It costs a toy retailer $10 to purchase a certain doll. He estimates that, if he charges x dollars per doll, he can sell 80 - 2x dolls per week. Find a function for his weekly profit.

Answer:

Price per doll to be sold = x.

Investment = $10 * by the number of dolls sold

Profit on 1 doll = $x-10$

Number of dolls sold per week = $80-2x$

The Profit will be: x * (number of dolls sold in one week) - 10 * (Number number of dolls sold in one week)

The function for his weekly profit will be: \[p(x)=x\cdot\left(80-2x\right)-10\cdot \left(80-2x\right)\] Extended: \[p(x)=-2x^2+100x-800\]

Given the following function: \[ f (x) = 8x^3\:\:+\:7x^2\:\:-\:5\]

f <- function(x){
  return(8*x^3  + 7*x^2  - 5)
}

Step 1. Find f (3).

f(3)

## [1] 274

The answer will be f(3) = 274.

Step 2. Find f (-2).

f(-2)

## [1] -41

The answer will be f(-2) = -41.

Step 3. Find f (x + c).

The answer will be \[f (x + c) =8x^3+24cx^2+24c^2x+8c^3+7x^2+14cx+7c^2-5\]

Use the graph to find the indicated limits. If there is no limit, state “Does not exist”.

Answer: The Limit Does Not Exits.

Step 1. Find \[ \lim _{x\to 1^-}f\left(x\right) \]

Answer: \[ \lim _{x\to 1^-}f\left(x\right)=2 \]

Step 2. Find \[ \lim _{x\to 1^+}f\left(x\right) \]

Answer: \[ \lim _{x\to 1^+}f\left(x\right)=-5 \]

Step 3. Find \[ \lim _{x\to 1}f\left(x\right) \]

Answer: Limit DOES NOT EXIST.

Find the derivative for the following function \[f(x)=-2x^3\]

library(Deriv) 
f=function(x) {(-2) * x ^ 3}
Deriv(f)

## function (x) 
## -(6 * x^2)

Answer: \[f'(x)=-6x^2\]

Find the derivative for the following function \[\:f\:\left(x\right)=\frac{-8}{x^2}\]

library(Deriv) 
f=function(x) {(-8)/(x^2)}
Deriv(f)

## function (x) 
## 16/x^3

Answer: \[f'(x)=\frac{16}{x^3}\]

Find the derivative for the following function \[g(x)=5\sqrt[3]{x}\]

library(Deriv) 
f=function(x) {(5) * x^(1/3)}
Deriv(f)

## function (x) 
## 1.66666666666667/x^0.666666666666667

Answer: \[g'(x)=\frac{5}{3x^{\frac{2}{3}}}\]

Find the derivative for the following function \[y=-2x^{\left(9/8\right)}\]

library(Deriv) 
f=function(x) {(-2) * x^(9/8)}
Deriv(f)

## function (x) 
## -(2.25 * x^0.125)

Answer: \[y'=-\frac{9x^{\frac{1}{8}}}{4}\]

Consider the graph of f (x). What is the average rate of change of f (x) from $x_1=0$ to $x_2=4$? Please write your answer as an integer or simplified fraction.

Answer:

Taking our points as follows: $x_1=0, y_1=40$ and $x_2=4,y_2=35$ We have as follows: \[m=\frac{y_2-y_1}{x_2-x_1}\] \[m=\frac{35-40}{4-0}\] The Average rate of change will be $m=\frac{-5}{4}$

The cost of producing $x$ baskets is given by $C(x) = 630 + 2.4x$. Determine the average cost function.

