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\\<br /> -3 & -2 & 4 & 0\\<br /> 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.

Leave a Reply

Your email address will not be published. Required fields are marked *