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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
$\begin{vmatrix}1&\text{a}&\text{a}^2\\\text{a}^2&1&\text{a}\\\text{a}&\text{a}^2&1\end{vmatrix}=(\text{a}^3-1)^2$
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Answer

Let us consider the L.H.S of the given equation.
Let $\begin{vmatrix}1&\text{a}&\text{a}^2\\\text{a}^2&1&\text{a}\\\text{a}&\text{a}^2&1\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_2 + C_3,$ we have,
$=\begin{vmatrix}1+\text{a}+\text{a}^2&\text{a}&\text{a}^2\\1+\text{a}+\text{a}^2&1&\text{a}\\1+\text{a}+\text{a}^2&\text{a}^2&1\end{vmatrix}$
Taking term$ (1 + a + a_2)$ common, we have,
$=(1+\text{a}+\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\1&1&\text{a}\\1&\text{a}^2&1\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1 $and$ R_3 \rightarrow R_3 - R_1$​​​​​​​, we have,
$=(1+\text{a}+\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\0&1-\text{a}&\text{a}(1-\text{a})\\0&-\text{a}(1-\text{a})&(1-\text{a})(1+\text{a})\end{vmatrix}$
Taking the term (1 - a) common from $R_2 $and $R_3$, we have,
$=(1+\text{a}+\text{a}^2)(1-\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\0&1&\text{a}\\0&-\text{a}&(1+\text{a})\end{vmatrix}$
$=(1+\text{a}+\text{a}^2)(1-\text{a}^2)(1+\text{a}+\text{a}^2)$
$=(1+\text{a}+\text{a}^2)(1-\text{a}^2)$
$=\big[(1+\text{a}+\text{a}^2)(1-\text{a}^2)\big]^2$
$=\big[(\text{a}^3-1)\big]^2$
$=\text{R.H.S}$

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