Answer:

Taking A(x) = Average Cost By definition: Average Cost = Total Cost / Number of Units \[A(x) =\frac{630+2.4x}{x}\]

Use the Product Rule or Quotient Rule to find the derivative \[f(x)=\left(-2x^{-2}+1\right)\left(-5x+9\right)\]

Answer:

library(Deriv) 
f=function(x) {(-2 * x^(-2) + 1) *(-5 * x + 9)}
Deriv(f)

## function (x) 
## 4 * ((9 - 5 * x)/x^3) - 5 * (1 - 2/x^2)

\[f'(x)=\frac{-10x+36}{x^3}-5\] 11. Use the Product Rule or Quotient Rule to find the derivative \[f(x)=\frac{5x^{\left(1/2\right)}+7}{-x^3+1}\]

Answer:

\[f'(x)=\frac{25x^3+42x^{\frac{5}{2}}+5}{2\sqrt{x}\left(-x^3+1\right)^2}\]

Find the derivative for the given function. Write your answer using positive and negative exponents and fractional exponents instead of radicals \[f(x)=\left(3x^{-3}-8x+6\right)^{\frac{4}{3}}\]

Answer:

library(Deriv) 
f=function(x) {(3 * x^(-3) -8 * x + 6)^(4/3)}
Deriv(f)

## function (x) 
## -(1.33333333333333 * ((3/x^3 + 6 - 8 * x)^0.333333333333333 * 
##     (8 + 9/x^4)))

\[f'(x)=\frac{4}{3}\left(-9x^{-4}-8\right)\left(3x^{-3}-8x+6\right)^{\frac{1}{3}}\]

After a sewage spill, the level of pollution in Sootville is estimated by $f (t) = \frac{550t^2}{\sqrt{t^2+15}}$, where $t$ is the time in days since the spill occurred. How fast is the level changing after 3 days? Round to the nearest whole number.

Answer:

f <- function(t){
  return(round((550 * t^2)/((t ^ 2 +15)^(1/2)),0))
}
f(3)

## [1] 1010

The level of pollution has changed $1010$ times in three days.

The average home attendance per week at a Class AA baseball park varied according to the formula $N(t) = 1000(6 + 0.1t)^(1/2)$ where $t$ is the number of weeks into the season $(0 \le t \le 14)$ and N represents the number of people.

Step 1. What was the attendance during the third week into the season? Round your answer to the nearest whole number.

N <- function(t){
  return(round(1000 * (6 + 0.1 * t)^(1/2),0))
}
N(3)

## [1] 2510

Answer: The attendance was 2510.

Step 2. Determine $N'(5)$ and interpret its meaning. Round your answer to the nearest whole number.

Answer: \[N'(t)=\frac{50}{\sqrt{6+0.1t}}\]

N1 <- function(t){
  return(round(50 / (6 + 0.1 * t)^(1/2),0))
}
N1(5)

## [1] 20

N(5)

## [1] 2550

Interpretation: The first derivative values indicate where the function is increasing or decreasing. When the first derivative is positive, the function is increasing. When the first derivative is negative, the function is decreasing. Hence since we got a positive number, the attendance has increased.

Consider the following function \[3x^3 + 4y^3 = 77\]

Step 1. Use implicit differentiation to find $\frac{dy}{dx}$

Answer: \[\frac{dy}{dx}=\frac{-3x^2}{4y^2}\]

Step 2. Find the slope of the tangent line at $(3,-1)$.

Answer: By evaluating $x=3$ and $y=-1$ in $\frac{-3x^2}{4y^2}$ we obtain that the slope of the tangent line is $-\frac{27}{4}$

Find the intervals on which the following function is increasing and on which it is decreasing \[f(x)=\frac{x+3}{x-8}\]

Answer:

By taking the denominator and solving the equation $x-8=0:\quad x=8$

\[\mathrm{Domain\:of\:\:f(x)=}\:\frac{x+3}{x-8}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<8\quad \mathrm{or}\quad \:x>8\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:8\right)\cup \left(8,\:\infty \:\right)\end{bmatrix}\] In order to find out if the function will be increasing or decreasing, we find $f'(x)$. \[f'(x)=-\frac{11}{\left(x-8\right)^2}\]

By Evaluating $f'(x)$ with $x<8$ the result will always be negative: $f'(x)<0$; hence $f(x)$ will be decreasing for $x<0$.

