(3 points) Which of the following statements are true? (May contain one or more true statements, or none.)
A.
The equation Ar = b is consistent if the augmented matrix [Ab] has a pivot position in eve

Question

(3 points) Which of the following statements are true? (May contain one or more true statements, or none.)
A.
The equation Ar = b is consistent if the augmented matrix [Ab] has a pivot position in every row.
B.
If A is an m x n matrix and if the equation Az b is inconsistent for some b in IR™, then A cannot have a pivot position in every row.
C.
If the columns of an mx n matrix A span R", then the equation Axb is consistent for each b in R.
D.
If A is an mx n matrix whose columns do not span R"", then the equation Az b is inconsistent for some b in R.
E. The first entry in the product Az is a sum of products.
F.
The equation Az b is referred to as a vector equation.

VIPSolutions · Accepted Answer

# Analysis of Statements

Let's examine each statement for correctness.

---

## Statement A

> The equation $ Ax = b $ is consistent if the augmented matrix $[A\ b]$ has a pivot position in every row.

**False.**  
If $[A\ b]$ has a pivot in every row, this means there are no rows like $[ \ldots  \mid d]$ with $d \neq $, which *would* indicate inconsistency. However, the correct statement is: the system is consistent **if and only if** there is **no** row in the reduced form of $[A\ b]$ of the form $[ \ldots  \mid d]$ with $d \neq $. The presence of a pivot in every row (including possibly the last column) is not a guarantee of consistency.

---

## Statement B

> If $A$ is an $m \times n$ matrix and if the equation $Ax = b$ is inconsistent for some $b \in \mathbb{R}^m$, then $A$ cannot have a pivot position in every row.

**True.**  
If $A$ has a pivot in every row, then the columns of $A$ span $\mathbb{R}^m$, which means that $Ax = b$ is consistent for every $b$. The contrapositive: If $Ax = b$ is inconsistent for some $b$, $A$ does **not** have a pivot in every row.

---

## Statement C

> If the columns of an $m \times n$ matrix $A$ span $\mathbb{R}^m$, then the equation $Ax = b$ is consistent for each $b \in \mathbb{R}^m$.

**True.**  
By definition, if the columns of $A$ span $\mathbb{R}^m$, any $b$ can be expressed as a linear combination of the columns, so $Ax = b$ is always consistent.

---

## Statement D

> If $A$ is an $m \times n$ matrix whose columns do not span $\mathbb{R}^m$, then the equation $Ax = b$ is inconsistent for some $b \in \mathbb{R}^m$.

**True.**  
If the columns do not span $\mathbb{R}^m$, there exists at least one vector $b$ in $\mathbb{R}^m$ that cannot be written as a linear combination of the columns of $A$; for this $b$, $Ax = b$ is inconsistent.

---

## Statement E

> The first entry in the product $Ax$ is a sum of products.

**True.**  
Matrix multiplication is defined such that each entry in $Ax$ (including the first) is a sum of products of entries from the first row of $A$ and entries from $x$.

---

## Statement F

> The equation $Ax = b$ is referred to as a vector equation.

**True.**  
The equation $Ax = b$ can be interpreted as a vector equation, where $Ax$ is the linear combination of the columns of $A$ equaling the vector $b$.

---

# Summary Table

| Statement | True/False | Reason |
|-----------|:----------:|--------|
| A         | False      | Misstates the pivot condition for consistency |
| B         | True       | Direct consequence of the spanning property |
| C         | True       | Follows from the definition of "span" |
| D         | True       | Contrapositive of statement C |
| E         | True       | Follows from definition of matrix-vector multiplication |
| F         | True       | Common terminology in linear algebra |

