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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Applying R1 → R2 + R2 + R3]
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Taking out 3(a + b) common from R1]
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix}$ [Applying C1 → C1 - C2 and C2 → C2 - C3]
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix}$ [Taking out b common from C1 and C2]
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$

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