Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

To Prove

$</p><p>\left|\begin{array}{ccc}</p><p>a&b&c\\</p><p>l&m&n\\</p><p>p&q&r\end{array}</p><p>\right|=\left|\begin{array}{ccc}</p><p>m&b&q\\</p><p>l&a&p\\</p><p>n&c&r\end{array}</p><p>\right|$

we know that Ri <-> Rj,or Ci <->Cj determinant |A| = -|A|

Apply C1 <-> C2

$</p><p>=> -\left|\begin{array}{ccc}</p><p>b&a&c\\</p><p>m&l&n\\</p><p>q&p&r\end{array}</p><p>\right|$

Apply R1 <-> R2

$</p><p>=> \left|\begin{array}{ccc}</p><p>m&l&n\\</p><p>b&a&c\\</p><p>q&p&r\end{array}</p><p>\right|$

We also know that |A|=|A'|

take transpose

$=> \left|\begin{array}{ccc}</p><p>m&b&q\\</p><p>l&a&p\\</p><p>n&c&r\end{array}</p><p>\right|$

Hence proved.

For

$</p><p>\left|\begin{array}{ccc}</p><p>a&b&c\\</p><p>l&m&n\\</p><p>p&q&r\end{array}</p><p>\right|=\left|\begin{array}{ccc}</p><p>l&m&n\\</p><p>p&q&r\\</p><p>a&b&c\end{array}</p><p>\right|$

we know that Ri <-> Rj,or Ci <->Cj determinant |A| = -|A|

Apply R1 <-> R2

$</p><p>-\left|\begin{array}{ccc}</p><p>l&m&n\\</p><p>a&b&c\\</p><p>q&p&r\end{array}</p><p>\right|$

Apply R2 <-> R3

$</p><p>\left|\begin{array}{ccc}</p><p>l&m&n\\</p><p>p&q&r\\</p><p>a&b&c\end{array}</p><p>\right|$

Hence proved.

Hope it helps you.

Read more on brainly https://brainly.in/question/13362792