Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

$\left|\begin{array}{ccc}</p><p>a+b&b+c&c+a\\l+m&m+n&n+l\\p+q&q+r&r+p\end{array}\right|=2\left|\begin{array}{ccc}</p><p>a&b&c\\l&m&n\\p&q&r\end{array}\right|$

We know that from the property of Determinant,we can split the determinant

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

Property : if we interchange two rows or two columns once sign of Determinant will change

So

In second Determinant

C1 <-> C3

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

C2<-> C3

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

$=2\left|\begin{array}{ccc}</p><p>a&b&c\\l&m&n\\p&q&r\end{array}\right|$

So,it is proved

$\left|\begin{array}{ccc}</p><p>a+b&b+c&c+a\\l+m&m+n&n+l\\p+q&q+r&r+p\end{array}\right|=2\left|\begin{array}{ccc}</p><p>a&b&c\\l&m&n\\p&q&r\end{array}\right|$

Hope it helps you.

Answer 2

Answer:

Step-by-step explanation:

\begin{gathered}\left|\begin{array}{ccc} < /p > < p > a+b&b+c&c+a\\l+m&m+n&n+l\\p+q&q+r&r+p\end{array}\right|=2\left|\begin{array}{ccc} < /p > < p > a&b&c\\l&m&n\\p&q&r\end{array}\right|\end{gathered}

∣

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</p><p>a+b

l+m

p+q

b+c

m+n

q+r

c+a

n+l

r+p

∣

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∣

∣

∣

∣

∣

=2

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</p><p>a

l

p

b

m

q

c

n

r

∣

∣

∣

∣

∣

∣

∣

We know that from the property of Determinant,we can split the determinant

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

∣

∣

∣

∣

∣

∣

∣

</p><p>a+b

l+m

p+q

b+c

m+n

q+r

c+a

n+l

r+p

∣

∣

∣

∣

∣

∣

∣

=

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</p><p>a

l

p

b

m

q

c

n

r

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+

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</p><p>b

m

q

c

n

r

a

l

p

∣

∣

∣

∣

∣

∣

∣

Property : if we interchange two rows or two columns once sign of Determinant will change

So

In second Determinant

C1 <-> C3

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

=

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</p><p>a

l

p

b

m

q

c

n

r

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−

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</p><p>a

l

p

c

n

r

b

m

q

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∣

C2<-> C3

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

=

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</p><p>a

l

p

b

m

q

c

n

r

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+

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</p><p>a

l

p

b

m

q

c

n

r

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\begin{gathered}=2\left|\begin{array}{ccc} < /p > < p > a&b&c\\l&m&n\\p&q&r\end{array}\right|\end{gathered}

=2

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</p><p>a

l

p

b

m

q

c

n

r

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So,it is proved

\begin{gathered}\left|\begin{array}{ccc} < /p > < p > a+b&b+c&c+a\\l+m&m+n&n+l\\p+q&q+r&r+p\end{array}\right|=2\left|\begin{array}{ccc} < /p > < p > a&b&c\\l&m&n\\p&q&r\end{array}\right|\end{gathered}

∣

∣

∣

∣

∣

∣

∣

</p><p>a+b

l+m

p+q

b+c

m+n

q+r

c+a

n+l

r+p

∣

∣

∣

∣

∣

∣

∣

=2

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</p><p>a

l

p

b

m

q

c

n

r

∣

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Hope it helps you.