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

$\begin{array}{1 1} \begin{bmatrix}-4 & 5\\1 & 6\end{bmatrix} \\\begin{bmatrix}4 & 4\\ 1 & 6\end{bmatrix} \\ \begin{bmatrix}-5 & 4\\1 & 6\end{bmatrix}\\\begin{bmatrix}-6 & 5\\0 & 6\end{bmatrix}\end{array}$

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 $1 \leq j \leq n.$
Given $A'=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix} \rightarrow$ $A=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}$
$B=\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$
$A+2B=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}+2\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}+\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-2-2 & 1+0\\3+2 & 2+4\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-4 & 1\\5 & 6\end{bmatrix}$
$(A+2B)'=\begin{bmatrix}-4 & 5\\1 & 6\end{bmatrix}$
edited Mar 13, 2013