Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 04 (b)/Solution 1

The columns of a linear transformation matrix, A, are populated by the transformed vectors of each of the basis vectors. In this case the first column of A will be $[1,0,0]\times \mathbf {a}$, the second, $[0,1,0]\times \mathbf {a}$, and the third, $[0,0,1]\times \mathbf {a}$. By computing these cross products, we get,

${\begin{aligned}\left[1,0,0\right]\times \mathbf {a} =[1,2,-1]&=[0,1,2]\\\left[0,1,0\right]\times \mathbf {a} =[1,2,-1]&=[-1,0,-1]\\\left[0,0,1\right]\times \mathbf {a} =[1,2,-1]&=[-2,1,0]\end{aligned}}.$

Therefore the linear transformation matrix A which corresponds to T is,

$A={\begin{bmatrix}0&-1&-2\\1&0&1\\2&-1&0\end{bmatrix}}$.

Notice if we wanted some sort of confirmation of our answer we can pick a random vector, say x=[3,4,7], compute $\mathbf {x} \times \mathbf {a}$ and compute Ax and make sure they are the same thing.

$[3,4,7]\times [1,2,-1]=[-18,10,2]$

and

$A{\begin{bmatrix}3\\4\\7\end{bmatrix}}={\begin{bmatrix}-18\\10\\2\end{bmatrix}}$

which is indeed the same result.