Question 13, Exercise 2.1

Solutions of Question 13 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $A$ is a square matrix of order $3$ then show that $A+A^t$ is symmetric.

$$A=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=( A+A^t )$$ $$A+A^t=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{matrix} \right]+\left[ \begin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \\ \end{matrix} \right]$$ $$A+A^t=\left[ \begin{matrix} a_{11}+a_{11} & a_{12}+a_{21} & a_{13}+a_{31} \\ a_{21}+a_{12} & a_{22}+a_{22} & a_{23}+a_{32} \\ a_{31}+ a_{13} & a_{32}+a_{23} & a_{33}+a_{33} \\ \end{matrix} \right]$$ $$( A+A^t )^t=\left[ \begin{matrix} a_{11}+a_{11} & a_{21}+a_{12} & a_{31}+a_{13} \\ a_{12}+a_{21} & a_{22}+a_{22} & a_{32}+a_{23} \\ a_{13}+a_{31} & a_{23}+a_{32} & a_{33}+a_{33} \\ \end{matrix} \right]$$ $$( A+A^t )^t=( A+A^t )$$

If $A$ is a square matrix of order $3$ then show that $A-{{A}^{t}}$ is skew symmetric.

$$A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]$$ $${{A}^{t}}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{21}} & {{a}_{31}} \\ {{a}_{12}} & {{a}_{22}} & {{a}_{32}} \\ {{a}_{13}} & {{a}_{23}} & {{a}_{33}} \\ \end{matrix} \right]$$ For skew-symmetric, we have, $${{\left( A-{{A}^{t}} \right)}^{t}}=-\left( A-{{A}^{t}} \right)$$ $$A-{{A}^{t}}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]-\left[ \begin{matrix} {{a}_{11}} & {{a}_{21}} & {{a}_{31}} \\ {{a}_{12}} & {{a}_{22}} & {{a}_{32}} \\ {{a}_{13}} & {{a}_{23}} & {{a}_{33}} \\ \end{matrix} \right]$$ $$A-{{A}^{t}}=\left[ \begin{matrix} {{a}_{11}}-{{a}_{11}} & {{a}_{12}}-{{a}_{21}} & {{a}_{13}}-{{a}_{31}} \\ {{a}_{21}}-{{a}_{12}} & {{a}_{22}}-{{a}_{22}} & {{a}_{23}}-{{a}_{32}} \\ {{a}_{31}}-{{a}_{13}} & {{a}_{32}}-{{a}_{23}} & {{a}_{33}}-{{a}_{33}} \\ \end{matrix} \right]$$ $${{\left( A-{{A}^{t}} \right)}^{t}}=\left[ \begin{matrix} {{a}_{11}}-{{a}_{11}} & {{a}_{21}}-{{a}_{12}} & {{a}_{31}}-{{a}_{13}} \\ {{a}_{12}}-{{a}_{21}} & {{a}_{22}}-{{a}_{22}} & {{a}_{32}}-{{a}_{23}} \\ {{a}_{13}}-{{a}_{31}} & {{a}_{23}}-{{a}_{32}} & {{a}_{33}}-{{a}_{33}} \\ \end{matrix} \right]$$ $${{\left( A-{{A}^{t}} \right)}^{t}}=-\left[ \begin{matrix} {{a}_{11}}-{{a}_{11}} & {{a}_{12}}-{{a}_{21}} & {{a}_{13}}-{{a}_{31}} \\ {{a}_{21}}-{{a}_{12}} & {{a}_{22}}-{{a}_{22}} & {{a}_{23}}-{{a}_{32}} \\ {{a}_{31}}-{{a}_{13}} & {{a}_{32}}-{{a}_{23}} & {{a}_{33}}-{{a}_{33}} \\ \end{matrix} \right]$$ $${{\left( A-{{A}^{t}} \right)}^{t}}=-\left( A-{{A}^{t}} \right)$$

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