---

# Final Answer

**The true statements are:**
- **B**
- **C**
- **D**
- **E**
- **F**

VIPSolutions · Answer

# Analysis of Statements on Matrix Equations ## Statement A > The equation $ Ax = b $ is consistent if the augmented matrix $[A\ b]$ has a pivot position in every row. **False.** A pivot in every row does not guarantee consistency; it must also hold that there are no contradictory rows of the form $[ \ldots \mid d]$ with $d \neq 0$. --- ## Statement B > If $A$ is an $m \times n$ matrix and if the equation $Ax = b$ is inconsistent for some $b \in \mathbb{R}^m$, then $A$ cannot have a pivot position in every row. **True.** If $A$ has a pivot in every row, it spans $\mathbb{R}^m$ and $Ax = b$ is consistent for all $b$. Thus, inconsistency implies $A$ lacks a pivot in every row. --- ## Statement C > If the columns of an $m \times n$ matrix $A$ span $\mathbb{R}^m$, then the equation $Ax = b$ is consistent for each $b \in \mathbb{R}^m$. **True.** By definition, spanning $\mathbb{R}^m$ means any vector $b$ can be expressed as a combination of the columns, ensuring the equation $Ax = b$ is consistent. --- ## Statement D > If $A$ is an $m \times n$ matrix whose columns do not span $\mathbb{R}^m$, then the equation $Ax = b$ is inconsistent for some $b \in \mathbb{R}^m$. **True.** If the columns do not span $\mathbb{R}^m$, there exists at least one vector $b$ that cannot be represented as a linear combination of the columns, making $Ax = b$ inconsistent for that $b$. --- ## Statement E > The first entry in the product $Ax$ is a sum of products. **True.** Matrix multiplication involves taking the dot product of the first row of $A$ with the vector $x$, resulting in a sum of products. --- ## Statement F > The equation $Ax = b$ is referred to as a vector equation. **True.** This equation is indeed referred to as a vector equation, where $Ax$ denotes a linear combination of the columns of $A$ equating to the vector $b$. --- # Summary Table | Statement | True/False | Reason | |-----------|:----------:|--------| | A | False | Misstates the pivot condition for consistency | | B | True | Direct consequence of the spanning property | | C | True | Follows from the definition of "span" | | D | True | Contrapositive of statement C | | E | True | Follows from definition of matrix-vector multiplication | | F | True | Common terminology in linear algebra | --- # Final Answer **The true statements are:** - **B** - **C** - **D** - **E** - **F** --- # Analysis of the Function We need to analyze $$ f(x) = x \sqrt{121 - x^2}, \quad -11 \leq x \leq 11. $$ ### Step 1: Domain The expression inside the square root must be non-negative: $$ 121 - x^2 \geq 0 \implies -11 \leq x \leq 11. $$ Thus, the interval is the full domain. --- ### Step 2: Differentiate Using the product rule: $$ f(x) = x (121 - x^2)^{1/2}. $$ Differentiating: $$ f'(x) = (1)(121 - x^2)^{1/2} + x \cdot \frac{1}{2}(121 - x^2)^{-1/2} \cdot (-2x). $$ Simplifying: $$ f'(x) = \sqrt{121 - x^2} - \frac{x^2}{\sqrt{121 - x^2}} = \frac{121 - 2x^2}{\sqrt{121 - x^2}}. $$ --- ### Step 3: Critical Points Setting the numerator to zero: $$ 121 - 2x^2 = 0 \implies x^2 = \frac{121}{2} \implies x = \pm \frac{11}{\sqrt{2}} \approx \pm 7.78. $$ Critical points: $x \approx -7.78, 7.78$. --- ### Step 4: Sign of Derivative The denominator $\sqrt{121 - x^2} > 0$ in $(-11, 11)$. The sign of $f'(x)$ is determined by $121 - 2x^2$: - If $x^2 < 60.5$, then $f'(x) > 0$. - If $x^2 > 60.5$, then $f'(x) < 0$. --- ### Step 5: Intervals - **Increasing** on $ (-\sqrt{60.5}, \sqrt{60.5}) $ or approximately $(-7.78, 7.78)$. - **Decreasing** on $[-11, -7.78) \cup (7.78, 11]$. --- ### Final Answer $$ \text{Increasing on } (-7.78, 7.78), \quad \text{Decreasing on } [-11, -7.78) \cup (7.78, 11]. $$

VIPSolutions · Answer

# Analysis of the Given Vectors

We are given the following column vectors in $\mathbb{R}^5$:

$$
v_1=\begin{pmatrix}1\0\-1\2\-3\end{pmatrix},\quad
v_2=\begin{pmatrix}0\1\-2\0\-3\end{pmatrix},\quad
v_3=\begin{pmatrix}0\-2\3\2\3\end{pmatrix},\quad
v_4=\begin{pmatrix}-1\-2\4\0\6\end{pmatrix}.
$$

We aim to determine the dimension of $\operatorname{Span}\{v_1,v_2,v_3,v_4\}$, which corresponds to the rank of these vectors.

