Question

F is a transfomation of R^3 to `R^3` so that f(x,y,z)=(x-y,x+y,z-x) How can i find the standart matrix of this transformation and how can i find the inverse of this transformation? I am not sure how to approach this problem

Answer 1

The matrix is `F=[(1,–1,0),(1,1,0),(–1,0,1)]`.

The inverse matrix is `F^(–1)=[(1/2,1/2,0),(–1/2,1/2,0),(1/2,1/2,1)]`.

Explanation:

A linear transformation such as

`f(x,y,z) = (x-y,"  "x+y,"  "z-x)`

can be expressed as matrix multiplication. Just like how the function `f` takes in the vector `stackrelrarrx = (x, y, z)` and gives back a new vector `(x-y,"  "x+y,"  "z-x),` we are looking for the `3xx3` matrix `F` that satisfies

`F xx stackrelrarrx=[(x-y),(x+y),(z-x)]`.

Let's remember how matrix multiplication works: when multiplying a `3xx3` matrix `[(a,b,c),(d,e,f),(g,h,i)]` with a `3xx1` vector `[(x),(y),(z)],` the result is the `3xx1` vector as follows:

`[(a,b,c),(d,e,f),(g,h,i)]_ color(red)(3 xx 3) xx [(x),(y),(z)]_ color(red)(3xx1) = [(ax+by+cz),(dx+ey+fz),(gx+hy+iz)]_ color(red)(3 xx 1)`

We want to find what `a,b,c,...,i` need to be so that the product vector (on the right) matches the output vector from the function `f.`

In other words, we want `[(ax+by+cz),(dx+ey+fz),(gx+hy+iz)]=[(x-y),(x+y),(z-x)].`

It may still look like a jumble of letters, but remember—we're going to choose what `a` through `i` are. Let's look at the first line of the equality:

`ax+by+cz=x-y`

What do `a, b,` and `c` have to be to make this true? Well, on the right we have `1x` and `–1y,` so to make the left side match, we should choose `a=1,` `b=–1,` and `c=0:`

`1x+(–1)y+0z=x-y`

The process is the same for the other two rows. The matrix is

`F=[(1,–1,0),(1,1,0),(–1,0,1)].`

Notice:

the first row contains the coefficients of `x-y,`
the second row contains the coefficients of `x+y,` and
the third row contains the coefficients of `z-x,`

all in `x,y,z` order, of course.

The inverse matrix `F^(–1)` is the matrix which "undoes" the action done by `F.` That is, just as `F` takes the input `stackrelrarrx` and gives `Fstackrelrarrx`, the matrix `F^(–1)` takes in `Fstackrelrarrx` as input and gives back just `stackrelrarrx:`

`F^(–1) xx Fstackrelrarrx = stackrelrarrx`

or

`F^(–1) xx [(x-y),(x+y),(z-x)] = [(x),(y),(z)].`

You can find this inverse the same way as before; this one's not too hard to visualize, though others may take some effort. You can also find it by computing the inverse of `F.` Instructions should be available in any introductory matrix algebra textbook.

`F^(–1)=[(1/2,1/2,0),(–1/2,1/2,0),(1/2,1/2,1)]`