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Which of the following is true about the function \( f(x) = x^5 + 3x^4 + 9x^3 - 23x^2 - 36 \)?
\begin{enumerate}
\item[I)] \( f(x) \) has five real roots.
\item[II)] \( f(x) \) has three imaginary roots.
\item[III)] \( f(x) \) has a double root.
\item[IV)] As \( x \to \infty, f(x) \to \infty \)
\end{enumerate}
\noindent Options:\\
A) I, II, IV\\
B) II and III\\
C) III and IV\\
D) II and IV | We know 4 is true based on the fact that this is a positive odd polynomial. Using Descartes' rule of signs we see that we can have maximum 1 positive real root and 4,2, or 0 negative real roots. By doing some factoring we find that we have a double root. | AI-Algebra6 | C |
Simplify the expression: $\frac{3 x^{2}}{2 x^{-1}} \cdot \frac{y^{2}}{6 x^{2} y}$
\begin{align*}
\text{A)}\ & \frac{x y}{4} &
\text{B)}\ & \frac{4 x}{y} \\
\text{C)}\ & \frac{y}{4 x} &
\text{D)}\ & \frac{4}{x y}\\
\end{align*} | Cancel out like terms with exponent rules. | Alg2-24 | A |
Solve the equation $-2(3 x+2)^{\frac{3}{2}}=-54$ for the variable $x$.
\begin{align*}
\text{A)}\ & x=9 &
\text{B)}\ & x=\frac{5}{6} \\
\text{C)}\ & x=\frac{23}{6} &
\text{D)}\ & x=\frac{7}{3} \\
\end{align*} | $(3 x+2)^{3 / 2}=27,3 x+2=9,3 x=7, x=7 / 3$ | Alg2-25 | D |
Where is the hole in the graph of the function $f(x)=\frac{x^{2}+x-6}{x+3}$?
\begin{align*}
\text{A)}\ & (3,1) &
\text{B)}\ & (-3,0) \\
\text{C)}\ & (-3,-6) &
\text{D)}\ & (-3,-5)\\
\end{align*} | Hole at $\mathrm{x}=-3$ because that value creates a division by zero. By dividing both sides by $x+3$ we are left with $x-2$, so plugging -3 results in a $y$ value of -5. | Alg2-26 | D |
What is the horizontal asymptote of the function $f(x)=\frac{2 x^{3}+2}{x^{2}-16}$?
\begin{align*}
\text{A)}\ & y=4 \ and \ y=4 &
\text{B)}\ & y=2\\
\text{C)}\ & y=0 &
\text{D)}\ & \text{ no horizontal asymptotes} \\
\end{align*} | Because the exponent in the numerator is larger than the one in the denominator there is no asymptote. | Alg2-27 | D |
Determine any domain restriction(s) given the expression $\frac{x^{2}-9}{x^{2}-3 x-18}$.
\begin{align*}
\text{A)}\ & x=3\ and \ x=-3 &
\text{B)}\ & x=6 \ and \ x=-3 \\
\text{C)}\ & x=6 &
\text{D)}\ & x=\frac{1}{2}\\
\end{align*} | $\frac{(x+3)(x-3)}{(x-6)(x+3)}$, $x$ cannot be -3 due to a hole and 6 due to an asymptote. | Alg2-28 | B |
Simplify the rational expression $\frac{n^2+2n-24}{n^2-11n+28}$
\begin{align*}
\text{A)}\ & \frac{n+6}{n-7} &
\text{B)}\ & \frac{n+6}{n-4}\\
\text{C)}\ & \frac{n+6}{n+7} &
\text{D)}\ & \frac{n-4}{n-7}\\
\end{align*} | $\frac{(n+6)(n-4)}{(n-7)(n-4)}=\frac{n+6}{n-7}$ | Alg2-29 | A |
Simplify the rational expression.
\[
\frac{\frac{x+8}{x^2-64}}{(x+8)(x-8)}
\]
\noindent Options:\\
A) \( x + 2 \)\\
B) \(\frac{4(x + 8)}{(x^2 - 64)^2}\)\\
C) \( \frac{x + 8}{4} \)\\
D) \( \frac{1}{4} \) | the numerator can first be simplified into \(\frac{1}{x-8}\). This allows the whole fraction to be simplified into \(\frac{x+8}{4}\). | AI-Algebra8 | C |
Simplify the expression: \(5\sqrt{2} - 3\sqrt{8} + 2\sqrt{18}\).
\noindent Options:\\
A) \(5\sqrt{2}\)\\
B) \(-\sqrt{2} + 6\sqrt{3}\)\\
C) \(5\sqrt{2} - 12\sqrt{2} + 6\sqrt{2}\)\\
D) \(-\sqrt{2}\) | \(5\sqrt{2} - 6\sqrt{2} + 6\sqrt{2} = 5\sqrt{2}\) | AI-Algebra7 | A |
Which expression is the simplest form of $4 \sqrt[3]{32}-\sqrt[3]{32}$ ?
\begin{align*}
\text{A)}\ & 3 \sqrt[3]{4} &
\text{B)}\ & 6 \sqrt[3]{4}\\
\text{C)}\ & 3 \sqrt[3]{32} &
\text{D)}\ & 16 \sqrt[3]{2}-4\\
\end{align*} | $\sqrt[3]{32}=2 \sqrt[3]{4}, 3 \times 2 \sqrt[3]{4}=6 \sqrt[3]{4}$ | Alg2-32 | B |
What is the simplified form of the expression $\sqrt{98 x^{3} y^{5} z}$ ?
