Mathematics, 15.01.2020 05:31, qudoniselmore0
Suppose that a and b are nonsingular matrices. then ab is also nonsingular. furthermore, a theorem from linear algebra then states that (ab)-1= b-1a-1 . (a) verify this theorem for 2× 2 matrices. (b) prove the theorem for any n × n matrix
Answers: 3
![A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]](/tpl/images/0455/6026/85210.png)
and ![B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]](/tpl/images/0455/6026/a8ff2.png)
![AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}a.e+b.g&a.f+b.h\\c.e+d.g&c.f+d.h\end{array}\right]](/tpl/images/0455/6026/7b5d7.png)
![(AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]](/tpl/images/0455/6026/03c47.png)
![A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]](/tpl/images/0455/6026/a7274.png)
![B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]](/tpl/images/0455/6026/9f641.png)
![B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]](/tpl/images/0455/6026/87bfc.png)
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Suppose that a and b are nonsingular matrices. then ab is also nonsingular. furthermore, a theorem f...
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