Question

What you mean by the term invertible in linear algebra?

Answer 1

A function is called invertible if there is another function that produces the inverse result. Basically, if \[y=f _{1}(x)\]
Then f1(x) is only invertible, if another function f2(x) exists, so that
\[f_{2}(y) = x\]
for any x. f2(x) in this case is called the inverse of f1(x).

For example, if \[f(x)=2x+4\]

then \[f(x)^{-1} = 1/2x - 2\]



Matrices are invertible if their determinant is non-zero. Non-square matrices don't have an inverse. The inverse of a matrix means that

\[M _{1} * M_{2} * M_{2}^{-1} = M_{1}\]

Or alternatively, \[M_{2}*M_{2}^{-1} = I\]
Where I is the identity matrix and M2^-1 stands for the inverse of M2.
.

This is particularly useful in computer graphics, where affine transformations are often represented by 4x4 matrices. To undo the transformation of any point in space, simply invert its transformation matrix. This is needed to transform points between spaces, or to properly transform normal vectors under non-uniform scaling (multiply them with the inverse transpose of the transformation matrix).

Interesting to note is that if a matrix M represents a rotation of an orthogonal coordinate system (the vectors that compose the matrix are perpendicular to each other), the transpose of M is also its inverse. For small matrices, efficient ways to invert exist for specific ranks of matrices. For arbitrarily sized matrices, Gaussian elimination can be used to calculate the inverse.

Another view of a matrix inverse is that, since a matrix can be used to represent a system of linear equations, the matrix inverse would represent the coefficients of the inverses of each of the functions.

Answer 2

Isn't f(x)^-1=1/2x+4/

Answer 3

I mean isn't that 1/2x+4?

Answer 4

Let's try i out. In the above example of f(x) for x=6, f(x) = 2*6+4 = 16.
For your inverse, 1/2*16 + 4 = 12, which is !=6 (the original input into f(x)) - so that can't be the inverse.
If we use 1/2x-2, we get 1/2*16-2, which is 6.

For a simple linear equation you can simply invert the calculation. In f(x), first we multiply by 2, then we add 4. So to get the inverse for any x, first we have to subtract 4, then divide by 2. That gives us
\[f(x) = 1/2*(x-4)\]
which we can multiply out to 1/2x-2 :)

Answer 5

I dnt stil get t. If u dnt mind cn u explain t with anothr example?

Answer 6

Sure, let's take a different function, say f(x) = 3x.
To solve, you simply multiply x by 3, so the inverse must be a division by 3. That means that the inverse of f(x)=3x is f(x)^-1 = x/3.

Now, if we add another operation such as addition that we do *after* the multiply, we need to do the inverse of it *before* in the inverse function.
So, let's take f(x) = 3x+1
and assign a variable y, to make it clearer: y = 3x+1
To get the inverse, we're trying to find a function that does
\[f(y) = x \]
Basically, the result (y) of f(x) is the parameter for the inverse.
For y=3x+1, that means we first have to subtract 1, then divide by 3:
\[f(x)^{-1} = (x-1)/3 \]

A more visual way to look at it would be, to graph f(x). f(x) tells you, for any given position on the x-axis of the graph, what the position along the y axis is that the plot of f(x) runs through.
The inverse, f(x)^-1, would tell you the opposite - for a given position along the y axis, it would tell you exactly where on the x axis the function plot is.

I hope that was a little clearer :)

Answer 7

I got t nw. Thanx a lot.

Answer 8

For the 1st exapmple you showed shouldn't f(x)^-1 be x-4/2

Answer 9

Yes! Or, more precisely (x-4)/2 - remember, we have to do the subtraction first, since the addition happened last in f(x).
But if we multiply out (x-4)/2, we get x/2 -4/2, which is 1/2x - 2 :)

Answer 10

Cn u explain the last multiplication part f u dnt mind

Answer 11

It's called the distributive property and can help you eliminate parentheses. If you have something like (a+b)*c - a sum in parentheses multiplied by a number - you can multiply each addend separately, then add the products. For example,
you could say that
(3-1)*4
is (2)*4 = 8
But you could also multiply each 3 and 1 by 4 first, then subtract:
(3-1)*4 = 3*4 - 1*4 = 12-4 = 8

So, when you have variables in parentheses, you can multiply them out.
(a-b)*c = a*c - b*c

And the same works for division as well as multiplication. Say you have an equation like

(x+5)*2 = 16

And want to solve for x:
x*2 + 5*2 = 16
2x + 10 = 16
2x = 16-10
2x = 6
x = 6/2
x = 3

Answer 12

That last equation is just an example of course, since it would be easier to solve by just dividing by 2 and subtracting 5, but you get the idea ;)

Answer 13

Another example:

\[(x+5x) * x = 24\]
\[x^2 + 5x^2 = 24\]
\[6x^2 = 24\]
\[x^2 = 4\]
\[x = 2\]

Answer 14

Ya I got it. i thought 1/(2x-2). that's y all dis confusion. So sorry abt t.

Answer 15

Thanx a lot 4 dis biggest help.

Answer 16

glad to help :)