Question

Determinants questions​

Answer 1

$\underline{\text{Proof :}}$

$\text{L.H.S.}=\left | \begin{array}{ccc}a&b&c\\a^{2}&b^{2}&c^{2}\\bc&ac&ab\end{array}\right |$

$\mathrm{C'_{1}\to aC_{1}\:,\:C'_{2}\to bC_{2}\:,\:C'_{3}\to cC_{3}}$

$=\frac{1}{abc}\left | \begin{array}{ccc}a^{2}&b^{2}&c^{2}\\a^{3}&b^{3}&c^{3}\\abc&abc&abc\end{array}\right |$

$\mathrm{Taking\:abc\:common\:from\:R_{3}}$

$=\frac{abc}{abc}\left | \begin{array}{ccc}a^{2}&b^{2}&c^{2}\\a^{3}&b^{3}&c^{3}\\1&1&1\end{array}\right |$

$\mathrm{Doing\:C'_{2}=C_{3}\to C_{2}}$

$=-\left | \begin{array}{ccc}a^{2}&b^{2}&c^{2}\\1&1&1\\a^{3}&b^{3}&c^{3}\end{array}\right |$

$\mathrm{Doing\:C'_{1}=C_{2}\to C_{1}}$

$=(-1)^{2}\left | \begin{array}{ccc}1&1&1\\a^{2}&b^{2}&c^{2}\\a^{3}&b^{3}&c^{3}\end{array}\right |$

$=\left | \begin{array}{ccc}1&1&1\\a^{2}&b^{2}&c^{2}\\a^{3}&b^{3}&c^{3}\end{array}\right |=\text{R.H.S.}$

$\text{Hence, proved.}$

Answer 2

Answer:

I don't know DETERMINANTS. I know DISCRIMINANT. D =√b² - 4ac

Step-by-step explanation: