The matrix $\left[ {\begin{array}{*{20}{c}}
0&5&{ - 7} \\
{ - 5}&0&{11} \\
7&{ - 11}&0
\end{array}} \right]$is
A. A skew - symmetric matrix
B. A symmetric matrix
C. A diagonal matrix
D. An upper triangular matrix
Answer
640.5k+ views
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.
Complete step-by-step answer:
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’ = $\left[ {\begin{array}{*{20}{c}}
0&{ - 5}&7 \\
5&0&{ - 11} \\
{ - 7}&{11}&0
\end{array}} \right]$
We can clearly see transpose is negative to the matrix itself that is M = -M’
So it is a skew-symmetric matrix.
So, the correct answer is “Option B”.
Note: Whenever we face such a type of question the key concept of solving the question is we have to first know about these terms like transpose, minor etc. and also remember properties. Here we need to know that if the matrix is symmetric then there is no change in the diagonal element is observed and when the matrix is skew-symmetric then the diagonal elements are changed in sign and not in magnitude we also need to know that all the elements other than diagonal elements in the diagonal matrix are zero. Knowing this will help you and will help to solve your problems.
Complete step-by-step answer:
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’ = $\left[ {\begin{array}{*{20}{c}}
0&{ - 5}&7 \\
5&0&{ - 11} \\
{ - 7}&{11}&0
\end{array}} \right]$
We can clearly see transpose is negative to the matrix itself that is M = -M’
So it is a skew-symmetric matrix.
So, the correct answer is “Option B”.
Note: Whenever we face such a type of question the key concept of solving the question is we have to first know about these terms like transpose, minor etc. and also remember properties. Here we need to know that if the matrix is symmetric then there is no change in the diagonal element is observed and when the matrix is skew-symmetric then the diagonal elements are changed in sign and not in magnitude we also need to know that all the elements other than diagonal elements in the diagonal matrix are zero. Knowing this will help you and will help to solve your problems.
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