If A and B are symmetric matrices of the same order, then


If A and B are symmetric matrices of the same order, then

(i) $A B-B A$ is a_________

(ii) $B A-2 B A$ is a________


It is given that, $A$ and $B$ are symmetric matrices of the same order.

$\therefore A^{T}=A$ and $B^{T}=B$


$(A B-B A)^{T}$

$=(A B)^{T}-(B A)^{T} \quad\left[(X+Y)^{T}=X^{T}+Y^{T}\right]$

$=B^{T} A^{T}-A^{T} B^{T} \quad\left[(X Y)^{T}=Y^{T} X^{T}\right]$

$=B A-A B$       [Using (1)]

$=-(A B-B A)$

Since $(A B-B A)^{T}=-(A B-B A)$, so the matrix $A B-B A$ is a skew-symmetric matrix.

$A B-B A$ is a _skew-symmetric matrix__


Disclaimer: The solution has been provided for the following question.

$B A-2 A B$ is a

$=(B A)^{T}-(2 A B)^{T} \quad\left[(X+Y)^{T}=X^{T}+Y^{T}\right]$

$=(B A)^{T}-2(A B)^{T} \quad\left[(k X)^{T}=k(X)^{T}\right]$

$=A^{T} B^{T}-2 B^{T} A^{T} \quad\left[(X Y)^{T}=Y^{T} X^{T}\right]$

$=A B-2 B A$     [Using (1)]

Since $(B A-2 A B)^{T} \neq B A-2 A B$ or $(B A-2 A B)^{T} \neq-(B A-2 A B)$, so the matrix $B A-2 A B$ is neither symmetric nor skew-symmetric matrix.

$B A-2 A B$ is a neither symmetric nor skew-symmetric matrix


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