Find the value (A power of T) power -1 where
A power -1 =
1 2 3
2 0 1
1 1 -1
the answer in the book is
1 2 1
2 0 1
3 1 -1
Not sure how they got this.
What you mean is (A^T)-1 and your trying to find the orginal matrix A. Well first find the determinant of [1 2 3] [2 0 1] [1 1 -1 ] which is 11. Multiply everything by 11. To get A^T which is [11 22 33][22 0 11][11 11 -11]. Then undo the transpose which would be [11 22 11][22 0 11][33 11 -11] and reduce the matrix by 1/11 which = [1 2 1][2 0 1][3 1 -1].
Yes, it is. You have already determined that the transpose of the matrix (not a "power") is $\displaystyle \begin{bmatrix}1 & 2 & 1 \\ 2 & 0 & 1 \\ 3 & 1 & -1 \end{bmatrix}$.
Now you can write that matrix with the identity matrix right beside it:
$\displaystyle \begin{bmatrix}1 & 2 & 1 \\ 2 & 0 & 1 \\ 3 & 1 & -1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$.
Do you know what "row operations" are? if so, row reduce the first matrix to the identity matrix and, at the same time, apply those row operations to the identity matrix. When the first matrix has been reduced to the identity matrix, the second matrix will be the inverse.
(It seems very strange to me that you would be asked to invert a matrix if you have been taught nothing about finding an inverse matrix!)
here is how to use row-reduction find find an inverse:
start with your matrix A, and "augment it" with the identity matrix I.
now row-reduce A, but apply the same row-reductions to I at the same time.
when you have row-reduced A to I, the results of what you have done to the identity matrix will be A^{-1}.
i will illustrate with a 2x2 matrix:
suppose
$\displaystyle A = \begin{bmatrix}2&3\\3&5 \end{bmatrix}$.
form this augmented matrix:
$\displaystyle \begin{bmatrix}2&3&|&1&0\\3&5&|&0&1 \end{bmatrix}$
first row operation: (-3/2)R_{1}+R_{2}:
$\displaystyle \begin{bmatrix}2&3&|&1&0\\0&\frac{1}{2}&|&-\frac{3}{2}&1 \end{bmatrix}$
second row operation: 2R_{2}:
$\displaystyle \begin{bmatrix}2&3&|&1&0\\0&1&|&-3&2 \end{bmatrix}$
third row operation: -3R_{2}+R_{1}:
$\displaystyle \begin{bmatrix}2&0&|&10&-6\\0&1&|&-3&2 \end{bmatrix}$
fourth row operation: (1/2)R_{1}:
$\displaystyle \begin{bmatrix}1&0&|&5&-3\\0&1&|&-3&2 \end{bmatrix}$
this tells us:
$\displaystyle A^{-1} = \begin{bmatrix}5&-3\\-3&2 \end{bmatrix}$, which you can verify by multiplying.