By Evaluating $f'(x)$ with $x>8$ the result will always be negative: $f'(x)<0$; hence $f(x)$ will be decreasing for $x>0$.

By Evaluating $f'(x)$ with $x=8$ the result will be always inconclusive since a division by zero is not defined.

A frozen pizza is placed in the oven at $t = 0$. The function $F(t) = 14+\frac{367t^2}{\left(t^2+100\right)}$ approximates the temperature (in degrees Fahrenheit) of the pizza at time $t$.

Step 1. Determine the interval for which the temperature is increasing and the interval for which it is decreasing. Please express your answers as open intervals.

First: Find $F'(t)$ \[F'(t)=\frac{73400t}{\left(t^2+100\right)^2}\] By solving the equation $F'(t)=0$, we obtain $t=0$. \[0=\frac{73400t}{\left(t^2+100\right)^2}\]

Second: Find $F"(t)$ \[F"(t)=\frac{73400\left(-3t^2+100\right)}{\left(t^2+100\right)^3}\] By Evaluating $t=0$ in our second derivative, we obtain: $F"(0)=\frac{367}{50}$

Since $F"(0)>0$, it means that $F(0)$ is a minimum with $t=0$.

Temperature Increase: \[F(t)>0\] for $t>0$.

Temperature Decrease: \[F(t)<0\] for $t<0$.

Step 2. Over time, what temperature is the pizza approaching?

We obtain the answer by solving: \[\lim _{t\to \infty }\left(14+\frac{367t^2}{\left(t^2+100\right)}\right)=381\].

Answer: Over time, the oven approaches a temperature of $381$ Fahrenheit degrees.

A study says that the package flow in the East during the month of November follows $f (x) = \frac{x^3}{3340000}-\frac{7x^2}{9475}+\frac{42417727x}{1265860000}+\frac{1}{33}$, where $1 \le x \le 30$ is the day of the month and $f (x)$ is in millions of packages. What is the maximum number of packages delivered in November? On which day are the most packages delivered? Round your final answer to the nearest hundredth.

Answer: \[f'(x)=\frac{3x^2}{3340000}-\frac{14x}{9475}+\frac{42417727}{1265860000}\] Solving: $f'(x)=0$ we obtain: $x=23$ and $x=\frac{1844249}{1137}$. We only will take $x=23$ since it belongs in the November Calendar. \[0=\frac{3x^2}{3340000}-\frac{14x}{9475}+\frac{42417727}{1265860000}\] \[f"(x)=\frac{3x}{1670000}-\frac{14}{9475}\] By solving $f"(23)$ we obtain \[f"(23)=-\frac{909049}{632930000}\] Since $f"(23)<0$ We conclude that the day with the most packages delivered is November 23rd.

$f(23)=0.41$

Answer: November 23rd was the day with most packages delivered. A total of $0.41$ million packages were delivered.

Use the Second Derivative Test to find all local extrema, if the test applies. Otherwise, use the First Derivative Test. Write any local extrema as an ordered pair. \[f(x)=7x^2+28x-35\]

Answer:

First derivative: $f'(x)=14x+28$: $f'(x)=0$: $0=14x+28$: $x=-2$

Second Derivative: $f"(x)=14$: $f"(-2)=14$

Since $f"(-2)>0$; we deduct that $(x=-2,f(-2)=-63)$ is a minimum.

Answer: Local Extrema = $(-2,-63)$

Use the Second Derivative Test to find all local extrema, if the test applies. Otherwise, use the First Derivative Test. Write any local extrema as an ordered pair. \[f(x)=-6x^3 + 27x^2 + 180x\]

Answer:

First derivative: $f'(x)=-18x^2+54x+180$: $f'(x)=0$: $0=-18x^2+54x+180$: $x=-2$ and $x=5$

Second Derivative: $f"(x)=-36x+54$

$f"(-2)=126$ Since $f"(-2)>0$; we deduct that $(x=-2,f(-2)=-204)$ is a minimum.
$f"(5)=-126$ Since $f"(5)<0$; we deduct that $(x=5,f(5)=825)$ is a maximum.

Answer: Local Extrema = $(-2,-204)$ and $(5,825)$.

A beauty supply store expects to sell 120 flat irons during the next year. It costs $1.60 to store one flat iron for one year. To reorder, there is a fixed cost of $6 , plus $4.50 for each flat iron ordered. In what lot size and how many times per year should an order be placed to minimize inventory costs?

Answer:

Let’s take as follows: $x$ = Number of orders per year. Lot size = 120 dived by the number of orders per year $\frac{120}{x}$. Storage fees = (Lot size)*1.60 Reorder fees = (Number of orders per year) * (6 + 4.5 * (Lot size))

The final function will be the addition or the Storage Fees plus the Reorder fees.

$f(x)=$ (Storage fees) + (Reorder fees) $f(x)=$ (Lot size)1.60 + (Number of orders per year) (6 + 4.5 * (Lot size)) \[f(x)=\frac{120}{x}\cdot 1.60+x\cdot \left(6\:+\:4.5\:\cdot \frac{120}{x}\right)\] Find $f'(x)$ \[f'(x)=-\frac{192}{x^2}+6\] Solve $f'(x)=0$ \[0=-\frac{192}{x^2}+6\] We obtain $x=5.65$, $x=-5.65$. We discard $x=-5.65$ since we can not consider it (The number of order are always positive).

Find $f"(x)$ \[f"(x)=\frac{384}{x^3}\]

And by evaluating $f"(5.65)$ we obtain $f"(5.65)=607.88$ and since $f"(5.65)>0$ we conclude that $x=5.65$ will bring a minimum point for $f(x)$.

Since we can not do decimal orders, we will round up to take $x=6$; and evaluating e obtain $f(6)=608$

Answer:

By placing $6$ orders during the year of $20$ items per lot, the beauty supply store will minimize the inventory costs.

A shipping company must design a closed rectangular shipping crate with a square base. The volume is 18432 ft^3. The material for the top and sides costs $3 per square foot and the material for the bottom costs $5 per square foot. Find the dimensions of the crate that will minimize the total cost of material.

Answer:

Let’s take: $x$ = dimension for square base of the crate. $y$ = height of the crate $Volume = 18432ft^3$ Cost of Material for sides + top = $3 * (4 * (x * y)) + 3 * (x * x)$ Cost of Material for base = $5 * ( x * x )$

The Total Cost = (Cost of Material for sides + top) + (Cost of Material for base)

\[C(x)=\left(3\:\cdot \left(4\:\cdot \left(x\:\cdot \:y\right)\right)\:+\:3\:\cdot \left(x\:\cdot \:x\right)\right)+\left(5\:\cdot \left(\:x\:\cdot \:x\:\right)\right)\]

By calculating the volume: $V=x*x*y$: $V=x^2y$:

By solving the equation in terms of $y$, we obtain:

$18432=x^2y$: $\frac{18432}{x^2}=y$

Now our cost function in terms of $x$ will be:

\[C(x)=\left(3\:\cdot \left(4\:\cdot \left(x\:\cdot \left(\frac{18432}{x^2}\right)\right)\right)\:+\:3\:\cdot \left(x\:\cdot \:x\right)\right)+\left(5\:\cdot \left(\:x\:\cdot \:x\:\right)\right)\] By simplifying we obtain: \[C(x)=\frac{221184}{x}+8x^2\]

Now, by finding the first and second derivatives and evaluating, we obtain the following results:

$x=24$ and $y=32$

Answer: The minimum cost of material can be achieved by having a square base of 24 ft per side and a height of 32 ft.

A farmer wants to build a rectangular pen and then divide it with two interior fences. The total area inside of the pen will be 1056 square yards. The exterior fencing costs $14.40 per yard and the interior fencing costs $12.00 per yard . Find the dimensions of the pen that will minimize the cost.

