# Matrices

• Nov 19th 2009, 09:44 AM
Matrices

Im trying to evaluate the following, but not sure if i'm going wrong,could someone please check this.

Evaluating 2A + 3B

Where

$\displaystyle A = \left(\begin{array}{cc}3&7\\4&-1\\-5&2\end{array}\right)$ $\displaystyle B = \left(\begin{array}{cc}-5&-3\\10&6\end{array}\right)$

To evaluate i use the scalar multiplication?

so:

2A becomes = $\displaystyle \left(\begin{array}{cc}6&14\\8&-2\\-10&4\end{array}\right)$

and

3B becomes= $\displaystyle \left(\begin{array}{cc}-15&9\\30&18\end{array}\right)$

So my dillema is, i cant add these together can i because the incorrect rows to columns - or am i thinking this through to much?

• Nov 19th 2009, 09:57 AM
masters
Quote:

Im trying to evaluate the following, but not sure if i'm going wrong,could someone please check this.

Evaluating 2A + 3B

Where

$\displaystyle A = \left(\begin{array}{cc}3&7\\4&-1\\-5&2\end{array}\right)$ $\displaystyle B = \left(\begin{array}{cc}-5&-3\\10&6\end{array}\right)$

To evaluate i use the scalar multiplication?

so:

2A becomes = $\displaystyle \left(\begin{array}{cc}6&14\\8&-2\\-10&4\end{array}\right)$

and

3B becomes= $\displaystyle \left(\begin{array}{cc}-15&9\\30&18\end{array}\right)$

So my dillema is, i cant add these together can i because the incorrect rows to columns - or am i thinking this through to much?

No, your thinking is just fine. You cannot add matrices with different dimensions.
• Nov 19th 2009, 09:59 AM
Thanks Masters, so is that evaluated as far as possible?
• Nov 19th 2009, 10:08 AM
masters
Quote: