab + 3bc + 3ac)$
$= 3(a + b + c) (ab + bc + ca)$
Hence, the given result is proved.

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