Answer:

Let’s take: $x$ = wide side of rectangle $y$ = length of the rectangle $Area = 1056yards^2$ Cost of Material for the exterior fence = $14.40 * (y+x+y+x)$ Cost of Material for the interior fence = $12 * ( x + x )$

The Total Cost for the fence = (Cost of Material for the exterior fence) + (Cost of Material for the interior fence)

\[C(x)=14.40*(2x+2y) + 12*(2x)\]

By calculating the area: $A=x*y$

By solving the equation in terms of $y$, we obtain:

$1056=x*y$: $\frac{1056}{x}=y$

Now our cost function in terms of $x$ will be:

\[C(x)=14.40\cdot \left(2x+2\cdot \left(\frac{1056}{x}\right)\right)\:+\:12\cdot \left(2x\right)\] By simplifying we obtain: \[C(x)=52.8x+\frac{30412.8}{x}\]

Now, by finding the first and second derivatives and evaluating, we obtain the following results:

$x=24$ and $y=44$

Answer: The minimum cost of material can be achieved by having a $x=24$ yards and $y=44$ yards.

It is determined that the value of a piece of machinery declines exponentially. A machine that was purchased 7 years ago for $67000 is worth $37000 today. What will be the value of the machine 9 years from now? Round your answer to the nearest cent.

Answer:

Since we have two set of points $(t_1,f(t_1))$, $(t_2,f(t_2))$ as follows:

$(0,67000)$, $(7,37000)$; we can find an exponential function $y=ab^t$.

After some equation solving we deduct that our exponential function is: \[f(t)=67000\left(\frac{37}{67}\right)^{\frac{t}{7}}\]

Now in order to calculate the future value we run $f(16)=17244.50$.

The value of the machine after 9 years will be $17244.50.

The demand function for a television is given by $p = D(x) = 23.2 - 0.4x$ dollars. Find the level of production for which the revenue is maximized.

Answer:

Let’s take as follows:

x = units $p(x)$ = Price = $23.2 - 0.4x$ $D(x)$ = Demand = $23.2 - 0.4x$ R = Revenue

Since Revenue = units * price

$R(x)=x*(23.2 - 0.4x)$: $R(x)=23.2x-0.4x^2$

Now, by finding the first and second derivatives and evaluating, we obtain the following results:

$x=29$

In order to maximize the revenue the level of production must be 29 units.

The amount of goods and services that costs $400 on January 1, 1995 costs $426.80 on January 1, 2006 . Estimate the cost of the same goods and services on January 1, 2017. Assume the cost is growing exponentially. Round your answer to the nearest cent.

Answer:

Let’s take as follows:

January 1, 1995 in terms of time let’s do $t=0$, $t$ in years. January 1, 2006 in terms of time let’s do $t=11$, $t$ in years. January 1, 2017 in terms of time let’s do $t=22$, $t$ in years.

Since we have two set of points $(t_1,f(t_1))$, $(t_2,f(t_2))$ as follows:

$(0,400)$, $(11,426.80)$; we can find an exponential function $y=ab^t$.

After some equation solving we deduct that our exponential function is: \[f(t)=400\cdot \left(\frac{426.80}{400}\right)^{\frac{t}{11}}\]

Now in order to calculate the future value we run $f(22)=455.40$.

The value of the machine after 22 years will be $455.40.

A manufacturer has determined that the marginal profit from the production and sale of x clock radios is approximately 380 - 4x dollars per clock radio.

Step 1. Find the profit function if the profit from the production and sale of 38 clock radios is $1700.

Preamble:

Since we have our marginal profit function:

$P'(x)=380-4x$; we conclude that our profit function is $P(x)= 380 x - 2x^2+C$

by evaluating $x=38$ and solving the equation \[1700=\:380\:\cdot \left(38\right)\:-\:2\:\cdot \left(38\right)^2+C\]

We conclude $C=-9852$.

