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Anni [7]
3 years ago
5

What is the sum of (3x-2)(2x-7)​

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
6 0

Answer:

6x^2 - 25x + 14

Step-by-step explanation:

(3x - 2) (2x - 7)

F.O.I.L.   (first, outside, inside, last)

(first) 3x * 2x = 6x^2

(outside) 3x * -7 = -21x

(inside) -2 * 2x = -4x

(last) -2 * -7 = 14

Doss [256]3 years ago
5 0

Answer:6x-9

Step-by-step explanation: I think so I’m not sure tho

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Suppose that A and B are nonsingular matrices. Then AB is also nonsingular. Furthermore, a theorem from linear algebra then stat
kherson [118]

Answer:

A) Verified

B) Proved

Step-by-step explanation:

a) Let's verify it for 2 x 2 matrix,

A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] and B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]

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]

(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]

A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]

B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]

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]

So it is proved that the results are same.

b) Now, let's prove it for any n x n matrix.

(AB)(AB)^{-1}=I\\\\A^{-1}(AB)(AB)^{-1}=A^{-1}I\\\\IB(AB)^{=1}=A^{-1}I\\\\B(AB)^{=1}=A^{-1}\\\\B^{-1}B(AB)^{=1}=B^{-1}A^{-1}\\\\I(AB)^{=1}=B^{-1}A^{-1}\\\\(AB)^{=1}=B^{-1}A^{-1}

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