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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
If $\triangle_1=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix},\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix},$ then:
  • ✓
    $\triangle_1+\triangle_2=0$
  • B
    $\triangle_1+2\triangle_2=0$
  • C
    $\triangle_1=\triangle_2$
  • D
    None of these.
✓

Answer

Correct option: A.
$\triangle_1+\triangle_2=0$
$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}\left[R_1, R_2, R_3\right.$ are multiplies by $a, b$ and $c$ respectively, therefore we divide by $abc]$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix} [$Taking $abc$ common from $\left.C _2\right]$
$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
We know that the value of a determinant remains unchanged if its rows and columns are interchanged. so,
$\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $
$=-\triangle_1$
$\triangle_1+\triangle_2=0$

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