Question

[tex] If A= \left[\begin{array}{ccc}6&7\\8&9\end{array}\right], and B⁻¹=\left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right],Find (AB)⁻¹. [\tex]

Answer 1

Step-by-step explanation:

We know that

${(AB)}^{ - 1} = {B}^{ - 1} \times {A}^{ - 1}$

In the problem B⁻¹ is already given,we have to find A⁻¹ from A.

${A}^{ - 1} = \frac{ {A}^{T} }{ |A| }$

$A= \left[\begin{array}{ccc}6&7\\8&9\end{array}\right]$

$A^T= \left[\begin{array}{ccc}6&8\\7&9\end{array}\right]$

|A|=6×9 -7×8 =54-56 =-2

$A^{-1}=\frac{1}{-2} \left[\begin{array}{ccc}6&8\\7&9\end{array}\right]$

Now B⁻¹ × A⁻¹

$\left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \times \frac{1}{-2} \left[\begin{array}{ccc}6&8\\7&9\end{array}\right]$

$\frac{1}{-2}\left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \times \left[\begin{array}{ccc}6&8\\7&9\end{array}\right]$

$\frac{1}{-2}\left[\begin{array}{ccc}30-14&40-18\\-42+21&-56+27\end{array}\right]$

$B^{-1}A^{-1}= (AB)^{-1}=\frac{1}{-2}\left[\begin{array}{ccc}16&22\\-21&-29\end{array}\right]$

$B^{-1}A^{-1}= (AB)^{-1}=\frac{-1}{2}.2\left[\begin{array}{ccc}8&11\\\frac{-21}{2}&\frac{-29}{2}\end{array}\right]$

$(AB)^{-1}=\left[\begin{array}{ccc}-8&-11\\\frac{21}{2}&\frac{29}{2}\end{array}\right]$

Hope it helps you.