## Step 1: Establish Linear Relations

We will check if $v_4$ can be expressed as a linear combination of $v_1$, $v_2$, and $v_3$:

$$
a v_1 + b v_2 + c v_3 = v_4.
$$

This leads to the following system of equations based on components:

1. $a = -1$  (from the first component)
2. $b - 2c = -2$  (from the second component)
3. $-a - 2b + 3c = 4$  (from the third component)
4. $2a + 2c = 0$  (from the fourth component)
5. $-3a - 3b + 3c = 6$  (from the fifth component)

## Step 2: Solve the System

From equation (1):

$$
a = -1.
$$

Substituting into equation (4):

$$
2(-1) + 2c = 0 \implies -2 + 2c = 0 \implies c = 1.
$$

Substituting $c = 1$ into equation (2):

$$
b - 2(1) = -2 \implies b - 2 = -2 \implies b = 0.
$$

Now we can check if these values satisfy equations (3) and (5):

- Substituting $a = -1$, $b = 0$, $c = 1$ into equation (3):

$$
-(-1) - 2(0) + 3(1) = 4 \implies 1 + 0 + 3 = 4 \quad \text{(True)}.
$$

- Substituting into equation (5):

$$
-3(-1) - 3(0) + 3(1) = 6 \implies 3 + 0 + 3 = 6 \quad \text{(True)}.
$$

Thus, we have found:

$$
v_4 = -v_1 + 0\cdot v_2 + v_3,
$$

indicating that $v_4$ is a linear combination of $v_1$ and $v_3$.

## Step 3: Determine Linear Independence

The vectors $v_1$, $v_2$, and $v_3$ remain to be checked for linear independence. Since $v_4$ is dependent on $v_1$ and $v_3$, we investigate the first three vectors:

- By performing row reduction on the matrix formed by $v_1$, $v_2$, and $v_3$, we can check if they are linearly independent.

The resulting rank of the matrix formed by $v_1$, $v_2$, and $v_3$ confirms that they produce three independent rows.

## Conclusion

Therefore, we conclude:

- The vectors $v_1$, $v_2$, $v_3$, and $v_4$ are linearly dependent, resulting in a maximum rank of 3.
- The vectors $v_1$, $v_2$, and $v_3$ are linearly independent.

### Final Answers to Questions

- **a**: Is it a 4-dimensional subspace in $\mathbb{R}^5$? — **No.**
- **b**: Is it a 3-dimensional subspace in $\mathbb{R}^5$? — **Yes.**
- **c**: Is it a 2-dimensional subspace in $\mathbb{R}^5$? — **No.**

VIPSolutions · Answer

# Analysis of the Given Vectors

We are given the following column vectors in $\mathbb{R}^5$:

$$
v_1=\begin{pmatrix}1\0\-1\2\-3\end{pmatrix},\quad
v_2=\begin{pmatrix}0\1\-2\0\-3\end{pmatrix},\quad
v_3=\begin{pmatrix}0\-2\3\2\3\end{pmatrix},\quad
v_4=\begin{pmatrix}-1\-2\4\0\6\end{pmatrix}.
$$

We aim to determine the dimension of $\operatorname{Span}\{v_1,v_2,v_3,v_4\}$, which corresponds to the rank of these vectors.

## Step 1: Establish Linear Relations

We will check if $v_4$ can be expressed as a linear combination of $v_1$, $v_2$, and $v_3$:

$$
a v_1 + b v_2 + c v_3 = v_4.
$$

This leads to the following system of equations based on components:

1. $a = -1$  (from the first component)
2. $b - 2c = -2$  (from the second component)
3. $-a - 2b + 3c = 4$  (from the third component)
4. $2a + 2c = 0$  (from the fourth component)
5. $-3a - 3b + 3c = 6$  (from the fifth component)

## Step 2: Solve the System

From equation (1):

$$
a = -1.
$$

Substituting into equation (4):

$$
2(-1) + 2c = 0 \implies -2 + 2c = 0 \implies c = 1.
$$

Substituting $c = 1$ into equation (2):

$$
b - 2(1) = -2 \implies b - 2 = -2 \implies b = 0.
$$

Now we can check if these values satisfy equations (3) and (5):

- Substituting $a = -1$, $b = 0$, $c = 1$ into equation (3):

$$
-(-1) - 2(0) + 3(1) = 4 \implies 1 + 0 + 3 = 4 \quad \text{(True)}.
$$

- Substituting into equation (5):

$$
-3(-1) - 3(0) + 3(1) = 6 \implies 3 + 0 + 3 = 6 \quad \text{(True)}.
$$

Thus, we have found:

$$
v_4 = -v_1 + 0\cdot v_2 + v_3,
$$

indicating that $v_4$ is a linear combination of $v_1$ and $v_3$.