\begin{align*}
\text{A)}\ & 2 x y z \sqrt{7 x y z} &
\text{B)}\ & 7 x^{2} y^{2} \sqrt{2 y z}
\\
\text{C)}\ & 7 x y^{2} \sqrt{2 x y z} &
\text{D)}\ &49 x y^{2} \sqrt{2 x y z} \\
\end{align*} | Take out all square factors. | Alg2-33 | C |
Evaluate each of the following expressions.
a) $\log _{4} \frac{1}{64}=?$
b) $\log _{5} 625=?$ | a. $4^{x}=1 / 64, x=3$, \
b. $5^{x}=625, x=4$ | Alg2-37 | a. 3, b. 4 |
Jasmine invests $\$ 2,658$ in a retirement account with a fixed annual interest rate of $9 \%$ compounded continuously. What will the account balance be after 15 years? | Using the compound interest formula we have $2658\times (1.09)^{15} = 9681.72$ | Alg2-41 | 9681.72 |
Remy invests $\$ 8,589$ in a retirement account with a fixed annual interest rate of $7 \%$ compounded continuously. How long will it take for the account balance to reach $\$ 21,337.85$ ? | Using the compound interest formula we have $8589 \times(1.07)^{x}=21337.85, x=13.45$ | Alg2-42 | 13.45 years |
Solve $\sqrt{x^{2}+2 x-6}=\sqrt{x^{2}-14}$ | Square both sides then rearrange to $2x=-8, x=-4$. | Alg2-44 | -4 |
In $\triangle \mathrm{ABC}, \mathrm{AB}=10 \mathrm{~cm}, \angle \mathrm{B}=90^{\circ}$, and $\angle \mathrm{C}=60^{\circ}$. Determine the length of $\mathrm{BC}$.
\begin{align*}
\text{A)}\ & 10 \mathrm{~cm} &
\text{B)}\ & 10 \sqrt{3} \mathrm{~cm}\\
\text{C)}\ & \frac{10 \sqrt{3}}{3} \mathrm{~cm} &
\text{D)}\ & 20 \mathrm{~cm}\\
\end{align*} | This is a special $30-60-90$ triangle where $\mathrm{AB}$ is the longer leg and $\mathrm{BC}$ is the shorter leg. To go from the longer leg to the shorter one we should multiply by $\frac{\sqrt{3}}{3}$ | PreCal-7 | C |
If the length of the shorter leg of a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle is $5 \sqrt{3}$, then the length of the longer leg is
\begin{align*}
\text{A)}\ & 10 &
\text{B)}\ & 10 \sqrt{3} \\
\text{C)}\ & 10 \sqrt{6} &
\text{D)}\ & 15\\
\end{align*} | To find the longer leg of a 30-60-90 triangle from the shorter leg, we must multiply by $\sqrt{3}$. This means our longer leg has a length of 15 | PreCal-9 | D |
If the sides of a triangle are 6,7 , and 9 ; then the triangle is
\begin{align*}
\text{A)}\ & \text{a} \ 45^{\circ}-45^{\circ}-90^{\circ} \ \text{triangle} &
\text{B)}\ & \text{an acute triangle}\\
\text{C)}\ & \text{an obtuse triangle} &
\text{D)}\ & \text{a right triangle}\\
\end{align*} | We can eliminate A because there are not two equal sides. Now compare $6^{2}+7^{2}$ and $9^{2}$. Since $6^{2}+7^{2}$ is larger, this is an acute triangle. | PreCal-10 | B |
The \_\_\_\_\_ ratio compares the length of the adjacent leg to the length of the hypotenuse
\begin{align*}
\text{A)}\ & sine &
\text{B)}\ & cosine \\
\text{C)}\ & tangent &
\text{D)}\ & \text{none of the above}\\
\end{align*} | Cosine compares adjacent and hypotenuse. | Geo12 | B |
Which of the following forms a right triangle?
\begin{align*}
\text{A)}\ & \sqrt{4},\sqrt{9},\sqrt{25} &
\text{B)}\ & 1,2,3 \\
\text{C)}\ & 5,11,13 &
\text{D)}\ & 3,4,5\\
\end{align*} | D is a 3-4-5 triangle which is a known right triangle. This problem can also be done by using the pythagorean theorem. | Geo13 | D |
Find the length of the diagonal of a square whose perimeter measures 28 cm.
\begin{align*}
\text{A)}\ & 7 \ cm &
\text{B)}\ & 7\sqrt{2} \ cm \\
\text{C)}\ & 7\sqrt{3} \ cm &
\text{D)}\ & 28\sqrt{2}\ cm\\
\end{align*} | A square has 4 equal sides so each side must be 7 cm. The length of the diagonal is found by a 45-45-90 triangle so the diagonal is $7\sqrt{2}$ cm. | Geo14 | B |
Which of the following transformations creates a figure that is similar (but not congruent) to the original figure?
\begin{align*}
\text{A)}\ & Dilation &
\text{B)}\ & Rotation \\
\text{C)}\ & Translation &
\text{D)}\ & Reflection\\
\end{align*} | B and C don't change the shape of the object and D results in a nonsimilar object. Dilation keeps similarity while changing size. | Geo16 | A |
What is the image of the point (4, –2) after a dilation of 3?
