For the matrix A= [ ll 1 & 5 6 & 7 ], verify that (A

Here, A= [ ll 1 & 5 6 & 7 ] and A^ = [ ll 1 & 6 5 & 7 ] subtracting A' from A, we get, A-A^ = [ ll 1 & 5 6 & 7 ]- [ ll 1 & 6 5 & 7 ] A-A^ = [ ll 1-1 & 5-6 6-5 & 7-7 ]= [ ll 0 & -1 1 & ~~~ 0 ] Clearly, (A-A^ )^ = [ cc 0 & 1 -1 & 0 ]~~~...(i) We can rewrite the above equation as (A-A^ )^ =(-1) [ cc 0 & -1 1 & 0 ]=(-1)(A-A^ ) ~~~~~... using (i) (A – A')' = -(A – A') Hence we can say that matrix A is a Skew-symmetric matrix.

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Question

For the matrix $A=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right]$, verify that (A – A′) is a skew-symmetric matrix

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Answer

Here, $A=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right] \text { and } A^{\prime}=\left[\begin{array}{ll} {1} & {6} \\ {5} & {7} \end{array}\right]$
subtracting A' from A, we get,
$A-A^{\prime}=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right]-\left[\begin{array}{ll} {1} & {6} \\ {5} & {7} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{ll} {1-1} & {5-6} \\ {6-5} & {7-7} \end{array}\right]=\left[\begin{array}{ll} {0} & {-1} \\ {1} & ~~~{0} \end{array}\right]$
Clearly, $\left(\mathrm{A}-\mathrm{A}^{\prime}\right)^{\prime}=\left[\begin{array}{cc} {0} & {1} \\ {-1} & {0} \end{array}\right]~~~...(i)$
We can rewrite the above equation as
$\left(\mathrm{A}-\mathrm{A}^{\prime}\right)^\prime=(-1)\left[\begin{array}{cc} {0} & {-1} \\ {1} & {0} \end{array}\right]=(-1)(A-A^\prime) ~~~~~... using (i)$
$\therefore$ (A – A')' = -(A – A')
Hence we can say that matrix A is a Skew-symmetric matrix.