## Step 3: Determine Linear Independence

The vectors $v_1$, $v_2$, and $v_3$ remain to be checked for linear independence. Since $v_4$ is dependent on $v_1$ and $v_3$, we investigate the first three vectors:

- By performing row reduction on the matrix formed by $v_1$, $v_2$, and $v_3$, we can check if they are linearly independent.

The resulting rank of the matrix formed by $v_1$, $v_2$, and $v_3$ confirms that they produce three independent rows.

## Conclusion

Therefore, we conclude:

- The vectors $v_1$, $v_2$, $v_3$, and $v_4$ are linearly dependent, resulting in a maximum rank of 3.
- The vectors $v_1$, $v_2$, and $v_3$ are linearly independent.

### Final Answers to Questions

- **a**: Is it a 4-dimensional subspace in $\mathbb{R}^5$? — **No.**
- **b**: Is it a 3-dimensional subspace in $\mathbb{R}^5$? — **Yes.**
- **c**: Is it a 2-dimensional subspace in $\mathbb{R}^5$? — **No.**

VIPSolutions · Answer

# Influence Lines for a Simply Supported Beam

## Structure Details

- **Beam Type:** Simply supported
- **Length:** 9 m
- **Supports:**
  - **Support A:** Pinned
  - **Support B:** Roller
- **Point C:** Located 3 m from A
- **Load:** Moving unit load at any position $ x \in [0, 9] $

---

## a. Influence Line for Vertical Reaction at A

**Objective:** Determine the influence line for the vertical reaction $ R_A $.

### Procedure

1. **Remove Support A:** 
   - Apply a unit vertical displacement upward at A.
2. **Effect on Beam:**
   - The beam rotates downward towards B due to its supports.
3. **Influence Line Shape:** 
   - Linear from 1 at A to 0 at B.

### Equation

For a position $ x $ from point A:

$$
IL_{R_A}(x) = 1 - \frac{x}{9}, \quad 0 \leq x \leq 9
$$

---

## b. Influence Line for Shear at Point C

**Objective:** Determine the influence line for internal shear force at point C (3 m from A).

### Procedure

1. **Make a Virtual Cut at C:**
   - Apply a unit shear force: upward on the left face, downward on the right.
2. **Discontinuity at C:**
   - The influence line will show a jump of 1 due to unit shear.

### Using Static Analysis

For a unit load at position $ x $:

- **Reaction at A:**
$$
R_A = \frac{9 - x}{9}
$$

- **Shear at C:**
$$
V_C(x) =
\begin{cases}
-\frac{x}{9}, & 0 \leq x < 3 \
1 - \frac{x}{9}, & 3 < x \leq 9
\end{cases}
$$

### Discontinuity at $ x = 3 $

- Left of C: $ V_C^- = -\frac{1}{3} $
- Right of C: $ V_C^+ = \frac{2}{3} $
- Jump: $ \frac{2}{3} - (-\frac{1}{3}) = 1 $

---

## c. Influence Line for Moment at Point C

**Objective:** Determine the influence line for bending moment at point C.

### Procedure

1. **Make a Virtual Cut at Point C:**
   - Apply a unit moment (clockwise) at C.
2. **Influence Line Shape:**
   - Takes a triangular shape, continuous with a peak at C.