\begin{align*}
\text{A)}\ & (12,-6) &
\text{B)}\ & (7,1) \\
\text{C)}\ & (1,-5) &
\text{D)}\ & (\frac{4}{3},-\frac{2}{3})\\
\end{align*} | A dilation of 3 means we should multiply each coordinate by 3. | Geo17 | A |
What is the center of the circle whose equation is $(x-1)^2+(y+3)^2=25$?
\begin{align*}
\text{A)}\ & (-1,3) &
\text{B)}\ & (3,-1) \\
\text{C)}\ & (1,-3) &
\text{D)}\ & (-3,1)\\
\end{align*} | The center is $(h, k)$ with circle equation of $(x-h)^2+(y-k)^2=r^2$ | Geo20 | C |
A \_\_\_\_\_ is a quadrilateral with two pairs of congruent adjacent sides and no congruent opposite sides.
\begin{align*}
\text{A)}\ & Rectangle &
\text{B)}\ & Rhombus \\
\text{C)}\ & Kite &
\text{D)}\ & Trapezoid\\
\end{align*} | Rectangles and Rhombus both have congruent opposite sides and a Trapezoid does not have two pairs of congruent adjacent sides. The Kite is the only thing that fits the description. | Geo21 | A |
Find the number of sides of a convex polygon if the measures of its interior angles have a sum of $2340^{\circ}$.
\begin{align*}
\text{A)}\ & 13 &
\text{B)}\ & 11 \\
\text{C)}\ & 15 &
\text{D)}\ & 7\\
\end{align*} | A convex polygon has interior angles summing to $180(n-2)$ where $n$ is the number of sides. Solving for $n$ results in 15 sides. | Geo23 | C |
The base of a square pyramid has sides of 10 and the slant height is 15. Find the surface area of the pyramid.
\begin{align*}
\text{A)}\ & 85 &
\text{B)}\ & 220 \\
\text{C)}\ & 310 &
\text{D)}\ & 400\\
\end{align*} | If the slant of the pyramid is 15 and the sides are 10, each side triangle has an area of 75 and the base has an area of 100. This comes to a sum of 400. | Geo39 | D |
Find the volume of a cone, to the nearest cubic inch, whose radius is 12 inches and whose height is 15 inches.
\begin{align*}
\text{A)}\ & 2827 &
\text{B)}\ & 2262 \\
\text{C)}\ & 565 &
\text{D)}\ & 188\\
\end{align*} | The area of a cone is $\frac{1}{3}bh$, since radius is 12 the base is $144\pi$. | Geo40 | B |
Find the volume of a hemisphere (half a sphere) whose radius is 10 feet. Round the answer to the nearest cubic foot.
\begin{align*}
\text{A)}\ & 419 &
\text{B)}\ & 1047 \\
\text{C)}\ & 2094 &
\text{D)}\ & 4189\\
\end{align*} | The formula for the volume of a sphere is $\frac{4}{3}\pi r^3$. | Geo41 | C |
The ratio of the volumes of two similar spheres is 8 : 27. If the larger sphere’s volume is $135 \ cm^3$, what is the volume of the smaller solid?
\begin{align*}
\text{A)}\ & 90\ {cm}^3 &
\text{B)}\ & 40\ {cm}^3 \\
\text{C)}\ & 80\ cm^3 &
\text{D)}\ & 50\ cm^3\\
\end{align*} | Since the larger sphere has a volume of 135, applying the ratio we can calculate the volume of a smaller solid by dividing by 27 and multiplying by 8. | Geo43 | B |
Two circles have areas of $49\pi \ in.^2$ and $144\pi\ in.^2$. What is the ratio of their radii?
\begin{align*}
\text{A)}\ & 49:144 &
\text{B)}\ & 49\pi:144\pi \\
\text{C)}\ & 7:12 &
\text{D)}\ & 343:1728\\
\end{align*} | We can determine the radius of each circle by dividing by pi then square-rooting the result. The smaller circle has radius of 7 and the larger one has radius of 12. | Geo44 | C |
Evaluate: $\log_5{125} = $
\begin{align*}
\text{A)}\ & 25 &
\text{B)}\ & 2 \\
\text{C)}\ & 3 &
\text{D)}\ & 1\\
\end{align*} | $5^3=125$ | PreCal-1 | C |
Express the logarithmic equation as an exponential equation and solve: \(\log_4 \frac{1}{64} = x\)
\noindent Options:\\
A) \( x^4 = \frac{1}{64}; x = -3 \)\\
B) \( 4^x = \frac{1}{64}; x = -3 \)\\
C) \( 64^x = \frac{1}{4}; x = -3 \)\\
D) \( \left(-\frac{1}{4}\right)^x = 64; x = -\frac{1}{3} \) | Definition of log. | AI-Algebra9 | B. |
Use the fact that $255^{\circ} = 210^{\circ} + 45^{\circ}$ to determine the \textit{\underline{exact}} value of $sin 255^{\circ}$.
\begin{align*}
\text{A)}\ & \frac{\sqrt{6}-\sqrt{2}}{4} &
\text{B)}\ & \frac{-\sqrt{2}-\sqrt{6}}{4} \\
\text{C)}\ & -\frac{1}{2} &
\text{D)}\ & \frac{1}{2} \\
\end{align*} | From $sin(a+b) = sin(a)cos(b)+cos(a)sin(b)$, we have
\begin{align*}
sin(255)& = sin(210)cos(45)+cos(210)sin(45)\\
&=-\frac{1}{2}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}
\end{align*} | PreCal-3 | B |
Find the \underline{exact} value for $sin2\theta$ given that $sin\theta = -\frac{12}{13}$ and $\pi\leq\theta\leq\frac{3\pi}{2}$ .
