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## Solution to Linear Algebra Done Wrong Exercise 1.3.1

Multiply

a) $\begin{pmatrix} 1 & 2 & 3\\ 4 &5 & 6\end{pmatrix}\begin{pmatrix} 1 \\ 3\\ 2 \end{pmatrix}$;

b) $\begin{pmatrix} 1 & 2 \\ 0 & 1\\ 2 & 0\end{pmatrix}\begin{pmatrix} 1 \\ 3 \end{pmatrix}$;

c) $\begin{pmatrix} 1 & 2 & 0 & 0\\ 0 & 1 & 2 & 0\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix} 1 \\ 2\\ 3 \\ 4 \end{pmatrix}$;

d) $\begin{pmatrix} 1 & 2 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\\ 0 & 0 &0\end{pmatrix}\begin{pmatrix} 1 \\ 2\\ 3 \\ 4 \end{pmatrix}$.

Solution:

a) $\begin{pmatrix} 1 & 2 & 3\\ 4 &5 & 6\end{pmatrix}\begin{pmatrix} 1 \\ 3\\ 2 \end{pmatrix}=\begin{pmatrix} 1\cdot 1+ 2\cdot 3+ 3 \cdot 2\\ 4 \cdot 1+5\cdot 3+6\cdot 2\end{pmatrix}=\begin{pmatrix} 11\\ 31\end{pmatrix}$;

b) $\begin{pmatrix} 1 & 2 \\ 0 & 1\\ 2 & 0\end{pmatrix}\begin{pmatrix} 1 \\ 3 \end{pmatrix}=\begin{pmatrix}1\cdot 1+2\cdot 3\\ 0\cdot 1+1\cdot 3\\ 2\cdot 1+0\cdot 3\end{pmatrix}=\begin{pmatrix}7\\ 3\\ 2\end{pmatrix}$;

c) $\begin{pmatrix} 1 & 2 & 0 & 0\\ 0 & 1 & 2 & 0\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3 \\4 \end{pmatrix}=\begin{pmatrix}1\cdot 1+2\cdot 2+0\cdot 3+0\cdot 4\\ 0\cdot 1+1\cdot 2+2\cdot 3+0\cdot 4\\ 0\cdot 1+0\cdot 2+1\cdot 3+2\cdot 4 \\ 0\cdot 1+0\cdot 2+0\cdot 3+1\cdot 4\end{pmatrix}=\begin{pmatrix}5\\ 8\\ 11\\4\end{pmatrix}$;

d) $\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3 \\4 \end{pmatrix}=\begin{pmatrix}1\cdot 1+2\cdot 2+0\cdot 3\\ 0\cdot 1+1\cdot 2+2\cdot 3\\ 0\cdot 1+0\cdot 2+1\cdot 3 \\ 0\cdot 1+0\cdot 2+0\cdot 3\end{pmatrix}=\begin{pmatrix}5\\ 8\\ 3\\0\end{pmatrix}$.

### This Post Has One Comment

1. a) should be [13 31]. 1+6+6 = 13