### Using Static Analysis

For a unit load at position $ x $:

- **Moment at C:**
$$
M_C(x) =
\begin{cases}
\frac{x}{3}, & 0 \leq x \leq 3 \
\frac{9 - x}{6}, & 3 < x \leq 9
\end{cases}
$$

### Maximum Value

At $ x = 3 $:
$$
M_C(3) = 1
$$

---

## Final Summary of Influence Line Equations

| Quantity                   | Influence Line Equation                                 |
|----------------------------|--------------------------------------------------------|
| Vertical Reaction at A     | $ IL_{R_A}(x) = 1 - \frac{x}{9} $ for $ 0 \leq x \leq 9 $ |
| Shear at C                | $ IL_{V_C}(x) = \begin{cases} -\frac{x}{9} & 0 \leq x < 3 \ 1 - \frac{x}{9} & 3 < x \leq 9 \end{cases} $ |
| Moment at C               | $ IL_{M_C}(x) = \begin{cases} \frac{x}{3} & 0 \leq x \leq 3 \ \frac{9 - x}{6} & 3 < x \leq 9 \end{cases} $ |

---

Would you like plotted diagrams for these influence lines?

VIPSolutions · Answer

# Analysis of the Piecewise Function We are given the piecewise function: $$ g(x) = \begin{cases} \dfrac{1}{|x-1|}, & 0 \leq x < 2,\[6pt] -x^{2} + 6x - 10, & 2 \leq x < 4,\[6pt] (x-4)^{3} - 2, & x \geq 4,\[6pt] -g(-x), & x < 0 \end{cases} $$ ### 1. Interval $0 \leq x < 2: \; g(x) = \dfrac{1}{|x-1|}$ - **Subinterval: $0 \leq x < 1$** - Here, $|x-1| = 1 - x$, thus $g(x) = \dfrac{1}{1-x}$. - Defined on $[0, 1)$, increases from $g(0) = 1$ to $+\infty$ as $x \to 1^{-}$. - **Subinterval: $1 < x < 2$** - Here, $|x-1| = x - 1$, thus $g(x) = \dfrac{1}{x-1}$. - Defined on $(1, 2)$, decreases from $+\infty$ as $x \to 1^{+}$ to $g(2^{-}) = 1$. - **Asymptotes and Limits** - **Vertical asymptote:** $x = 1$ - $\lim_{x \to 1^{-}} g(x) = +\infty, \quad \lim_{x \to 1^{+}} g(x) = +\infty$. - Value at $x = 0$: $g(0) = 1$. - Left- and right-hand limit at $x = 2$: $\lim_{x \to 2^{-}} g(x) = 1$. --- ### 2. Interval $2 \leq x < 4: \; g(x) = -x^{2} + 6x - 10$ - **Simplification:** - Rewrite as $g(x) = -(x-3)^{2} - 1$. - This is a downward-opening parabola with vertex at $x = 3$, $g(3) = -1$. - **Endpoint values:** - $g(2) = -4 + 12 - 10 = -2$. - $\lim_{x \to 2^{-}} g(x) = 1$ (from previous piece), indicating a jump discontinuity at $x = 2$: left limit $= 1$ but $g(2) = -2$. - $g(4^{-}) = -16 + 24 - 10 = -2$. - **Parabola Characteristics:** - Symmetric about $x = 3$ and attains maximum on $[2, 4)$ at vertex $g(3) = -1$. - Values on $[2, 4)$ range from $-2$ to $-1$. --- ### 3. Interval $x \geq 4: \; g(x) = (x-4)^{3} - 2$ - **Cubic Function Characteristics:** - A cubic shifted right by 4 and down 2. At $x = 4$: $g(4) = -2$. - **Continuity:** - Matches left-hand value at $x = 4$: $g(4^{-}) = -2$, thus $g$ is continuous at $x = 4$. - **Slope Analysis:** - Derivative at $x = 4$: $3(x-4)^{2}$, giving a slope of 0. - Left-hand slope at $x = 4$ from the parabola: $(-2x + 6)|_{x=4} = -2$. - Slopes differ, so $g$ is **not differentiable** at $x = 4$. - **Behavior:** - For $x > 4$, the cubic increases without bound as $x \to \infty$. --- ### 4. Interval $x < 0: \; g(x) = -g(-x)$ - **Odd Function Definition:** - Reflects $g$ for $x < 0$ as $g$ becomes an odd function: $g(-x) = -g(x)$. - **Consequences:** - **Vertical asymptote:** at $x = -1$ (mirror of $x = 1$). - $g(-0) = -g(0) = -1$; left-hand value at $x = 0$ is $-1$, right-hand value is $1$. - **Jump at $x = 0$**: left-hand value $-1$, right-hand value $1$. - **Example Values:** - $g(-2) = -g(2) = -(-2) = 2$. - $g(-3) = -g(3) = -(-1) = 1$. - **Behavior at Infinity:** - As $x \to -\infty$, $g(x) \to -\infty$ by oddness and cubic growth on the right. --- ### 5. Summary of Continuity, Discontinuities, and Asymptotes - **Vertical Asymptotes:** $x = 1$ and $x = -1$ (blow-up to $\pm \infty$). - **Jump Discontinuities:** - At $x = 2$ (left limit = 1, value = -2). - At $x = 0$ (left value = -1, right value = 1). - **Continuous Points:** - Continuous on intervals defined by single formulas; continuous at $x = 4$. - **Differentiability:** - Fails at $x = 1$ (vertical asymptote), at $x = 0, 2$ (jumps), and at $x = 4$ (corner/slope mismatch). Smooth elsewhere. --- ### 6. Key Exact Values to Plot - $g(0) = 1$, $g(0^{-}) = -1$ (left-hand value). - $g(2) = -2$, $\lim_{x \to 2^{-}} g(x) = 1$. - $g(3) = -1$ (parabola vertex). - $g(4) = -2$ and the cubic increases for $x > 4$. - Vertical asymptotes at $x = \pm 1$ (blow-up to $\infty$). --- ### 7. Sketching Instructions 1. **Mark asymptotes:** Dashed lines at $x = 1$ and $x = -1$. 2. **For $0 \leq x < 1$:** Draw $y = \frac{1}{1-x}$ rising from $1$ to $+\infty$ as $x \to 1^{-}$. 3. **For $1 < x < 2$:** Draw $y = \frac{1}{x-1}$ falling from $+\infty$ at $x = 1^{+}$ to $1$ at $x = 2^{-}$. 4. **At $x = 2$:** Place a filled dot at $(2, -2)$ (parabola value). Note the open point at $(2, 1)$ for the jump. 5. **For $2 \leq x < 4$:** Draw the parabola $g(x) = -(x-3)^{2} - 1$, with vertex $(3, -1)$ and values $-2$ at $2$ and $4^{-}$. 6. **For $x \geq 4$:** Draw the cubic $g(x) = (x-4)^{3} - 2$ starting at $(4, -2)$ and increasing steeply. 7. **For negative $x$:** Reflect the positive side across the origin, plotting points using $g(-x) = -g(x)$ and include vertical asymptote at $x = -1$. --- This structured analysis provides a comprehensive understanding of the function $g(x)$ for sketching and evaluating its behavior across different intervals.

