# Recitation 2B

Example 1: Given $u=\begin{bmatrix}1\\-2\end{bmatrix}$ and $v=\begin{bmatrix}2\\-5\end{bmatrix}$, find $4\mathbf{u}, (-3)\mathbf{v}$ and $4\mathbf{u}+(-3)\mathbf{v}$.

Solution: $4\mathbf{u}=\begin{bmatrix}4\\-8\end{bmatrix}$, $(-3)\mathbf{v}=\begin{bmatrix}-6\\15\end{bmatrix}$ and $4\mathbf{u}+(-3)\mathbf{v}=\begin{bmatrix}-2\\7\end{bmatrix}$.

Example 2: Let $\mathbf{a_1}=\begin{bmatrix}1\\-2\\-5\end{bmatrix}$, $\mathbf{a_2}=\begin{bmatrix}2\\5\\6\end{bmatrix}$ and $\mathbf{b}=\begin{bmatrix}7\\4\\-3\end{bmatrix}$. Determine whether $\mathbf{b}$ can be generated as a linear combination of $\mathbf{a_1}$ and $\mathbf{a_2}$. That is, determine whether weights $x_1$ and $x_2$ exist such that $$x_1\mathbf{a_1} + x_2\mathbf{a_2} = \mathbf{b}.$$ If the vector equation has a solution, find it.

Solution: The following augmented matrix corresponds to the vector equation $x_1\mathbf{a_1}+x_2\mathbf{a_2}=\mathbf{b}$. $$\begin{bmatrix}1 & 2 & 7\\-2 & 5 & 4\\ -5 & 6 & -3\end{bmatrix}$$ Reduce this matrix to its echelon form $$\begin{bmatrix}1 & 2 & 7\\0 & 1 & 2\\ 0 & 0 & 0\end{bmatrix}$$ which tells us $$x_1 = 3, x_2=2.$$

Example 3: For $\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}$ in $\mathbb{R}^m$, write the linear combination $3\mathbf{v_1}-5\mathbf{v_2}+7\mathbf{v_3}$ as a matrix times a vector.

Solution: Suppose $A$ is an $m\times 3$ matrix and $\mathbf{x}$ is in $\mathbb{R}$

Example 4: Compute $A\mathbf{x}$, where $A=\begin{bmatrix}2 & 3 & 4\\ -1 & 5 & -3\\ 6 & -2 & 8\end{bmatrix}$ and $x=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$.

Solution: Use the Row-Vector Rule, we have $$\begin{bmatrix}2 & 3 & 4\\ -1 & 5 & -3\\ 6 & -2 & 8\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}2x_1+3x_2+4x_3\\-x_1+5x_2-3x_3\\6x_1-2x_2+8x_3\end{bmatrix}.$$

Example 5: Determine if the following homogeneous system has a nontrivial solution. Then describe the solution set. \begin{aligned}3x_1 + 5x_2 - 4x_3 &=& 0\\-3x_1 - 2x_2 + 4x_3 &=& 0\\6x_1 + x_2 - 8x_3 &=& 0\end{aligned}

Solution: Apply row reduction on the correspondent augmented matrix $$\begin{bmatrix}3 & 5 & -4 & 0\\
-3 & -2 & 4 & 0\\
6 & 1 & -8 & 0\end{bmatrix}.$$
We obtain $$\begin{bmatrix}3 & 0 & -4 & 0\\0 & 3 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}$$
which tells us that the solution set is $x_1 = 4x_3/3, x_2 = 0$ and $x_3$ is a free variable. For instance, when $x_3=3$, $x_1=4, x_2=0, x_3=3$ is a nontrivial solution.

Example 6: Describe all solutions of $A\mathbf{x}=\mathbf{b}$, where $$A=\begin{bmatrix}3 & 5 & -4\\-3 & -2 & 4\\6 & 1 & -8\end{bmatrix}$$ and $$\mathbf{b}=\begin{bmatrix}7\\-1\\-4\end{bmatrix}.$$

Solution: Apply row reduction on the correspondent augmented matrix $$\begin{bmatrix}3 & 5 & -4 & 7\\ -3 & -2 & 4 & -1\\ 6 & 1 & -8 & -4\end{bmatrix}.$$ We obtain $$\begin{bmatrix}3 & 5 & -4 & 7\\ 0 & 3 & 0 & 6\\ 0 & -9 & 0 & -18\end{bmatrix}$$ and then $$\begin{bmatrix}3 & 0 & -4 & -3\\ 0 & 1 & 0 & 2\\ 0 & 0 & 0 & 0\end{bmatrix},$$
which tells us $x_1=-1+4x_3/3, x_2=2$ and $x_3$ is a free variable.