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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}=(\text{a}-1)^3$
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Answer

We have to prove,
$\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}=(\text{a}-1)^3$
$\therefore\ \text{L.H.S}=\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2+2\text{a}-2\text{a}-1&2\text{a}+1-\text{a}-2&0\\2\text{a}+1-3&\text{a}+2-3&0\\3&3&1\end{vmatrix}$
$[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_2\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3]$
$=\begin{vmatrix}(\text{a}-1)(\text{a}+1)&(\text{a}-1)&0\\2(\text{a}-1)&(\text{a}-1)&0\\3&3&1\end{vmatrix}$
$=(\text{a}-1)^2\begin{vmatrix}(\text{a}+1)&1&0\\2&1&0\\3&3&1\end{vmatrix}$
$[\text{Taking (a}-1)\text{ common from R}_1\text{ and R}_2\text{ each}]$
$=(\text{a}-1)^2\big[1(\text{a}+1)-2\big]=(\text{a}-1)^3$
$=\text{R.H.S}$
Hence proved.

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