QQuestion
If
A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\quad B=\begin{bmatrix}1&2\\2&4\end{bmatrix},\quad C=\begin{bmatrix}4&1\\1&5\end{bmatrix},\quad I=\begin{bmatrix}1&0\\0&1\end{bmatrix}find A(B + C) − 14I.
✓Answer
First, add B and C:
B+C=\begin{bmatrix}1&2\\2&4\end{bmatrix}+\begin{bmatrix}4&1\\1&5\end{bmatrix} B+C=\begin{bmatrix}5&3\\3&9\end{bmatrix}Now,
A(B+C)-14I=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}5&3\\3&9\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}= \begin{bmatrix}1(5)+3(3)&1(3)+3(9)\\2(5)+4(3)&2(3)+4(9)\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}
= \begin{bmatrix}5+9&3+27\\10+12&6+36\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}
= \begin{bmatrix}14&30\\22&42\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}
= \begin{bmatrix}0&30\\22&28\end{bmatrix}
Therefore, A(B + C) − 14I = \begin{bmatrix}0&30\\22&28\end{bmatrix}.
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