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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Prove that:
$\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
✓

Answer

$\text{L.H.S}=\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
Multiply $R_1, R_2$ and $R_3$ by a, b and c respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}-\text{abc}&\text{ab}^2+\text{abc}&\text{ac}^2+\text{abc}\\\text{a}^2\text{b}+\text{abc}&-\text{abc}&\text{bc}^2+\text{abc}\\\text{a}^2\text{c}+\text{abc}&\text{b}^2\text{c}+\text{abc}&-\text{abc}\end{vmatrix}$
Take a, b and c common from $C_1, C_2$ and $C_3$​​​​​​​ respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}-\text{bc}&\text{ab}+\text{ac}&\text{ac}+\text{ab}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
Apply: $R_1 → R_1 + R_2 + R_3$​​​​​​​
$=\begin{vmatrix}\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}1&1&1\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}0&1&0\\\text{a}\text{b}+\text{bc}+\text{ac}&-\text{ac}&\text{bc}+\text{ab}+\text{ac}\\0&\text{b}\text{c}+\text{ac}&-\text{ab}-\text{bc}-\text{ac}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3\begin{vmatrix}0&1&0\\0&-\text{ac}&1\\0&\text{b}\text{c}+\text{ac}&1\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
$=\text{R.H.S}$

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