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If $A$ and $B$ are symmetric matrices of the same order, then which one of the following is not true?
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Verified Answer
The correct answer is:
$A B-B A$ is symmetric
Given, $A$ and $B$ are symmetric matrix of same order.
$$
\Rightarrow \quad A=A^{\prime}, B=B^{\prime}
$$
(i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}=A+B \quad$ [from Eq. (i)] $\Rightarrow(A+B)$ is symmetric.
(ii) $(A-B)^{\prime}-\Lambda^{\prime}-B^{\prime}-\Lambda-B$ [from Eq. (i)] $\Rightarrow(A-B)$ is symmetric.
$$
\text { (iii) } \begin{aligned}
(A B+B A)^{\prime} &=(A B)^{\prime}+(B A)^{\prime} \\
&=B^{\prime} A^{\prime}+A^{\prime} B^{\prime} \\
&=B A+A B \quad\lfloor\text { from Eq. (i) }] \\
&=(A B+B A)
\end{aligned}
$$
$\Rightarrow(A B+B A)$ is symmetric.
$$
\text { (iv) } \begin{aligned}
(A B-B A)^{\prime} &=(A B)^{\prime}-(B A)^{\prime} \\
&=B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \\
&=B A-A B \quad \text { [from Eq. (i)] } \\
&=-(A B-B A)
\end{aligned}
$$
$\Rightarrow(A B-B A)$ is skew-symmetric matrix.
$$
\Rightarrow \quad A=A^{\prime}, B=B^{\prime}
$$
(i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}=A+B \quad$ [from Eq. (i)] $\Rightarrow(A+B)$ is symmetric.
(ii) $(A-B)^{\prime}-\Lambda^{\prime}-B^{\prime}-\Lambda-B$ [from Eq. (i)] $\Rightarrow(A-B)$ is symmetric.
$$
\text { (iii) } \begin{aligned}
(A B+B A)^{\prime} &=(A B)^{\prime}+(B A)^{\prime} \\
&=B^{\prime} A^{\prime}+A^{\prime} B^{\prime} \\
&=B A+A B \quad\lfloor\text { from Eq. (i) }] \\
&=(A B+B A)
\end{aligned}
$$
$\Rightarrow(A B+B A)$ is symmetric.
$$
\text { (iv) } \begin{aligned}
(A B-B A)^{\prime} &=(A B)^{\prime}-(B A)^{\prime} \\
&=B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \\
&=B A-A B \quad \text { [from Eq. (i)] } \\
&=-(A B-B A)
\end{aligned}
$$
$\Rightarrow(A B-B A)$ is skew-symmetric matrix.
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