VIPSolutions · Answer

# Analysis of Vector Spaces

Let $ V $ be a vector space and $ A, B \subseteq V $ be nonempty subsets.

## (1) Proof that $ \text{span}(A \cap B) \subseteq \text{span}(A) \cap \text{span}(B) $

### Definitions

- The **span** of a set of vectors is the set of all linear combinations of those vectors.
- For subsets $ A $ and $ B $:
  - $ \text{span}(A) $ is the set of all linear combinations of vectors in $ A $.
  - $ \text{span}(B) $ is the set of all linear combinations of vectors in $ B $.

### Proof Steps

1. **Let $ v \in \text{span}(A \cap B) $**:
   - By definition, $ v $ can be expressed as a linear combination of vectors in $ A \cap B $.
   - That is, $ v = c_1 a_1 + c_2 a_2 + \ldots + c_n a_n $ where $ a_i \in A \cap B $ for $ i = 1, 2, \ldots, n $ and $ c_i \in \mathbb{R} $ (or the underlying field).

2. **Since $ a_i \in A \cap B $**:
   - Each vector $ a_i $ is in both $ A $ and $ B $.
   - Therefore, $ v $ is also a linear combination of vectors in $ A $, implying $ v \in \text{span}(A) $.

3. **Similarly**, since each $ a_i $ is also in $ B $:
   - It follows that $ v \in \text{span}(B) $.

4. **Conclusion**:
   - Since $ v \in \text{span}(A) $ and $ v \in \text{span}(B) $, we have $ v \in \text{span}(A) \cap \text{span}(B) $.