\begin{align*}
\text{A)}\ & \frac{119}{169} &
\text{B)}\ & -\frac{119}{169} \\
\text{C)}\ & \frac{120}{169} &
\text{D)}\ & -\frac{120}{169}\\
\end{align*} | \begin{align*}
sin(2x) & = 2sin(x)cos(x)\\
& = 2\times (-12/13)\times(-5/13)
& = \frac{120}{169}
\end{align*} | PreCal-4 | C |
Solve $7sinx+15=6sinx+14$ where $0\leq x \leq 2\pi$.
\begin{align*}
\text{A)}\ & 0 &
\text{B)}\ & \frac{\pi}{2} \\
\text{C)}\ & \pi &
\text{D)}\ & \frac{3\pi}{2}\\
\end{align*} | $sinx = -1$, so $x=\frac{3\pi}{2}$ | PreCal-5 | D |
Solve \( \cos^2\theta - 3\cos\theta - 4 = 0 \) where \( 0 \leq \theta < 2\pi \).
\noindent Options:\\
A) \( 0 \)\\
B) \( \frac{\pi}{2} \)\\
C) \( \pi \)\\
D) \( \frac{3\pi}{2} \) | \( (\cos(\theta) - 4)(\cos(\theta) + 1) = 0 \). Only second part can produce a solution so \( \theta = \pi \). | AI-Trigonometry1 | C |
Which of the following is \underline{not} a type of discontinuity?
\begin{align*}
\text{A)}\ & jump &
\text{B)}\ & hole \\
\text{C)}\ & \text{horizontal asymptote}&
\text{D)}\ & \text{vertical asymptote}\\
\end{align*} | Horizontal asymptote does not interrupt the function. | PreCal-7 | C |
Determine any points of discontinuity for $f(x)=\frac{x(x-5)}{(x-3)(x-5)}$
\begin{align*}
\text{A)}\ & 0 &
\text{B)}\ & 3 \\
\text{C)}\ & 3,5 &
\text{D)}\ & 0,3,5\\
\end{align*} | $x=5$ is a hole, $x=3$ is an infinite discontinuity. | PreCal-8 | C |
The synthetic division problem below proves which fact about $f(x)=x^4-3x^3+7x^2-60x-130$?\\
\[
\begin{array}{r|rrrrr}
15& 1 & -3 & 7 & -60 & -130 \\
& & 5 & 10 & 85 & 125\\
\hline
& 1 & 2 & 17 & 25 & -5\\
\end{array}
\]
\begin{align*}
\text{A)}\ & \text{ 5 is a root of f(x)} &
\text{B)}\ & \text{x-5 is a factor of f(x)}\\
\text{C)}\ & f(5)=-5 &
\text{D)}\ & x^3+2x^2+17x+25 \text{ is a factor of } f(x)\\
\end{align*} | Seeing as there is a remainder of -5 when the function is divided by $x-5$, plugging in 5 as $x$ must result in a $y$ of -5. | PreCal-9 | C |
Find the domain of $f(x)=log(x-5)$
\begin{align*}
\text{A)}\ & x>0 &
\text{B)}\ & x<5\\
\text{C)}\ & x>5 &
\text{D)}\ & \text{all real numbers}\\
\end{align*} | Log cannot be taken of negative numbers or zero. | PreCal-11 | C |
Identify the x and y –intercepts, if any, of the equation $y=\frac{-1}{x+1}+4$
\begin{align*}
\text{A)}\ & \text{x-int: }-1, \text{y-int: None} &
\text{B)}\ & \text{x-int: None, y-int: }3\\
\text{C)}\ & \text{x-int: }-\frac{3}{4}, \text{y-int: } 3 &
\text{D)}\ & \text{x-int:-1, y-int: }4\\
\end{align*} | Plug in zero for x to find y intercept and do opposite for x intercept. | PreCal-12 | C |
Find the first term and the common difference of the arithmetic sequence described: $8^{th}$ term $= 8$ ; $20^{th}$ term $= 44$
\begin{align*}
\text{A)}\ & a_1=-13; d=3 &
\text{B)}\ & a_1=-10, d=3 \\
\text{C)}\ & a_1=-13, d=-3 &
\text{D)}\ & a_1=-16, d=-3\\
\end{align*} | With a 12 term difference there is a $44-8 = 36$ unit jump, this means the common difference is 3. Subtracting $7\times 3 = 21$ from 8 results in a first term of -13. | PreCal-13 | A |
Which of the following is the equation of the horizontal asymptote of the graph of the function \( f(x) = \frac{4x^2}{x^3 - 5} \)?
