Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

Q2(i)

$\left|\begin{array}{ccc}</p><p>43&1&3\\35&7&2\\17&3&1\end{array}\right|=0$

C2->C2-14C3

$\left|\begin{array}{ccc}</p><p>43-42&1&3\\35-28&7&2\\17-14&3&1\end{array}\right|=0$

$\left|\begin{array}{ccc}</p><p>1&1&3\\7&7&2\\3&3&1\end{array}\right|=0$

Here C1 =C3

according to property of Determinant ;if two rows and two columns are identical than Determinant will be zero

so

$\left|\begin{array}{ccc}</p><p>1&1&3\\7&7&2\\3&3&1\end{array}\right|=0$

Q2(ii)$\left|\begin{array}{ccc}</p><p>1&a&a^2-bc\\1&b&b^2-ca\\1&c&c^2-ab\end{array}\right|=0$

R1->R1-R3

R2->R2-R3

$\left|\begin{array}{ccc}</p><p>0&a-c&a^2-bc-c^2+ab\\0&b-c&b^2-ca-c^2+ab\\0&c&c^2-ab\end{array}\right|=0$

$\left|\begin{array}{ccc}</p><p>0&a-c&a^2-c^2+ab-bc\\0&b-c&b^2-c^2+ab-ac\\0&c&c^2-ab\end{array}\right|=0$

$\left|\begin{array}{ccc}</p><p>0&a-c&(a-c)(a+c)+b(a-c)\\0&b-c&(b-c)(b+c)+a(b-c)\\0&c&c^2-ab\end{array}\right|=0$

take(a-c) common from R1

and (b-c) from R2

$(a-c)(b-c)\left|\begin{array}{ccc}</p><p>0&1&a+ b+c\\0&1&a+b+c\\0&c&c^2-ab\end{array}\right|=0$

we know that if two rows are identical,determinant will be zero

so

(a-c)(b-c)(0)

=0

Answer 2

Step-by-step explanation:

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

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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}

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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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=

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</p><p>a

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</p><p>b

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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

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</p><p>a

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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

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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

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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}

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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

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Hope it helps you.