Thus, we conclude that:

$$
\text{span}(A \cap B) \subseteq \text{span}(A) \cap \text{span}(B).
$$

---

## (2) Example where $ V = \mathbb{R}^3 $ and $ \text{span}(A \cap B) \neq \text{span}(A) \cap \text{span}(B) $

### Choose Sets

Let:

- $ A = \{ (1, 0, 0), (0, 1, 0) \} $
- $ B = \{ (0, 1, 0), (0, 0, 1) \} $

### Calculate Intersections and Spans

1. **Intersection**:
   - $ A \cap B = \{ (0, 1, 0) \} $

2. **Span of Intersection**:
   - $ \text{span}(A \cap B) = \text{span}(\{(0, 1, 0)\}) = \{ k(0, 1, 0) : k \in \mathbb{R} \} $

3. **Spans of $ A $ and $ B $**:
   - $ \text{span}(A) = \text{span}(\{(1, 0, 0), (0, 1, 0)\}) $ which is the entire plane spanned by the x-axis and y-axis.
   - $ \text{span}(B) = \text{span}(\{(0, 1, 0), (0, 0, 1)\}) $ which is the plane spanned by the y-axis and z-axis.

4. **Intersection of Spans**:
   - $ \text{span}(A) \cap \text{span}(B) $ includes all vectors of the form $ (0, k, l) $ for $ k, l \in \mathbb{R} $ because both spans contain the y-axis.

### Result

- **Span of Intersection**:
  $$
  \text{span}(A \cap B) = \text{span}(\{(0, 1, 0)\}) = \{ (0, k, 0) : k \in \mathbb{R} \}.
  $$

- **Intersection of Spans**:
  $$
  \text{span}(A) \cap \text{span}(B) = \text{span}(\{(0, 1, 0), (0, 0, 1)\}) = \{ (0, k, l) : k, l \in \mathbb{R} \}.
  $$

### Conclusion

Thus:

$$
\text{span}(A \cap B) \neq \text{span}(A) \cap \text{span}(B).
$$

This shows that $ \text{span}(A \cap B) $ is a line while $ \text{span}(A) \cap \text{span}(B) $ is a plane.

VIPSolutions · Answer

To calculate the accumulated amount after 7 years for an investment of $16,000 at an interest rate of 2.5% compounded in different ways, we can use the formula for compound interest:

$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$

where:
- $A$ = the amount of money accumulated after n years, including interest.
- $P$ = the principal amount (the initial amount of money).
- $r$ = annual interest rate (decimal).
- $n$ = number of times that interest is compounded per year.
- $t$ = number of years the money is invested for.

**Given:**
- $P = 16000$
- $r = 0.025$
- $t = 7$

Now we will compute the accumulated amount for each compounding frequency.

### (a) Compounded Annually ($n = 1$)

$$
A = 16000 \left(1 + \frac{0.025}{1}\right)^{1 \cdot 7} = 16000 \left(1 + 0.025\right)^{7} = 16000 \left(1.025\right)^{7}
$$

Calculating:

$$
A \approx 16000 \times 1.187686 = 18999.00
$$

### (b) Compounded Semiannually ($n = 2$)

$$
A = 16000 \left(1 + \frac{0.025}{2}\right)^{2 \cdot 7} = 16000 \left(1 + 0.0125\right)^{14} = 16000 \left(1.0125\right)^{14}
$$

Calculating:

$$
A \approx 16000 \times 1.183673 = 18937.73
$$

### (c) Compounded Quarterly ($n = 4$)

$$
A = 16000 \left(1 + \frac{0.025}{4}\right)^{4 \cdot 7} = 16000 \left(1 + 0.00625\right)^{28} = 16000 \left(1.00625\right)^{28}
$$

Calculating:

$$
A \approx 16000 \times 1.181660 = 18906.56
$$

### (d) Compounded Monthly ($n = 12$)

$$
A = 16000 \left(1 + \frac{0.025}{12}\right)^{12 \cdot 7} = 16000 \left(1 + 0.00208333\right)^{84} = 16000 \left(1.00208333\right)^{84}
$$

Calculating:

$$
A \approx 16000 \times 1.180348 = 18885.57
$$

### Summary of Results

- (a) Compounded Annually: **$18,999.00**
- (b) Compounded Semiannually: **$18,937.73**
- (c) Compounded Quarterly: **$18,906.56**
- (d) Compounded Monthly: **$18,885.57**

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