\noindent Options:\\
A) \( x = \frac{2}{5} \)\\
B) \( x = 5 \)\\
C) \( y = 0 \)\\
D) \( y = 4 \) | Since the power in the denominator is greater than the numerator the asymptote will be 0. | AI-Calculus1 | C |
Simplify: \( \log_3 2 + \log_3 4 - 3\log_3 5 \)
\noindent Options:\\
A) \( \log_3 (-119) \)\\
B) \( \log_3 \left(\frac{8}{25}\right) \)\\
C) \( \log_3 \left(\frac{2}{5}\right) \)\\
D) non-real answer | \( \log_3 2 + \log_3 4 - \log_3 5^3 = \log_3 8 - \log_3 125 = \log_3 \left(\frac{8}{125}\right) \) | AI-Algebra10 | B |
Find: $\tan ^{-1}\left[\tan \left(\frac{2 \pi}{3}\right)\right] \tan ^{-1}\left[\tan \left(\frac{2 \pi}{3}\right)\right]$
\begin{align*}
\text{A)}\ & \frac{2 \pi}{3} &
\text{B)}\ & -\frac{\pi}{3} \\
\text{C)}\ & \frac{\pi}{3} &
\text{D)}\ & \text{undefined}\\
\end{align*} | Since the range of tangent is only from -pi/2 to pi/2, subtract pi from $2 \mathrm{pi} / 3$ | PreCal-16 | B |
Find: $\sin \left[\sin ^{-1}(-2)\right]$
\begin{align*}
\text{A)}\ & 2 &
\text{B)}\ & -2 \\
\text{C)}\ & -\frac{1}{2} &
\text{D)}\ & \text{undefined}\\
\end{align*} | No inverse sin of -2. | PreCal-17 | D |
$\sin ^{2} x-1=$\_\_\_\_\_
\begin{align*}
\text{A)}\ & \cos ^{2} x &
\text{B)}\ & -\cos ^{2} x\\
\text{C)}\ & \csc ^{2} x &
\text{D)}\ & -\csc ^{2} x \\
\end{align*} | $\cos ^{\wedge} 2+\sin ^{\wedge} 2=1$ | PreCal-18 | B |
$\cos ^{-1}\left(\frac{1}{2}\right)=$
\begin{align*}
\text{A)}\ & \frac{\pi}{6} &
\text{B)}\ & \frac{\pi}{4} \\
\text{C)}\ & \frac{\pi}{3} &
\text{D)}\ & \frac{\pi}{2}\\
\end{align*} | Unit circle. | PreCal-19 | A |
Over the interval $[0,2 \pi)$, solve: $2 \sin x-\sqrt{3}=0$
\begin{align*}
\text{A)}\ & \frac{\pi}{6} &
\text{B)}\ & \frac{\pi}{6}, \frac{11 \pi}{6}\\
\text{C)}\ & \frac{\pi}{3} &
\text{D)}\ & \frac{\pi}{3}, \frac{11 \pi}{3}\\
\end{align*} | $\sin x=\frac{\sqrt{3}}{2}$, use unit circle | PreCal-20 | B |
$\tan ^{-1}\left[\tan \left(\frac{5 \pi}{4}\right)\right]=$
\begin{align*}
\text{A)}\ & \frac{5 \pi}{4} &
\text{B)}\ & \frac{\pi}{4} \\
\text{C)}\ & 1 &
\text{D)}\ & -1 \\
\end{align*} | Subtract pi to get within inverse tangent range. | PreCal-21 | B |
What is the exact value of $\cos 22.5^{\circ}$ ?
\begin{align*}
\text{A)}\ & \frac{\sqrt{2+\sqrt{2}}}{2} &
\text{B)}\ & -\frac{\sqrt{2+\sqrt{2}}}{2} \\
\text{C)}\ & \frac{\sqrt{2-\sqrt{2}}}{2} &
\text{D)}\ & -\frac{\sqrt{2-\sqrt{2}}}{2}\\
\end{align*} | $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}, \cos 22.5=\sqrt{\frac{1+\cos 45}{2}}$
change $\sin^2 x$ to $1-\cos^2 x$ then factor for $(\cos x-2)(\cos x-1)=0$, only right part can be solved for. | PreCal-22 | A |
Which one is a solution to the equation: $\sqrt{3} \tan x+1=0$
\begin{align*}
\text{A)}\ & \frac{-\pi}{3} &
\text{B)}\ & \frac{\pi}{6}\\
\text{C)}\ & \frac{5 \pi}{6} &
\text{D)}\ & \frac{2 \pi}{3}\\
\end{align*} | $\tan x= -\frac{1}{\sqrt{3}}$ | PreCal-24 | C |
Calculate the coefficient of $x^2$ in the expansion of
$(x-3)^{5}$
\begin{align*}
\text{A)}\ & 270 &
\text{B)}\ & 90 \\
\text{C)}\ & -17 &
\text{D)}\ & -270\\
\end{align*} | Each factor with a coefficient of $x^2$ will be multiplied by -3 3 times. This means each factor has a coefficient of -27. There will be 10 such factors. | PreCal-25 | D |
What is the expansion of the polynomial $(x-2)^{4}$ ?
\begin{align*}
\text{A)}\ & x^{4}+16 &
\text{B)}\ & x^{4}-8 x^{3}+24 x^{2}-32 x+16 \\
\text{C)}\ & x^{4}-16 x^{3}+32 x^{2}-32 x+16 &
\text{D)}\ & x^{4}+4 x^{3}+6 x^{2}+4 x+1\\
\end{align*} | Use polynomial expansion formula. | PreCal-26 | B |
Find the sum. \( \sum_{n=1}^{10} 4n - 5 \)
\noindent Options:\\
A) \( 235 \)\\
B) \( 35 \)\\
C) \( 36 \)\\
D) \( 170 \) | Since this is an arithmetic sequence we can sum the first and last term then multiply by the amount of terms and divide by 2 for the sum. \( (-1 + 35) \times \frac{10}{2} \) | AI-Algebra11 | D |
Find the sum. \( \sum_{k=1}^{\infty} 6 \left(-\frac{2}{3}\right)^{k-1} \)
\noindent Options:\\
A) \( 0.6 \)\\
B) \( -0.6 \)\\
C) \( 3.6 \)\\
D) \( -3.6 \) | Since this is an infinite geometric sequence: \( \frac{a}{1-r} = \frac{6}{1 + \frac{2}{3}} = \frac{18}{5} = 3.6 \) | AI-Series1 | C |
Given $\triangle A B C$, where $\angle \mathrm{A}=41^{\circ}, \angle \mathrm{B}=58^{\circ}$, and $\mathrm{c}=19.7 \mathrm{~cm}$, determine the measure of side $\mathrm{b}$.
