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Recitation 2B

Example 1: Given u=[1−2]u=\begin{bmatrix}1\\-2\end{bmatrix} and v=[2−5]v=\begin{bmatrix}2\\-5\end{bmatrix}, find 4u,(−3)v4\mathbf{u}, (-3)\mathbf{v} and 4u+(−3)v4\mathbf{u}+(-3)\mathbf{v}.

Solution: 4u=[4−8]4\mathbf{u}=\begin{bmatrix}4\\-8\end{bmatrix}, (−3)v=[−615](-3)\mathbf{v}=\begin{bmatrix}-6\\15\end{bmatrix} and 4u+(−3)v=[−27]4\mathbf{u}+(-3)\mathbf{v}=\begin{bmatrix}-2\\7\end{bmatrix}.

Example 2: Let a1=[1−2−5]\mathbf{a_1}=\begin{bmatrix}1\\-2\\-5\end{bmatrix}, a2=[256]\mathbf{a_2}=\begin{bmatrix}2\\5\\6\end{bmatrix} and b=[74−3]\mathbf{b}=\begin{bmatrix}7\\4\\-3\end{bmatrix}. Determine whether b\mathbf{b} can be generated as a linear combination of a1\mathbf{a_1} and a2\mathbf{a_2}. That is, determine whether weights x1x_1 and x2x_2 exist such that x1a1+x2a2=b.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 x1a1+x2a2=bx_1\mathbf{a_1}+x_2\mathbf{a_2}=\mathbf{b}. [127−254−56−3]\begin{bmatrix}1 & 2 & 7\\-2 & 5 & 4\\ -5 & 6 & -3\end{bmatrix} Reduce this matrix to its echelon form [127012000]\begin{bmatrix}1 & 2 & 7\\0 & 1 & 2\\ 0 & 0 & 0\end{bmatrix} which tells us x1=3,x2=2.x_1 = 3, x_2=2.

Example 3: For v1,v2,v3\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} in Rm\mathbb{R}^m, write the linear combination 3v1−5v2+7v33\mathbf{v_1}-5\mathbf{v_2}+7\mathbf{v_3} as a matrix times a vector.

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

Example 4: Compute AxA\mathbf{x}, where A=[234−15−36−28]A=\begin{bmatrix}2 & 3 & 4\\ -1 & 5 & -3\\ 6 & -2 & 8\end{bmatrix} and x=[x1x2x3]x=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}.

Solution: Use the Row-Vector Rule, we have [234−15−36−28][x1x2x3]=[2x1+3x2+4x3−x1+5x2−3x36x1−2x2+8x3].\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. 3x1+5x2−4x3=0−3x1−2x2+4x3=06x1+x2−8x3=0\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 [35−40−3−24061−80].\begin{bmatrix}3 & 5 & -4 & 0\\ -3 & -2 & 4 & 0\\ 6 & 1 & -8 & 0\end{bmatrix}. We obtain [30−4003000000]\begin{bmatrix}3 & 0 & -4 & 0\\0 & 3 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}

which tells us that the solution set is x1=4x3/3,x2=0x_1 = 4x_3/3, x_2 = 0 and x3x_3 is a free variable. For instance, when x3=3x_3=3, x1=4,x2=0,x3=3x_1=4, x_2=0, x_3=3 is a nontrivial solution.

Example 6: Describe all solutions of Ax=bA\mathbf{x}=\mathbf{b}, where A=[35−4−3−2461−8]A=\begin{bmatrix}3 & 5 & -4\\-3 & -2 & 4\\6 & 1 & -8\end{bmatrix} and b=[7−1−4].\mathbf{b}=\begin{bmatrix}7\\-1\\-4\end{bmatrix}.

Solution: Apply row reduction on the correspondent augmented matrix [35−47−3−24−161−8−4].\begin{bmatrix}3 & 5 & -4 & 7\\ -3 & -2 & 4 & -1\\ 6 & 1 & -8 & -4\end{bmatrix}. We obtain [35−4703060−90−18]\begin{bmatrix}3 & 5 & -4 & 7\\ 0 & 3 & 0 & 6\\ 0 & -9 & 0 & -18\end{bmatrix} and then [30−4−301020000],\begin{bmatrix}3 & 0 & -4 & -3\\ 0 & 1 & 0 & 2\\ 0 & 0 & 0 & 0\end{bmatrix},

which tells us x1=−1+4x3/3,x2=2x_1=-1+4x_3/3, x_2=2 and x3x_3 is a free variable.