Question

Anyone answer this..

Answer 1

Given Question :-

If

$\rm :\longmapsto\:A = \: \: \: \begin{gathered}\sf \left [ \begin{array}{ccc}a&b&c\\x&y&z\\p& q&r\end{array}\right ] \end{gathered}$

and

$\rm :\longmapsto\:B= \: \: \: \begin{gathered}\sf \left [ \begin{array}{ccc}q& - b&y\\ - p&a& -x\\r& - c&z\end{array}\right ] \end{gathered}$

and if A is invertible, which of the following is not true?

$\rm \: \: \: \: \: \: (A) \: \: |A| = |B|$

$\rm \: \: \: \: \: \: (B) \: \: |A| = - \: |B|$

$\rm \: \: \: \: \: \: (C) \: \: |adjA| \: = \: |adjB|$

$\rm \: \: \: \: \: \: (D) \: \: A \: is \: invertible \: iff \: B \: is \: invertible$

$\large\underline{\sf{Solution-}}$

Given matrices are

$\rm :\longmapsto\:A = \: \: \: \begin{gathered}\sf \left [ \begin{array}{ccc}a&b&c\\x&y&z\\p& q&r\end{array}\right ] \end{gathered}$

and

$\rm :\longmapsto\:B= \: \: \: \begin{gathered}\sf \left [ \begin{array}{ccc}q& - b&y\\ - p&a& -x\\r& - c&z\end{array}\right ] \end{gathered}$

Now, Consider,

$\rm :\longmapsto\: |B|$

$\: \: \rm \:  =  \: \: \begin{gathered}\sf \left | \begin{array}{ccc}q& - b&y\\ - p&a& -x\\r& - c&z\end{array}\right | \end{gathered}$

$\red{ \boxed{ \sf{ \:OP \: C_2 \: \to \: - 1(C_2)}}}$

$\: \: \rm \:  =  \: - \: \begin{gathered}\sf \left | \begin{array}{ccc}q&b&y\\ - p& - a& -x\\r&c&z\end{array}\right | \end{gathered}$

$\red{ \boxed{ \sf{ \:OP \: R_2 \: \to \: - 1(R_2)}}}$

$\: \: \rm \:  =  \: ( - 1)( - 1) \: \begin{gathered}\sf \left | \begin{array}{ccc}q&b&y\\ p&a&x\\r&c&z\end{array}\right | \end{gathered}$

$\: \: \rm \:  =  \:\: \begin{gathered}\sf \left | \begin{array}{ccc}q&b&y\\ p&a&x\\r&c&z\end{array}\right | \end{gathered}$

$\red{ \boxed{ \sf{ \:OP \: R_2 \leftrightarrow \: R_1}}}$

$\: \: \rm \:  =  \: - \: \begin{gathered}\sf \left | \begin{array}{ccc}p&a&x\\ q&b&y\\r&c&z\end{array}\right | \end{gathered}$

$\red{ \boxed{ \sf{ \:OP \: C_2 \leftrightarrow \: C_1}}}$

$\: \: \rm \:  =  \: ( - 1)( - 1) \: \begin{gathered}\sf \left | \begin{array}{ccc}a&p&x\\ b&q&y\\c&r&z\end{array}\right | \end{gathered}$

$\: \: \rm \:  =  \:\: \begin{gathered}\sf \left | \begin{array}{ccc}a&p&x\\ b&q&y\\c&r&z\end{array}\right | \end{gathered}$

We know,

$\red{ \boxed{ \sf{ \: |A| = | {A}^{T} | }}}$

$\: \: = \: \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b&c\\p&q&r\\x& y&z\end{array}\right | \end{gathered}$

$\red{ \boxed{ \sf{ \:OP \: R_2 \leftrightarrow \: R_1}}}$

$\: \: = \: - \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b&c\\x&y&z\\p& q&r\end{array}\right | \end{gathered}$

$\rm \:  =  \: - |A|$

$\bf\implies \: |B| \: = - \: |A|$

• So, option (A) is not true.

Reason :- Why other Options are correct.

Option C is correct.

As,

$\rm :\longmapsto\:adjA = { |A| }^{n - 1} = { |A| }^{2} = {( - |B|) }^{2} = { |B| }^{2} = adjB$

Option D is correct

As given that, A is invertible

$\bf\implies \: |A| \ne \: 0$

$\bf\implies \: |B| \ne \: 0$

$\bf\implies \:B \: is \: invertible$

and vice - versa.

Additional Information :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

Answer 2

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