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# The matrix $\left[ {\begin{array}{*{20}{c}} 0&5&{ - 7} \\ { - 5}&0&{11} \\ 7&{ - 11}&0 \end{array}} \right]$isA. A skew - symmetric matrixB. A symmetric matrixC. A diagonal matrixD. An upper triangular matrix

Last updated date: 20th Jun 2024
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Hint: In order to solve this problem we need to find the transpose of the given matrix and then observe which matrix we get. We need to know that symmetric matrix is the matrix whose transpose is same, skew-symmetric matrix is the matrix whose transpose is negative of the matrix itself and diagonal matrix is the matrix in which all other element than diagonal are zero. Knowing this will solve your problem.

As we know that for a matrix to be a symmetric matrix the transpose of it should be equal to the matrix itself. That is $A = {A^T}$
We also know that for a matrix to be a skew-symmetric matrix the transpose of it should be equal to the negative of the matrix itself. That is $A = - {A^T}$
And we have to know that the transpose of a matrix is an operator which flips a matrix over its diagonal, which is it switches the row and column indices of the matrix i.e. ${R_{ij}} \Leftrightarrow {C_{ij}}$.
Here matrix is M = $\left[ {\begin{array}{*{20}{c}} 0&5&{ - 7} \\ { - 5}&0&{11} \\ 7&{ - 11}&0 \end{array}} \right]$
Its transpose is M’ = $\left[ {\begin{array}{*{20}{c}} 0&{ - 5}&7 \\ 5&0&{ - 11} \\ { - 7}&{11}&0 \end{array}} \right]$