How Do You Find the Inverse of a Non-Square Matrix?

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I thought that I could find the inverse of the coefficient matrix, but it's originally 2x3, so I redacted the linearly dependent row and found the 2x2 A inverse. I'm not sure what to do after that.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110625_165538.jpg

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110625_170257.jpg
 
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Hi Shackleford! :smile:

The problem is not asking you to determine the inverse of T. The problem is asking you to calculate T^{-1}(1,11) which is the set of all triples (a,b,c) such that

T(a,b,c)=(1,11)

There will NOT be a unique (a,b,c) that satisfies this (in general). We will expect a set of triples as answer.

The equation brings us to a system of equations that you need to solve:

\left\{\begin{array}{c} a+b=1\\ 2a-c=11\\ \end{array}\right.
 
micromass said:
Hi Shackleford! :smile:

The problem is not asking you to determine the inverse of T. The problem is asking you to calculate T^{-1}(1,11) which is the set of all triples (a,b,c) such that

T(a,b,c)=(1,11)

There will NOT be a unique (a,b,c) that satisfies this (in general). We will expect a set of triples as answer.

The equation brings us to a system of equations that you need to solve:

\left\{\begin{array}{c} a+b=1\\ 2a-c=11\\ \end{array}\right.

Oh. Well, I quickly misread that problem. It's no problem now. I'll do it in the morning. Thanks!
 
The strategy is to find the set of solutions to the homogeneous equation and then find a particular solution.

a + b = 1
2a - c = 11

a + b = 0
2a - c = 0

Implies a = a, b = -a, c = 2a.

KH = a(1, -1, 2)

Solving the system yields

a -(1/2)c = 11/2
b -(1/2)c = -9/2

The book's answer sets c = 0 and gives the particular solution as (11/2, -9/2, 0). Not too difficult a problem. It's a good problem to test your fundamental understanding of the theory and technique.
 
my daughter is studying in Class XI in a AP board school with BiPC.She is struggling to find the value of Tan inverse 4. Pl help.
 
Do you mean arctan(4) or tan(1/4)? I can read your question either way. Also: what angular units are used (degrees? radians?).

RGV
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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