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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 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

$\triangle=\begin{vmatrix} \text{a}^2+2\text{a}& 2\text{a}+1 & 1\$0.3em] 2\text{a}+1 &\text{a}+2 &1 \$0.3em] 3 & 3 & 1 \end{vmatrix}$
$R_1 → R_1 – R_2$ and $R_2 → R_2– R_3$
$\triangle=\begin{vmatrix} \text{a}^2+1& \text{a}-1 & 0\$0.3em] 2(\text{a}-1) &\text{a}-1 &0 \$0.3em] 3 & 3 & 1 \end{vmatrix}$
$\triangle=(\text{a}-1)^2\begin{vmatrix} \text{a}+1& 1 & 0\$0.3em] 2 &1 &0 \$0.3em] 3 & 3 & 1 \end{vmatrix}$
Expanding
$(a – 1)^2.(a – 1) = (a – 1)^3$.

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