# Invertible matrix

• Nov 20th 2012, 11:10 AM
tinabaker
Invertible matrix
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.
• Nov 20th 2012, 11:23 AM
bonfire09
Re: Invertible matrix
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].
• Nov 20th 2012, 11:25 AM
tinabaker
Re: Invertible matrix
I did not study a determinant yet.

Is it possible to calulate without determinant?
• Nov 20th 2012, 11:26 AM
tinabaker
Re: Invertible matrix
what is (A^T)
of
1 2 3
2 0 1
1 1 -1
?

Thank you!
• Nov 20th 2012, 11:47 AM
tinabaker
Re: Invertible matrix
what is (A^T)
of
1 2 3
2 0 1
1 1 -1

is
1 2 1
2 0 1
3 1 (-1)
• Nov 20th 2012, 11:50 AM
fkf
Re: Invertible matrix
To find the transpose of a given matrix. Make first row become first column, and the first column to be the first row etc.

A =
[1 2]
[3 4]

A^T =
[1 3]
[2 4]
• Nov 20th 2012, 01:19 PM
HallsofIvy
Re: Invertible matrix
Quote:

Originally Posted by tinabaker
I did not study a determinant yet.

Is it possible to calulate without determinant?

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!)
• Nov 21st 2012, 08:20 AM
tinabaker
Re: Invertible matrix
I am doing my homework ahead of class...well trying. Thank for your help!
• Nov 21st 2012, 08:43 AM
tinabaker
Re: Invertible matrix
yes, I know row operations.
1 2 1
2 0 1
3 1 -1
will be
1 0 0
0 1 0
0 0 1.

How can I use it after that?

the answer in the book is

1 2 1
2 0 1
3 1 -1
• Nov 21st 2012, 12:41 PM
Deveno
Re: Invertible matrix
here is how to use row-reduction find find an inverse:

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)R1+R2:

$\displaystyle \begin{bmatrix}2&3&|&1&0\\0&\frac{1}{2}&|&-\frac{3}{2}&1 \end{bmatrix}$

second row operation: 2R2:

$\displaystyle \begin{bmatrix}2&3&|&1&0\\0&1&|&-3&2 \end{bmatrix}$

third row operation: -3R2+R1:

$\displaystyle \begin{bmatrix}2&0&|&10&-6\\0&1&|&-3&2 \end{bmatrix}$

fourth row operation: (1/2)R1:

$\displaystyle \begin{bmatrix}1&0&|&5&-3\\0&1&|&-3&2 \end{bmatrix}$

this tells us:

$\displaystyle A^{-1} = \begin{bmatrix}5&-3\\-3&2 \end{bmatrix}$, which you can verify by multiplying.
• Nov 21st 2012, 03:07 PM
tinabaker
Re: Invertible matrix
thank you.