\begin{align*}
\text{A)}\ & \text{not possible} &
\text{B)}\ & 16.91 \mathrm{~cm} \\
\text{C)}\ & 0.89 \mathrm{~cm} &
\text{D)}\ & 12.94 \mathrm{~cm}\\
\end{align*} | Using law of $\operatorname{sines} \frac{\sin 58}{x}=\frac{\sin 81}{19.7}$ | PreCal-29 | B |
Given $\triangle A B C$, where $\mathrm{a}=9, \mathrm{~b}=12$, and $\mathrm{c}=16$, determine the measure of angle $\mathrm{B}$. Round to the nearest tenth.
\begin{align*}
\text{A)}\ & \text{not possible} &
\text{B)}\ & \quad 132.1^{\circ} \\
\text{C)}\ & 47.9^{\circ} &
\text{D)}\ & 1^{\circ}\\
\end{align*} | Use law of cosines | PreCal-30 | C |
In $\triangle A B C, A=47^{\circ}, B=56^{\circ}$, and $c=14$, find $b$.
\begin{align*}
\text{A)}\ & 77&
\text{B)}\ & 7.9 \\
\text{C)}\ & 10.5&
\text{D)}\ & 11.9\\
\end{align*} | Use law of sines | PreCal-31 | D |
Evaluate \( \tan(\alpha - \beta) \) given: \( \tan\alpha = -\frac{4}{3}, \frac{\pi}{2} < \alpha < \pi \) and \( \cos\beta = \frac{1}{2}, 0 < \beta < \frac{\pi}{2} \).
\noindent Options:\\
A) \( \frac{25\sqrt{3} + 48}{39} \)\\
B) \( -\frac{25\sqrt{3} + 48}{39} \)\\
C) \( \frac{16 + 7\sqrt{3}}{47} \)\\
D) \( -\frac{16 + 7\sqrt{3}}{47} \) | \( \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} = \frac{-\frac{4}{3} - \sqrt{3}}{1 - \frac{4}{3}\sqrt{3}} \) | AI-Trigonometry2 | A |
Evaluate $p(x)=x^{3}+x^{2}-11 x+12$ for $x=2$. | Plug in 2. | PreCal-33 | 2 |
Write $\ln \frac{x^{2}\left(y^{2}-z\right)^{3}}{\sqrt{y+1}}$ as the sum and/or difference of logarithms. Express powers as factors. | Split using $\log$ rules $2 \ln x+3 \ln \left(y^{2}-z\right)-\frac{1}{2} \ln (y+1)$. | PreCal-35 | 2 \ln x+3 \ln \left(y^{2}-z\right)-\frac{1}{2} \ln (y+1) |
Rewrite the following as the log of a single expression and simplify.
$$
\frac{1}{3} \log 125+2 \log (x-1)-3 \log (x+4)
$$ | first move all powers then combine: $\log 125^{1 / 3}+\log (x-1)^{2}-\log (x+4)^{3}$
$$
\log \left(5 *(x-7)^{2} /(x+4)^{3}\right)
$$ | PreCal-36 | \log \left(5 \times (x-7)^{2} /(x+4)^{3}\right) |
Solve $27^{3 x}=81$ for $x$. | $27^{\frac{4}{3}}=81,3 x=\frac{4}{3}, x=\frac{4}{9}$ | PreCal-37 | 4/9 |
Use long division to divide $f(x)=6 x^{3}-x^{2}-5 x+2$ by $3 x-2$. | $\frac{6 x^{3}-x^{2}-5 x+2}{3 x-2}=2 x^{2}+\frac{3 x^{2}-5 x+2}{3 x-2}=2 x^{2}+x+\frac{-3 x+2}{3 x-2}=2 x^{2}+x-1$ | PreCal-38 | 2 x^{2}+x-1 |
A culture of bacteria obeys the law of uninhibited growth. If 500 bacteria are present initially and there are 800 after 1 hour, how many will be present after 5 hours? | The rate of growth is $800 / 500$ per hour so $500 \times \frac{8^5}{5}=5242.88$
% | PreCal-40 | 5242.88 |
%
% \begin{align*}
% \text{A)}\ & &
% \text{B)}\ & \\
% \text{C)}\ & &
% \text{D)}\ & \\
% \end{align*}
% | % | PreCal- | |
% | PreCal- | ||
If \( f(x) \) is the function given by \( f(x) = e^{3x} + 1 \), at what value of \( x \) is the slope of the tangent line to \( f(x) \) equal to 2?
\noindent Options:\\
A) \( -0.173 \)\\
B) \( 0 \)\\
C) \( -0.135 \)\\
D) \( -0.366 \)\\
E) \( 0.231 \) | \( f'(x) = 3e^{3x} = 2. \) \( \ln(2) = 3x, x = -0.135 \) | AI-Calculus2 | C |
Which of the following is an equation for a line tangent to the graph of \( f(x) = e^{3x} \) when \( f'(x) = 9 \)?
