M240:Linear Algebra Spring 2023
Unit One
Homework Help


Section 1.1

8. "Describe the solution set" means find the solution. It will be a unique soltion. Scale row three before using it in row replace of row two.
14. The augmented matrix is $\left[\begin{array}{rrr}1 & -3 & 0 & | & 5 \\ -1 & 1 & 5 & | & 2 \\ 0 & 1 & 1 & | & 0 \end{array}\right]$.
22. Use the augmented matrix for this system if you prefer.
T/F Two are False.


Section 1.2

4. Three rows means there can be, at most, 3 pivot columns.
8. Two equation and three variable create the possibiliity of a free variable because not every variable column can basic.
14. There are two free variables.
20. One of these systems has infinitely many solutions.
T/F. Two are True.


Section 1.3

8. For example, $\vec{w} = 2\vec{v} - \vec{u}$.
Is every vector in $\mathbb{R}^2$ a linear cmbination of $\vec{v}$ and $\vec{u}$? Remember that the coefficients of the vectors can be any real number (fractions).
12. Set up the augmented matrix and reduce it until you can determine if the last column is a pivot column.
14. Is the last column a pivot column or not?
16. You are free to choose any real number values for weights $x_1$ and $x_2$ in the linear combination expression $x_1\vec{v}_1 + x_2\vec{v}_2$. Therefore, choose simple pairs such as $(x_1, x_2) = (0, 0), (1, 0), ...$
18. Augmented matrix.
T/F. Both are true or both are false.
34. (a) Use an augmented matrix.

(b) What is a simple solution for $x_1$, $x_2$, and $x_3$ if you want $x_1\vec{v}_1 + x_2\vec{v}_2 + x_3\vec{v}_3 = \vec{v}_3$?.


Section 1.4

12. The solution is unique (that is, there is only one vector $\vec{x} = (x_1, x_2, x_3)$ that solves this system.
14. Use an augmented matrix. Is the last column a pivot column?
T/F. Just one statement is false.


Section 1.5

2. The associated matrix A for the homogeneous system must have a free variable (non-pivot column) for a non-trivial solution.
4. The associated matrix A for the homogeneous system has more columns than rows, so there must be at least one free variable.
22. The number of free variables determines the dimension of the solution set. One solution set is a plane through the origin. How is the solution set to the other equation related to the first solution set?
T/F. Either all but one is true, or all but one is false.
36. Does $A\vec{x} = \vec{b}$ have a solution?
46. Is one column a multiple of the other?


Section 1.6

2. The free variable $p_s$ allows you to choose any positive value for $p_s$. Once $p_s$ is chosen, the values for the other variables can be determined.
3. (b)The matrix version for the case is (not all of the values are shown)
$\left[\begin{array}{rrr}0.8 & -0.8 & * \\ * & 0.9 & -0.4 \\ -0.5 & -0.1 & * \end{array}\right]$
(c) You will find that the price (value) of machinery, represented by $p_M$, is a free variable. Then, $p_C = 1.417p_m$ and $p_F = 0.917p_m$. Set $p_m$ equal to 100.
6. There is one free variable.
12. The are four nodes, so there will be four equations. The minimum value for $x_1$ is a integer between 35 and 50.


Section 1.7

2. Use an augmented matrix.
6. Are there any free variables?
12. Use an augmeted matrix.
16. Are the column vectors a multiple of each other?
18. Consider the number of vectors versus the dimension of the vectors (the number of entries in a vector).
20. Apply one of the theorems from this section.


Section 1.8

10. You may use technology to reduce the matrix. There is one free variable, so there are infinitely many solutions. You may use technology to reduce the matrix.
14. This transformation represents a contraction of the vector.
20. Use the basic definition of $A\vec{x}$ to determine the matrix A.
T/F. Two are true and two are false.
32. Spanning means any vector in $\mathbb{R}^n$ can be writtten as a linear combination of the vectors $\vec{v}_1, \vec{v}_2, ... \vec{v}_p$. Then, use Equation (5), page 70.
Suppose $\vec{u}$ is an arbitrary vector from $\mathbb{R}^n$. Therefore, we know there exists scalars $c_1, c_2, ... c_p$ such that $\vec{u} = c_1\vec{v}_1 + c_2\vec{v}_2 + ... + c_p\vec{v}_p$. Then \begin{align} T(\vec{u}) &= T(c_1\vec{v}_1 + c_2\vec{v}_2 + ... + c_p\vec{v}_p)\\ &= c_1T(\vec{v}_1) + c_2T(\vec{v}_2) + ... + c_pT(\vec{v}_p)) \hspace{0.25in} (\mbox{using (5) on page 70)}\\ &= c_1\vec{0} + c_2\vec{0} + ... c_p\vec{0} \hspace{0.25in} (\mbox{given that} T(\vec{v}_i) = \vec{0})\\ &= \vec{0} \\ \label{eq:vector_subtraction} \end{align}
44. Use either the definition of a linear transformation or Equations (3) and (4) on page 70 to establish that $T$ so defined is a linear transformation. The steps in using the definition are shown below.
\begin{align} T(\vec{u} + \vec{v}) &= T((u_1, u_2, u_3) + (v_1, v_2, v_3))\\ &= T(u_1 + v_1, u_2+v_2, u_3+v_3) \hspace{0.25in} (\mbox{def of vector addition})\\ &= (u_1+v_1, 0, u_3+v_3) \hspace{0.25in} (\mbox{def of the transformation})\\ &= (u_1, 0, u_3) + (v_1, 0, v_3) \hspace{0.25in} (\mbox{def of vector addition}) \\ &= T(\vec{u}) + T(\vec{v}) \hspace{0.25in} (\mbox{def of the transformation}) \label{eq:vector sum} \end{align}
\begin{align} T(c\vec{u}) &= T(c(u_1, u_2, u_3))\\ &= T(cu_1, cu_2, cu_3) \hspace{0.25in} (\mbox{scalar-vector multiplication})\\ &= (cu_1, 0, cu_3) \hspace{0.25in} (\mbox{def of the transformation})\\ &= c(u_1, 0, u_3) \hspace{0.25in} (\mbox{scalar-vector multiplication})\\ &= cT(\vec{u}) \hspace{0.25in} (\mbox{def of the transformation})\\ \label{eq:scalar multiplication} \end{align}


Section 1.9

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44. Use either the definition of a linear transformation or Equations (3) and (4) on page 70 to establish that $T$ so defined is a linear transformation. The steps in using (3) and (4) are shown below.
\begin{align} T(S(\vec{0})) &= T(\vec{0}) \hspace{0.25in} (\mbox{because } S \mbox{ is a linear transformation})\\ &= \vec{0} \hspace{0.25in} (\mbox{because } T \mbox{ is a linear transformation}) \end{align}
\begin{align} T(S(c\vec{u} + d\vec{v})) &= T(cS(\vec{u} + dS(\vec{v})) \hspace{0.25in} (\mbox{because } S \mbox{ is a linear transformation})\\ &= cT(S(\vec{u})) + dT(S(\vec{v})) \hspace{0.25in} (\mbox{because } T \mbox{ is a linear transformation}) \end{align}