# If $$A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$$ and $$B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$$ then find $( A + 2B)'.$

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$j $\leq$ n.
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
Step1:
Given:
$A'=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix}$
We know that (A')'=A
The transpose of a matrix can be obtained by changing the rows & the column.
$A=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}$
$B=\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$
$2B=2\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$
$2B=\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}$
Step2:
$A+2B=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}+\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}$
$\qquad\;\;\;\;=\begin{bmatrix}-4 & 1\\5 & 6\end{bmatrix}$
The transpose of a matrix can be obtained by changing the rows & the column.
$(A+2B)'=\begin{bmatrix}-4 & 5\\1 & 6\end{bmatrix}$

edited Dec 24, 2013