Problem with matrices question, finding A^2

The question is this one below:

Let

A =

(1 -3)

2 0

How am I supposed to find A^2?

Also Let

B =

(-6 3 )

8 20

How do I find Matrix X such that 5X-A=B ?

Full explanations would be nice, someone tried to explain it to me briefly but I didn't quite understand it.

Re: Problem with matrices question, finding A^2

Quote:

Originally Posted by

**DaveWolfgang** How am I supposed to find A^2?

$\displaystyle A^2 = A\cdot A = \left[\begin{array}{cc}1&-3\\2&0\end{array}\right]\cdot\left[\begin{array}{cc}1&-3\\2&0\end{array}\right]$

Quote:

Originally Posted by

**DaveWolfgang** How do I find Matrix X such that 5X-A=B ?

You know that $\displaystyle X$ has to be a 2×2 matrix for the subtraction to be valid. If we let

$\displaystyle X = \left[\begin{array}{cc}a&b\\c&d\end{array}\right],$

then

$\displaystyle 5X - A = B$

$\displaystyle \Rightarrow5\left[\begin{array}{cc}a&b\\c&d\end{array}\right] - \left[\begin{array}{cc}1&-3\\2&0\end{array}\right] = \left[\begin{array}{cc}-6&3\\8&20\end{array}\right]$

$\displaystyle \Rightarrow\left[\begin{array}{cc}5a-1&5b+3\\5c-2&5d\end{array}\right] = \left[\begin{array}{cc}-6&3\\8&20\end{array}\right]$

Now equate the corresponding entries.

Re: Problem with matrices question, finding A^2

Quote:

Originally Posted by

**DaveWolfgang** How do I find Matrix X such that 5X-A=B ?

Or just X = (1/5)(B + A).