# For the matrix $A = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right)$,verify that

(i) $\left( {A + A'} \right)$is a symmetric matrix.

(ii) $\left( {A - A'} \right)$is a skew symmetric matrix.

Last updated date: 26th Mar 2023

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Answer

Verified

311.4k+ views

Hint: Use the property of symmetric and skew symmetric matrices directly on the given matrix expression.

Given the matrix $A = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right)$

We know that the transpose of a matrix is obtained by switching the rows with its columns

$ \Rightarrow A' = \left( {\begin{array}{*{20}{c}}

1&6 \\

5&7

\end{array}} \right)$

A Symmetric matrix is the one in which the matrix is equal to the transpose of itself.

\[ \Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }}A = A'\]

We need to prove $\left( {A + A'} \right)$is a symmetric matrix.

$A + A' = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

1&6 \\

5&7

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

2&{11} \\

{11}&{14}

\end{array}} \right){\text{ (1)}}$

Also,${\left( {A + A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}

2&{11} \\

{11}&{14}

\end{array}} \right){\text{ (2)}}$

From equations $(1)$and$(2)$ , we get ${\left( {A + A'} \right)^\prime } = \left( {A + A'} \right)$, which satisfies the above-mentioned condition of symmetric matrices.

Hence $\left( {A + A'} \right)$is a symmetric matrix.

A Skew symmetric matrix is the one in which the negative of the matrix is equal to the transpose of itself.

\[ \Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }} - A = A'\]

We need to prove $\left( {A - A'} \right)$is a skew symmetric matrix.

$A - A' = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

1&6 \\

5&7

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

0&{ - 1} \\

1&0

\end{array}} \right){\text{ (3)}}$

Also,${\left( {A - A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}

0&1 \\

{ - 1}&0

\end{array}} \right){\text{ (4)}}$

From equations $(3)$and$(4)$ , we get ${\left( {A - A'} \right)^\prime } = - \left( {A - A'} \right)$, which satisfies the above-mentioned condition of skew symmetric matrices.

Hence $\left( {A - A'} \right)$is a skew symmetric matrix verified.

Note: The above-mentioned results are true for all square matrices. Similarly using above results, it can be proved that a square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.

Given the matrix $A = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right)$

We know that the transpose of a matrix is obtained by switching the rows with its columns

$ \Rightarrow A' = \left( {\begin{array}{*{20}{c}}

1&6 \\

5&7

\end{array}} \right)$

A Symmetric matrix is the one in which the matrix is equal to the transpose of itself.

\[ \Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }}A = A'\]

We need to prove $\left( {A + A'} \right)$is a symmetric matrix.

$A + A' = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

1&6 \\

5&7

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

2&{11} \\

{11}&{14}

\end{array}} \right){\text{ (1)}}$

Also,${\left( {A + A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}

2&{11} \\

{11}&{14}

\end{array}} \right){\text{ (2)}}$

From equations $(1)$and$(2)$ , we get ${\left( {A + A'} \right)^\prime } = \left( {A + A'} \right)$, which satisfies the above-mentioned condition of symmetric matrices.

Hence $\left( {A + A'} \right)$is a symmetric matrix.

A Skew symmetric matrix is the one in which the negative of the matrix is equal to the transpose of itself.

\[ \Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }} - A = A'\]

We need to prove $\left( {A - A'} \right)$is a skew symmetric matrix.

$A - A' = \left( {\begin{array}{*{20}{c}}

1&5 \\

6&7

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

1&6 \\

5&7

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

0&{ - 1} \\

1&0

\end{array}} \right){\text{ (3)}}$

Also,${\left( {A - A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}

0&1 \\

{ - 1}&0

\end{array}} \right){\text{ (4)}}$

From equations $(3)$and$(4)$ , we get ${\left( {A - A'} \right)^\prime } = - \left( {A - A'} \right)$, which satisfies the above-mentioned condition of skew symmetric matrices.

Hence $\left( {A - A'} \right)$is a skew symmetric matrix verified.

Note: The above-mentioned results are true for all square matrices. Similarly using above results, it can be proved that a square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.

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