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Solution to Linear Algebra Done Wrong Exercise 1.1.1

Let ${\bf x} = (1,2,3)^T$, ${\bf y} = (y_1,y_2,y_3)^T$, ${\bf z}= (4,2,1)^T$. Compute $2{\bf x}$, $3{\bf y}$, ${\bf x}+2{\bf y}-3{\bf z}$.

Solution: We have $$ 2{\bf x}=\begin{pmatrix}2\\\\ 4\\\\ 6\end{pmatrix},\qquad 3{\bf y}=\begin{pmatrix}3y_1\\\\ 3y_2\\\\ 3y_3\end{pmatrix}, $$ $$ {\bf x}+2{\bf y}-3{\bf z}=\begin{pmatrix}2y_1-11\\\\ 2y_2-4\\\\ 2y_3\end{pmatrix}. $$

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