# Given A=((-1, 2), (3, 4)) and B=((-4, 3), (5, -2)), how do you find A-2B?

Apr 19, 2016

Follow the order of operations to find:
$A - 2 B = \left(\begin{matrix}- 1 & 2 \\ 3 & 4\end{matrix}\right) - \left(\begin{matrix}- 8 & 6 \\ 10 & - 4\end{matrix}\right) = \left(\begin{matrix}7 & - 4 \\ - 7 & 8\end{matrix}\right)$

#### Explanation:

To solve a matrix equation we follow the normal order of operations with the added restriction that multiplication and division need to happen in the order that they are written, since for matrices, $A B \ne B A$ in general (there are special cases where this is true).

So for our equation, $A - 2 B$, we need to start with the multiplication $2 B$. Multiplying a scalar, $2$, by a matrix, $B$, has the effect of multiplying each of the matrix elements by the scalar, therefore,

$2 B = 2 \cdot \left(\begin{matrix}- 4 & 3 \\ 5 & - 2\end{matrix}\right) = \left(\begin{matrix}- 8 & 6 \\ 10 & - 4\end{matrix}\right)$

subtracting two matrices requires that each matrix have the same dimensions - this is true in our case since $A$ and $B$ are both $2 \times 2$ matrices. Subtracting matrices results in the subtraction of each element from the corresponding element in the other matrix, i.e.

$\left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right) - \left(\begin{matrix}{b}_{11} & {b}_{12} \\ {b}_{21} & {b}_{22}\end{matrix}\right) = \left(\begin{matrix}{a}_{11} - {b}_{11} & {a}_{12} - {b}_{12} \\ {a}_{21} - {b}_{21} & {a}_{22} - {b}_{22}\end{matrix}\right)$

in our case

$A - 2 B = \left(\begin{matrix}- 1 & 2 \\ 3 & 4\end{matrix}\right) - \left(\begin{matrix}- 8 & 6 \\ 10 & - 4\end{matrix}\right) = \left(\begin{matrix}7 & - 4 \\ - 7 & 8\end{matrix}\right)$