# Show that the matrix $k{\text{A}}$ is symmetric or skew symmetric according as A is symmetric or skew symmetric.

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Hint- Here, we will be using the conditions for a matrix to be a symmetric or skew symmetric matrix.

Let A be a matrix and A’ be the transpose of the matrix A

So, matrix A is said to be symmetric matrix if ${\text{A = A'}}$ and matrix A is said to be skew symmetric matrix if ${\text{A}} = - {\text{A'}}$.

Let there be any matrix B such that ${\text{B}} = k{\text{A}}$ where $k$ is any constant.

Now for matrix B to be symmetric, \[{\text{B}} = {\text{B'}} \Rightarrow k{\text{A}} = \left( {k{\text{A}}} \right)'\]

Since for any constant $k$, \[\left( {k{\text{A}}} \right)' = k{\text{A'}} \Rightarrow k{\text{A}} = k{\text{A'}} \Rightarrow {\text{A}} = {\text{A'}}\] which is the condition for matrix A to be symmetric matrix i.e., for matrix ${\text{B}} = k{\text{A}}$ to be symmetric, matrix A should be a symmetric matrix

Now for matrix B to be skew symmetric, \[{\text{B}} = - {\text{B'}} \Rightarrow k{\text{A}} = - \left( {k{\text{A}}} \right)'\]

Since for any constant $k$, \[\left( {k{\text{A}}} \right)' = k{\text{A'}} \Rightarrow k{\text{A}} = - k{\text{A'}} \Rightarrow {\text{A}} = - {\text{A'}}\] which is the condition for matrix A to be skew symmetric matrix i.e., for matrix ${\text{B}} = k{\text{A}}$ to be skew symmetric, matrix A should be a skew symmetric matrix.

Therefore, matrix ${\text{B}} = k{\text{A}}$ is symmetric or skew symmetric according as matrix A is symmetric or skew symmetric.

Note- These types of problems can be solved by using the conditions for symmetric and skew symmetric matrices. The condition to be proved is to be simplified as much as possible

Let A be a matrix and A’ be the transpose of the matrix A

So, matrix A is said to be symmetric matrix if ${\text{A = A'}}$ and matrix A is said to be skew symmetric matrix if ${\text{A}} = - {\text{A'}}$.

Let there be any matrix B such that ${\text{B}} = k{\text{A}}$ where $k$ is any constant.

Now for matrix B to be symmetric, \[{\text{B}} = {\text{B'}} \Rightarrow k{\text{A}} = \left( {k{\text{A}}} \right)'\]

Since for any constant $k$, \[\left( {k{\text{A}}} \right)' = k{\text{A'}} \Rightarrow k{\text{A}} = k{\text{A'}} \Rightarrow {\text{A}} = {\text{A'}}\] which is the condition for matrix A to be symmetric matrix i.e., for matrix ${\text{B}} = k{\text{A}}$ to be symmetric, matrix A should be a symmetric matrix

Now for matrix B to be skew symmetric, \[{\text{B}} = - {\text{B'}} \Rightarrow k{\text{A}} = - \left( {k{\text{A}}} \right)'\]

Since for any constant $k$, \[\left( {k{\text{A}}} \right)' = k{\text{A'}} \Rightarrow k{\text{A}} = - k{\text{A'}} \Rightarrow {\text{A}} = - {\text{A'}}\] which is the condition for matrix A to be skew symmetric matrix i.e., for matrix ${\text{B}} = k{\text{A}}$ to be skew symmetric, matrix A should be a skew symmetric matrix.

Therefore, matrix ${\text{B}} = k{\text{A}}$ is symmetric or skew symmetric according as matrix A is symmetric or skew symmetric.

Note- These types of problems can be solved by using the conditions for symmetric and skew symmetric matrices. The condition to be proved is to be simplified as much as possible

Last updated date: 01st Oct 2023

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