Question 16 & 17, Exercise 2.2
Solutions of Questions 16 & 17 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.
Question 16
Let $A=\begin{bmatrix}3 & -1 \\4 & 2\end{bmatrix}$. Show that $|A^{-1}|=\dfrac{1}{|A|}$.
Solution
Given
$$A=\left[ \begin{matrix}
3 & -1 \\
4 & 2 \\
\end{matrix} \right]$$
$$|A|=6+4$$
$$\Rightarrow |A|=10\ldots (1)$$
$$A^{-1}=\dfrac{1}{|A|}AdjA$$
$$AdjA=\left[ \begin{matrix}
2 & 1 \\
-4 & 3 \\
\end{matrix} \right]$$
$$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix}
2 & 1 \\
-4 & 3 \\
\end{matrix} \right]$$
$$=\left[ \begin{matrix}
\dfrac{2}{10} & \dfrac{1}{10} \\
-\dfrac{4}{10} & \dfrac{3}{10} \\
\end{matrix} \right]$$
$$A^{-1}=\left[ \begin{matrix}
\dfrac{1}{5} & \dfrac{1}{10} \\
-\dfrac{2}{5} & \dfrac{3}{10} \\
\end{matrix} \right]$$
$$|A^{-1}|=\dfrac{3}{50}+\dfrac{2}{50}$$
$$|A^{-1}|=\dfrac{1}{10}$$
By using (1), above expression gives,
$$|A^{-1}|=\dfrac{1}{|A|}$$
Question 17
Verify that $( AB )^{-1}=B^{-1}A^{-1}$. If $A=\begin{bmatrix}2 & 3 \\1 & 0\end{bmatrix}$,$B= \begin{bmatrix} -1 & 1 \\2 & 3\end{bmatrix}$.
Solution
Given $$A=\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]$$, $$B=\left[ \begin{matrix} -1 & 1 \\ 2 & 3 \\ \end{matrix} \right]$$ $$|A|=-3$$ $$AdjA=\left[ \begin{matrix} 0 & -3 \\ -1 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{-3}\left[ \begin{matrix} 0 & -3 \\ -1 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \begin{matrix} 0 & 1 \\ \dfrac{1}{3} & -\dfrac{2}{3} \\ \end{matrix} \right]$$ $$|B|=-5$$ $$AdjB=\left[ \begin{matrix} 3 & -1 \\ -2 & -1 \\ \end{matrix} \right]$$ $$B^{-1}=\dfrac{1}{|B|}AdjB$$ $$B^{-1}=\dfrac{1}{-5}\left[ \begin{matrix} 3 & -1 \\ -2 & -1 \\ \end{matrix} \right]$$ $$B^{-1}=\left[ \begin{matrix} -\dfrac{3}{5} & \dfrac{1}{5} \\ \dfrac{2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]$$ $$B^{-1}A^{-1}=\left[ \begin{matrix} -\dfrac{3}{5} & \dfrac{1}{5} \\ \dfrac{2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 \\ \dfrac{1}{3} & -\dfrac{2}{3} \\ \end{matrix} \right]$$ $$B^{-1}A^{-1}=\left[ \begin{matrix} \dfrac{1}{15} & \dfrac{-11}{15} \\ \dfrac{1}{15} & \dfrac{4}{15} \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 \\ 2 & 3 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} -2+6 & 2+9 \\ -1 & 1 \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 4 & 11 \\ -1 & 1 \\ \end{matrix} \right]$$ $$\Rightarrow |AB|=4+11$$ $$\Rightarrow \,\,|AB|=15$$ $$AdjAB=\left[ \begin{matrix} 1 & -11 \\ 1 & 4 \\ \end{matrix} \right]$$ $$( AB )^{-1}=\dfrac{1}{|AB|}AdjAB$$ $$( AB )^{-1}=\dfrac{1}{15}\left[ \begin{matrix} 1 & -11 \\ 1 & 4 \\ \end{matrix} \right]$$ $$\Rightarrow ( AB )^{-1}=\left[ \begin{matrix} \dfrac{1}{15} & -\dfrac{11}{15} \\ \dfrac{1}{15} & \dfrac{4}{15} \\ \end{matrix} \right]$$ $$\Rightarrow ( AB )^{-1}=B^{-1}A^{-1}$$
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