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Mathematics Question on Matrices

If A=[3−4 1−1 ],A=\left[ \begin{matrix} 3 & -4 \\\ 1 & -1 \\\ \end{matrix} \right], then (A−A′)(A-A') is equal to (where, A′A' is transpose of matrix AA)

A

null matrix

B

identity matrix

C

symmetric

D

skew-symmetric

Answer

skew-symmetric

Explanation

Solution

The correct option is(D): skew-symmetric.

Given, A=[3−4 1−1 ]A=\left[ \begin{matrix} 3 & -4 \\\ 1 & -1 \\\ \end{matrix} \right]
Then, A′=[31 −4−1 ]A'=\left[ \begin{matrix} 3 & 1 \\\ -4 & -1 \\\ \end{matrix} \right]
Now, A−A′=[3−4 1−1 ]−[31 −4−1 ]A-A'=\left[ \begin{matrix} 3 & -4 \\\ 1 & -1 \\\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 1 \\\ -4 & -1 \\\ \end{matrix} \right]
⇒\Rightarrow A−A′=[0−5 50 ]A-A'=\left[ \begin{matrix} 0 & -5 \\\ 5 & 0 \\\ \end{matrix} \right] .. (i)
Now, we have A′−A=[31 −4−1 ]−[3−4 1−1 ]=[05 −50 ]A'-A=\left[ \begin{matrix} 3 & 1 \\\ -4 & -1 \\\ \end{matrix} \right]-\left[ \begin{matrix} 3 & -4 \\\ 1 & -1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 5 \\\ -5 & 0 \\\ \end{matrix} \right]
⇒\Rightarrow (A′−A)′=[0−5 50 ]=(A−A′)(A'-A)'=\left[ \begin{matrix} 0 & -5 \\\ 5 & 0 \\\ \end{matrix} \right]=(A-A')
[From E (i)] which represent that (A−A′)(A-A') is skew-symmetric matrix.