\noindent Options:\\
A) \( y = 3x + 2.633 \)\\
B) \( y = 9x - 0.366 \)\\
C) \( y = 9x - 0.295 \)\\
D) \( y = 3x - 0.295 \)\\
E) None of these | \( f'(x) = 3e^{3x} = 9, x = 0.336, y - 3 = 9(x - 0.336) \) | AI-Calculus3 | C |
If $f^{\prime}(x)=\ln x-x+2$, at which of the following values of $x$ does $f$ have a relative maximum value?
\begin{align*}
\text{A)}\ & 3.146 &
\text{B)}\ & 0.159 \\
\text{C)}\ & 1.000&
\text{D)}\ & 4.505\\
\text{E)}\ & \text{None of these}\\
\end{align*} | Check when function crosses $\mathrm{x}$ axis by graphing | PreCal-3 | A |
\( \int \frac{4x}{16+x^4} \,dx = \)
\noindent Options:\\
A) \( \frac{1}{4} \sec^{-1} \frac{x^2}{4} + C \)\\
B) \( \frac{1}{2} \tan^{-1} \frac{x^2}{4} + C \)\\
C) \( \frac{1}{8} \sec^{-1} \frac{x^2}{4} + C \)\\
D) \( 2 \tan^{-1} \frac{x^2}{4} + C \)\\
E) None of these | Use inverse tangent derivatives | AI-Calculus4 | B |
If $f(x)=3 x^{2}-x$, and $g(x)=f^{-1}(x)$ over the domain $[0, \infty)$, then $g^{\prime}(10)$ could be which of the following?
\begin{align*}
\text{A)}\ & 59 &
\text{B)}\ & \frac{1}{59} \\
\text{C)}\ & \frac{1}{10} &
\text{D)}\ & 11\\
\text{E)}\ & \frac{1}{11}&
\end{align*} | $10=3 x^{2}-x,(3 x+5)(x-2)=0, x=2, f^{\prime}(2)=11, g^{\prime}(10)=\frac{1}{11}$ | PreCal-5 | E |
Find the distance traveled in the first four seconds for a particle whose velocity is given by $v(t)=7 e^{-t^{2}}$, where $t$ stands for time.
\begin{align*}
\text{A)}\ & 0.976 &
\text{B)}\ & 6.204 \\
\text{C)}\ & 6.359 &
\text{D)}\ & 12.720\\
\text{E)}\ & 7.000 &
\end{align*} | Take the integral of velocity for distance. | PreCal-6 | B |
Find $\lim _{x \rightarrow 0}-\frac{\sin (5 x)}{\sin (4 x)}$
\begin{align*}
\text{A)}\ & 0 &
\text{B)}\ & 1\\
\text{C)}\ & -5/4 &
\text{D)}\ & 5/4 \\
\text{E)}\ & \text{None of these}&
\end{align*} | Use lhopitals rule to take derivative of numerator and denominator of limit. | PreCal-7 | C |
Find the area $\mathrm{R}$ bounded by the graphs of $y=\sqrt{x}$ and $y=x^{2}$
\begin{align*}
\text{A)}\ & 0.333 &
\text{B)}\ & -0.333 \\
\text{C)}\ & 1.000 &
\text{D)}\ & -1.000\\
\text{E)}\ & \text{None of these}&
\end{align*} | $\int_{0}^{1} \sqrt{\chi}-x^{2} d x=0.333$ | PreCal-8 | A |
\[
\frac{d}{dx} \int_{0}^{3x} \cos(t) \, dt =
\]
\noindent Options:\\
A) \( \sin 3x \)\\
B) \( -\sin 3x \)\\
C) \( \cos 3x \)\\
D) \( 3 \sin 3x \)\\
E) \( 3 \cos 3x \) | Integrate to \(\sin\) then chain rule out a 3. | AI-Calculus5 | E |
The average value of the function $f(x)=(x-1)^{2}$ on the interval [1,5] is:
\begin{align*}
\text{A)}\ & -\frac{16}{3} &
\text{B)}\ & \frac{16}{3} \\
\text{C)}\ & \frac{64}{5} &
\text{D)}\ & \frac{66}{3}\\
\text{E)}\ & \frac{256}{3} &
\end{align*} | $\frac{\int_{7}^{5}(x-7)^{2} d x}{4}=\frac{16}{3}$ | PreCal-10 | B |
Write the following expression as a logarithm of a single quantity: $\ln x-12 \ln \left(x^{2}-1\right)$
\begin{align*}
\text{A)}\ & \ln \left(\frac{x}{\left(x^{2}-1\right)^{-12}}\right)&
\text{B)}\ & \ln \left(\frac{x}{12\left(x^{2}-1\right)}\right)\\
\text{C)}\ & \ln \left(x-12\left(x^{2}-1\right)\right) &
\text{D)}\ & \ln \left(\frac{x}{\left(x^{2}-1\right)^{12}}\right)\\
\text{E)}\ & \text{None of these}&
\end{align*} | Use log rules to turn 12 into an exponent and change the subtraction into one log division | PreCal-11 | D |
Find an equation of the tangent line to the graph of $y=\ln \left(x^{2}\right)$ at the point $(1,0)$.
