Question

58. If A is square matrix, then which of the following is not true ?
(A) AA’ is symmetric
(B) A- A’ is skew symmetric
(C) A+ A’ is symmetric
(D) A2 is symmetric

Answer 1

Answer:

A is not true

may be it's right

Answer 2

Answer:

D.$A^2$ is symmetric

Explanation:

We are given that A is square matrix .

We have to find which is not true in give options.

Symmetric , If A'=A

Skew symmetric: If A'=-A

Suppose

A=$\left[\begin{array}{ccc}1&1\\0&0\end{array}\right]$

$A'=\left[\begin{array}{ccc}1&0\\1&0\end{array}\right]$

A.AA' is symmetric

AA'=$\left[\begin{array}{ccc}1&1\\0&0\end{array}\right]\cdot\left[\begin{array}{ccc}1&0\\1&0\end{array}\right]$

$AA'=\left[\begin{array}{ccc}2&0\\0&0\end{array}\right]$

$(AA')'=\left[\begin{array}{ccc}2&0\\0&0\end{array}\right]$

Hence, it is true.

B.A-A' is skew symmetric

$A-A'=\left[\begin{array}{ccc}1&1\\0&0\end{array}\right]-\left[\begin{array}{ccc}1&0\\1&0\end{array}\right]$

$A-A'=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]$

$(A-A')'=\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]$

$(A-A')'=-\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]$

Hence, A-A' is skew symmetric .

C.A+A' is symmetric

$A+A'=\left[\begin{array}{ccc}1&1\\0&0\end{array}\right]+\left[\begin{array}{ccc}1&0\\1&0\end{array}\right]$

$A+A'=\left[\begin{array}{ccc}2&1\\1&0\end{array}\right]$

$(A+A')'=\left[\begin{array}{ccc}2&1\\1&0\end{array}\right]$

Hence, A+A' is symmetric .

D.$A^2$ is symmetric

$A^2=A\cdot A=\left[\begin{array}{ccc}1&1\\0&0\end{array}\right]\cdot \left[\begin{array}{ccc}1&1\\0&0\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1&1\\0&0\end{array}\right]$

$(A^2)'=\left[\begin{array}{ccc}1&0\\1&0\end{array}\right]\neq A$

Hence, $A^2$ is not symmetric.

Answer:D.$A^2$ is symmetric