Answer: The profit function will be: $P(x)= 380 x - 2x^2-9852$

Step 2. What is the profit from the sale of 56 clock radios?

P=function(x) {380 * x - 2 * x^2 - 9852}
P(56)

## [1] 5156

The profit for selling 56 radios will be $5156.

Use integration by substitution to solve the integral below. \[\int \frac{-5\left(ln\left(y\right)\right)^3}{y}dy\]

Answer:

By applying integral substitution $\int f(g(x))*g'(x)dx = \int f(u)du$ having $u=g(x)$

$u=ln(y)$ $\frac{du}{dy}=\frac{1}{y}$: $du=\frac{1}{y}dy$

$-5\int u^3du=\frac{-5}{4}u^4+C$

And by reversing substitution we obtain or final result: $-\frac{5}{4}ln(y)^4+C$.

It was discovered that after t years a certain population of wild animals will increase at a rate of $P´(t) = 75 - 9t^\frac{1}{2}$ animals per year. Find the increase in the population during the first 9 years after the rate was discovered. Round your answer to the nearest whole animal.

Answer:

Since we have $P´(t) = 75 - 9t^\frac{1}{2}$ we can deduct that our population function is: $P(t)=75t-6t^{\frac{3}{2}}+C$.

P=function(t) {75 * t - 6 * t^1.5}
P(9) - P(0)

## [1] 513

The increase in the population after the first 9 years after the rate was discovered is 513 animals.

Find the area of the region bounded by the graphs of the given equations. $y\:=\:6x^2$, $y\:=\:6\sqrt{x}$

Answer:

By solving the equation \[6x^2\:=\:6\sqrt{x}\] we can find our points of intersection.

\[6x^2 = 6\sqrt{x}\] \[x^2 = \sqrt{x}\] \[x^4 = x\] $x^4 - x=0$

$x(x^3 - 1)=0$

Our points of intersection will be $x=0$ and $x=1$

Then by calculating definite integrals we can find as follows:

\[\int _0^16\sqrt{x}\:dx - -\int _0^16x^2dx=2\] Enter your answer below: The area of the bounded region by the graphs of the given equations is 2.

31 Solve the differential equation given below. \[\frac{dy}{dx\:}=x^3\:y\] Steps:

\[\frac{dy}{dx}=x^3y\] \[dy=x^3ydx\] \[\frac{dy}{y}=x^3dx\] \[\int\frac{dy}{y}=\int x^3dx\] \[ln(y)+C_1=\frac{x^4}{4}+C_2\] \[ln(y)=\frac{x^4}{4}+C_2-C_1\] \[ln(y)=\frac{x^4}{4}+C\] \[y=e^\frac{x^4}{4}+C\] Answer: \[y=e^{\frac{x^4}{4}+c_1}\]

Use integration by parts to evaluate the definite integral below. \[\int _{-7}^2\:x\sqrt{x+7}dx\]

Answer: \[\int _{-7}^2\:x\sqrt{x+7}dx=-\frac{144}{5}\]

The following can be answered by finding the sum of a finite or infinite geometric sequence. Round the solution to 2 decimal places. A rubber ball is dropped from a height of 46 meters, and on each bounce it rebounds up 22 % of its previous height.

Step 1. How far has it traveled vertically at the moment when it hits the ground for the $20^th$ time?

Answer:

By defining our function as follows:

x = number of bounces $f(x) = 46*(0.22)^x$

$\sum _{x=1}^{20}f(x)$

\[\sum _{x=1}^{20}\:\left(46\cdot \left(0.22\right)^x\right)=12.97\] Since on very bounce the, we obtain the resulting height at the x bounce, we have to multiply by 2 since the ball goes up and down plus we have to add the 46 meters from were the ball was dropped.

Distance traveled vertically = $46+2*12.97=71.95$

Step 2. If we assume it bounces indefinitely, what is the total vertical distance traveled?

