Question 3, Exercise 2.2

Solutions of Question 3 of Exercise 2.2 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.

Let $A$ be square matrix of order $3,$ then verify that $|A^t|=|A|$.

Let $$A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix}$$ Then \begin{align}|A|&=a_{11} \left( a_{22} a_{33}-a_{23} a_{32} \right)-a_{12}\left( a_{21}a_{33}-a_{23}a_{31} \right)+a_{13}\left( a_{21}a_{32}-a_{22}a_{31} \right)\\ &=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}\\ &=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31} \ldots (1) \end{align} Now $$ {{A}^{t}}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{21}} & {{a}_{31}} \\ {{a}_{12}} & {{a}_{22}} & {{a}_{32}} \\ {{a}_{13}} & {{a}_{23}} & {{a}_{33}} \\ \end{matrix} \right]\\ $$ Then \begin{align} |A^t|&=a_{11}\left( a_{22}a_{33}-a_{32}a_{23} \right)-a_{21}\left( a_{12}a_{33}-a_{32}a_{13} \right)+a_{31}\left( a_{12}a_{23}-a_{22}a_{13} \right)\\ &=a_{11}a_{22}a_{33}-a_{11}a_{32}a_{23}-a_{21}a_{12}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}-a_{31}a_{22}a_{13}\\ &=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31} \ldots (2) \end{align} Now comparing (1) and (2), we have $$|A|=|{{A}^{t}}|.$$

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