# Thread: Matrix (I think) question

1. ## 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

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

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.

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

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

where,

$det(A)=ad-bc$

So once you use that plug-and-go formula you just multiply the two matrices... $AA^{-1}$ if you multiply in opposite order $A^{-1}A$ it will be the same, which will be $I= \left[\begin{array}{ c c }1& 0 \\0 & 1 \end{array} \right]$ which is what they're asking you to demonstrate.

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

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

where,

$det(A)=ad-bc$

So once you use that plug-and-go formula you just multiply the two matrices... $AA^{-1}$ if you multiply in opposite order $A^{-1}A$ it will be the same, which will be $I= \left[\begin{array}{ c c }1& 0 \\0 & 1 \end{array} \right]$ which is what they're asking you to demonstrate.
Great! Thanks!