Question

Given

$A=\left(\begin{array}{rr}-2&2\\2&-4\end{array}\right)\quad \textrm{and} \quad B=\left(\begin{array}{rr}2&3\\2&-1\end{array}\right).$

Select the option(s) below which represent $\left(AB^{-1}\right)^{-1}.$

1. $\displaystyle \left(\begin{array}{rr}-3.5&-2.5\\-1.5&-0.5\end{array}\right).$
2. $\displaystyle \left(\begin{array}{rr}3.5&2.5\\1.5&0.5\end{array}\right).$
3. A^-1B
4. B^-1A^-1
5. BA^-1

Answer 1

Let's find ${B^{ - 1}}$ :

$\left( {\left. {\begin{matrix} 2&3\\ 2&{ - 1} \end{matrix}} \right|\begin{matrix} 1&0\\ 0&1 \end{matrix}} \right)\begin{matrix} {}\\ { - I} \end{matrix} \Leftrightarrow \left( {\left. {\begin{matrix} 2&3\\ 0&{ - 4} \end{matrix}} \right|\begin{matrix} 1&0\\ { - 1}&1 \end{matrix}} \right) \Leftrightarrow \left( {\left. {\begin{matrix} 2&3\\ 0&1 \end{matrix}} \right|\begin{matrix} 1&0\\ {\frac{1}{4}}&{ - \frac{1}{4}} \end{matrix}} \right)\begin{matrix} { - 3II}\\ {} \end{matrix} \Leftrightarrow \left( {\left. {\begin{matrix} 2&0\\ 0&1 \end{matrix}} \right|\begin{matrix} {\frac{1}{4}}&{\frac{3}{4}}\\ {\frac{1}{4}}&{ - \frac{1}{4}} \end{matrix}} \right) \Leftrightarrow \left( {\left. {\begin{matrix} 1&0\\ 0&1 \end{matrix}} \right|\begin{matrix} {\frac{1}{8}}&{\frac{3}{8}}\\ {\frac{1}{4}}&{ - \frac{1}{4}} \end{matrix}} \right)$

Then

${B^{ - 1}} = \left( {\begin{matrix} {\frac{1}{8}}&{\frac{3}{8}}\\ {\frac{1}{4}}&{ - \frac{1}{4}} \end{matrix}} \right)$

$A{B^{ - 1}} = \left( {\begin{matrix} { - 2}&2\\ 2&{ - 4} \end{matrix}} \right)\left( {\begin{matrix} {\frac{1}{8}}&{\frac{3}{8}}\\ {\frac{1}{4}}&{ - \frac{1}{4}} \end{matrix}} \right) = \left( {\begin{matrix} { - \frac{1}{4} + \frac{1}{2}}&{ - \frac{3}{4} - \frac{1}{2}}\\ {\frac{1}{4} - 1}&{\frac{3}{4} + 1} \end{matrix}} \right) = \left( {\begin{matrix} {\frac{1}{4}}&{ - \frac{5}{4}}\\ { - \frac{3}{4}}&{\frac{7}{4}} \end{matrix}} \right)$

Let's find ${\left( {A{B^{ - 1}}} \right)^{ - 1}}$ :

${\left( {A{B^{ - 1}}} \right)^{ - 1}} = \left( {\left. {\begin{matrix} {\frac{1}{4}}&{ - \frac{5}{4}}\\ { - \frac{3}{4}}&{\frac{7}{4}} \end{matrix}} \right|\begin{matrix} 1&0\\ 0&1 \end{matrix}} \right) \Leftrightarrow \left( {\left. {\begin{matrix} 1&{ - 5}\\ { - 3}&7 \end{matrix}} \right|\begin{matrix} 4&0\\ 0&4 \end{matrix}} \right)\begin{matrix} {}\\ { + 3I} \end{matrix} \Leftrightarrow \left( {\left. {\begin{matrix} 1&{ - 5}\\ 0&{ - 8} \end{matrix}} \right|\begin{matrix} 4&0\\ {12}&4 \end{matrix}} \right) \Leftrightarrow \left( {\left. {\begin{matrix} 1&{ - 5}\\ 0&{ - 2} \end{matrix}} \right|\begin{matrix} 4&0\\ 3&1 \end{matrix}} \right)\begin{matrix} { - \frac{5}{2}II}\\ {} \end{matrix} \Leftrightarrow \left( {\left. {\begin{matrix} 1&0\\ 0&{ - 2} \end{matrix}} \right|\begin{matrix} { - \frac{7}{2}}&{ - \frac{5}{2}}\\ 3&1 \end{matrix}} \right) \Leftrightarrow \left( {\left. {\begin{matrix} 1&0\\ 0&1 \end{matrix}} \right|\begin{matrix} { - \frac{7}{2}}&{ - \frac{5}{2}}\\ { - \frac{3}{2}}&{ - \frac{1}{2}} \end{matrix}} \right)$

Then

${\left( {A{B^{ - 1}}} \right)^{ - 1}} = \left( {\begin{matrix} { - 3.5}&{ - 2.5}\\ { - 1.5}&{ - 0.5} \end{matrix}} \right)$

By the properties of the inverse matrix

${\left( {A{B^{ - 1}}} \right)^{ - 1}} = {\left( {{B^{ - 1}}} \right)^{ - 1}}{A^{ - 1}} = B{A^{ - 1}}$

Answer: 1. ${\left( {A{B^{ - 1}}} \right)^{ - 1}} = \left( {\begin{matrix} { - 3.5}&{ - 2.5}\\ { - 1.5}&{ - 0.5} \end{matrix}} \right)$ , 5. $B{A^{ - 1}}$