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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES4 Marks
Question
Using properties of determinants, prove that
$\begin{vmatrix} \text{a}^2+2\text{a}& 2\text{a}+1 & 1\\[0.3em] 2\text{a}+1 & \text{a}+2 & 1 \\[0.3em] 3 & 3 & 1 \end{vmatrix}=(\text{a}-1)^3$
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Answer

$\triangle=\begin{vmatrix} \text{a}^2+2\text{a}& 2\text{a}+1 & 1\\[0.3em] 2\text{a}+1 &\text{a}+2 &1 \\[0.3em] 3 & 3 & 1 \end{vmatrix}$
R1 → R1 – R2 and R2 →  R2 – R3
$\triangle=\begin{vmatrix} \text{a}^2+1& \text{a}-1 & 0\\[0.3em] 2(\text{a}-1) &\text{a}-1 &0 \\[0.3em] 3 & 3 & 1 \end{vmatrix}$
$\triangle=(\text{a}-1)^2\begin{vmatrix} \text{a}+1& 1 & 0\\[0.3em] 2 &1 &0 \\[0.3em] 3 & 3 & 1 \end{vmatrix}$
Expanding
(a – 1)2.(a – 1) = (a – 1)3.

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