![A^{-1}={\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]}](/tpl/images/0571/1145/0c4cd.png)
![\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right]](/tpl/images/0571/1145/46618.png)
, its inverse is calculated using the formula:![\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] ^{-1}=\frac{1}{det\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]}\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]](/tpl/images/0571/1145/7ba71.png)
![\frac{1}{det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right]}\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]\\\\\\\\{det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right]}=1\\\\\\\\\therefore =\frac{1}{1}{\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]}\\\\\\\\={\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]}](/tpl/images/0571/1145/f22b3.png)
Consider the matrix shown below:
...
Answers: 2
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