We resolve this by solving as follows:

Total distance = \[d(x)=46+2*\sum _{x=1}^{\infty }\:\left(46\cdot \left(0.22\right)^x\right)=71.95\]

Find the Taylor polynomial of degree 5 near x = 4 for the following function \[y\:=\:3e^{\left(5x\:-\:3\right)}\]

Answer:

\[P(x)=3e^{17}\:+\:15e^{17}\:\left(x-4\right)\:+\:\frac{72}{2}\:e^{17}\:\left(x-4\right)^2\:+\:\frac{125}{2}\:e^{17}\:\left(x-4\right)^3\:+\:\frac{625}{8}\:e^{17}\:\left(x-4\right)^4\:+\:\frac{625}{8}\:e^{17}\:\left(x-4\right)^5\]

Linear Algebra

Using matrix operations, describe the solutions for the following family of equations: $x + 2y - 3z = 5$ $2x + y - 3z = 13$ $-x + y + 2z= -8$

Find the inverse of the above 3x3 (non-augmented) matrix.

Let’s define \[A\:=\:\begin{pmatrix}1&2&-3\\ 2&1&-3\\ -1&1&2\end{pmatrix}\:\:\:X\:=\:\begin{pmatrix}x\\ y\\ z\end{pmatrix}\:\:\:B\:=\:\begin{pmatrix}5\\ 13\\ -8\end{pmatrix}\] Since we have $A$, we need to find $A^{-1}$.

\[A^{-1}=\begin{pmatrix}-\frac{5}{6}&\frac{7}{6}&\frac{1}{2}\\ \frac{1}{6}&\frac{1}{6}&\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\end{pmatrix}\]

Solve for the solution using R.

For this, we need to arrange as follows:

$AX=B$ $X=A^{-1}B$

So by solving $A^{-1}B$ we solve the system of equations.

VA <- c(1,2,-1,2,1,1,-3,-3,2)
A <- matrix(VA, 3,3)
VB <- c(5,13,-8)
B <- matrix(VB, 3,1)
A1 <- solve(A)
X <- A1 %*% B
X

##      [,1]
## [1,]    7
## [2,]   -1
## [3,]    0

Answer from the above we conclude that $x=7$, $y=-1$ and $z=0$.

Modify the 3x3 matrix such that there exists only one non-zero variable in the solution set.

\[A\:=\:\begin{pmatrix}1&2&5\\ 2&1&13\\ 1&1&-8\end{pmatrix}\]

VA <- c(1,2,1,2,1,1,5,13,-8)
A <- matrix(VA, 3,3)
VB <- c(5,13,-8)
B <- matrix(VB, 3,1)
A1 <- solve(A)
X <- round(A1 %*% B,0)
X

##      [,1]
## [1,]    0
## [2,]    0
## [3,]    1

Consider the matrix, q=matrix(c(3,1,4,4,3,3,2,3,2),nrow=3). Let b=c(1,4,5). Use Cramer’s rule and R to determine the solution, x, to qx=b, if one exists. Show all determinants.

Answer:

q <- matrix(c(3,1,4,4,3,3,2,3,2),nrow=3)
b <- c(1,4,5)

xv <- matrix(c(1,4,5,4,3,3,2,3,2),nrow=3)
xy <- matrix(c(3,1,4,1,4,5,2,3,2),nrow=3)
xz <- matrix(c(3,1,4,4,3,3,1,4,5),nrow=3)
deter <- det(q)
x <- det(xv)/deter
y <- det(xy)/deter
z <- det(xz)/deter
x

## [1] 1.461538

## [1] -2.538462

## [1] 3.384615

Answer: \[x=\begin{pmatrix}\frac{19}{13}\\\frac{-33}{13}\\\frac{44}{13}\end{pmatrix}\]

CUNY Bridge - Math Week 1 Assignment

Duubar Villalobos Jimenez

January 11, 2017

Linear Algebra