# Matrix (I think) question

• Jul 21st 2013, 09:12 AM
Paze
Matrix (I think) question
I'm reading and learning through a finals exam for what I will be learning next year (I'm usually a year ahead) and there's a question:

A). Find the transpose of the matrix A

http://i.imgur.com/MEWDtOI.png

B). Show that $\displaystyle A^{-1}\cdot A=A\cdot A^{-1}=I$

I think this is related to matrices but I'm not sure what the question begs at all. I'm not sure if that's a I or l on the end of the equation either. No values seem to be given for this I/l.
• Jul 21st 2013, 12:40 PM
Re: Matrix (I think) question
Do you know the definitions of transpose and inverse matrices? These are just simple operations, for the transpose you just interchange rows and columns.

If $\displaystyle A=\left[\begin{array}{ c c }a & b \\c & d \end{array} \right]$ then you can find the inverse using the formula:

$\displaystyle A^{-1}=\frac{1}{detA}\left[\begin{array}{ c c }d &- b \\-c & a \end{array} \right]$

where,

$\displaystyle det(A)=ad-bc$

So once you use that plug-and-go formula you just multiply the two matrices...$\displaystyle AA^{-1}$ if you multiply in opposite order $\displaystyle A^{-1}A$ it will be the same, which will be $\displaystyle I= \left[\begin{array}{ c c }1& 0 \\0 & 1 \end{array} \right]$ which is what they're asking you to demonstrate.
• Jul 21st 2013, 06:10 PM
Paze
Re: Matrix (I think) question
Quote:

If $\displaystyle A=\left[\begin{array}{ c c }a & b \\c & d \end{array} \right]$ then you can find the inverse using the formula:
$\displaystyle A^{-1}=\frac{1}{detA}\left[\begin{array}{ c c }d &- b \\-c & a \end{array} \right]$
$\displaystyle det(A)=ad-bc$
So once you use that plug-and-go formula you just multiply the two matrices...$\displaystyle AA^{-1}$ if you multiply in opposite order $\displaystyle A^{-1}A$ it will be the same, which will be $\displaystyle I= \left[\begin{array}{ c c }1& 0 \\0 & 1 \end{array} \right]$ which is what they're asking you to demonstrate.