\begin{align*}
\text{A)}\ & y=x-2 &
\text{B)}\ & \mathrm{y}=2(x+1) \\
\text{C)}\ & y=2(x-1) &
\text{D)}\ & y=x-1 \\
\text{E)}\ & \text{None of these}&
\end{align*} | $y^{\prime}=\frac{2 x}{x^{2}}=\frac{2}{x}, y^{\prime}(7)=2$ | PreCal-12 | C |
Find the area $\mathrm{R}$ bounded by the graphs of $y=x$ and $y=x^{2}$
\begin{align*}
\text{A)}\ & \frac{1}{6} &
\text{B)}\ & \frac{1}{2} \\
\text{C)}\ & \frac{-1}{6} &
\text{D)}\ & \frac{-1}{2}\\
\text{E)}\ & \text{None of these}&
\end{align*} | $\int_{0}^{1} x-x^{2} d x$ | PreCal-13 | A |
Find the indefinite integral: $\int \frac{x}{-2 x^{2}+3} d x$
\begin{align*}
\text{A)}\ & \frac{1}{-4 x}+C &
\text{B)}\ & \ln \left|-2 x^{2}+3\right|+C\\
\text{C)}\ & \frac{-1}{4} \ln \left|-2 x^{2}+3\right|+C &
\text{D)}\ & \frac{\ln \left|-2 x^{2}+3\right|}{-2 x^{2}+3}+C\\
\text{E)}\ & \text{None of these}&
\end{align*} | $u$ substitution with $u=-2 x^{2}+3$ | PreCal-14 | C |
Find the indefinite integral: $\int x \ln (x) d x$
\begin{align*}
\text{A)}\ & \frac{(\ln x)^{2}}{x}+C &
\text{B)}\ & \frac{x^{2} \ln (x)}{2}-\frac{x^{2}}{4}+C \\
\text{C)}\ & x \ln (x)+C &
\text{D)}\ & \ln (x)+1+C\\
\text{E)}\ & \text{None of these}&
\end{align*} | Integration by parts. | PreCal-15 | B |
Using the substitution $u=2 x+1, \int_{0}^{2} \sqrt{2 x+1} d x$ is equivalent to which of the following?
\begin{align*}
\text{A)}\ & \frac{1}{2} \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{u} &
\text{B)}\ & \frac{1}{2} \int_{0}^{2} \sqrt{u} d u \\
\text{C)}\ & \frac{1}{2} \int_{1}^{5} \sqrt{u} d u &
\text{D)}\ & \int_{0}^{2} \sqrt{u} d u\\
\text{E)}\ & \text{None of these}&
\end{align*} | Find new limits by plugging in 0 and 2 to the equation to get 1 and 5. | PreCal-16 | C |
Region $R$ is the area bounded by the graphs of $y=x$ and $y=x^{3}$. Find the volume of the solid generated when $R$ is revolved about the $x$-axis.
\begin{align*}
\text{A)}\ & \frac{\pi}{3} &
\text{B)}\ & \frac{21 \pi}{4}\\
\text{C)}\ & \frac{4 \pi}{21} &
\text{D)}\ & 3 \pi\\
\text{E)}\ & \text{None of these}&
\end{align*} | $\int_{0}^{1} \Pi\left(x^{2}-x^{6}\right) d x$ | PreCal-17 | C |
Find the indefinite integral: $\int x e^{2 x} d x$
\begin{align*}
\text{A)}\ & \frac{e^{2 x}}{x}+\frac{x}{e^{2 x}}+C &
\text{B)}\ & \frac{\ln (x)}{e}+C\\
\text{C)}\ & \frac{x}{e^{2 x}}+C &
\text{D)}\ & \frac{x e^{2 x}}{2 x}-\frac{e^{2 x}}{4}+C\\
\text{E)}\ & \text{None of these}&
\end{align*} | Integration by parts | PreCal-18 | E |
$\int x \sqrt{x+3} d x=$
\begin{align*}
\text{A)}\ & \frac{2}{3} x^{\frac{3}{2}}+6 x^{\frac{1}{2}}+C &
\text{B)}\ & \frac{2(x+3)^{\frac{3}{2}}}{3}+C\\
\text{C)}\ & \frac{3(x+3)^{\frac{3}{2}}}{2}+C &
\text{D)}\ & \frac{4 x^{2}(x+3)^{\frac{3}{2}}}{3}+C\\
\text{E)}\ & \frac{2}{5}(x+3)^{\frac{5}{2}}-2(x+3)^{\frac{3}{2}}+C &
\end{align*} | Integration by parts. | PreCal-19 | E |
Consider the differential equation $\frac{dy}{dx} = \frac{y-1}{x^3}$, where $x\neq 0$. Find the general solution $y=f(x)$ to the differential equation. | $\frac{1}{y-1} d y=\frac{1}{x^{2}} d x, \ln (y-1)=-x^{-1}+c, y=c e^{-\frac{1}{x}}+1$. | y = ce^{-\frac{1}{x}}+1 |
|
Compute the determinant of the matrix
\[
B = \begin{pmatrix}
3 & 0 & 2 \\
2 & 0 & -2 \\
0 & 1 & 1
\end{pmatrix}.
\] | Using the definition of the determinant,
\[
\det(B) = 3(0 \cdot 1 - (-2) \cdot 1) - 0 + 2(2 \cdot 1 - 0 \cdot (-2)) = 10.
\]
\bigskip | 10. |
|
Let
\[
A = \begin{pmatrix}
a &0 & c &b\\
1 & 0 &1 & 3\\
2 & 1 & -1 & 4\\
0 & 1 & 1& 5
\end{pmatrix}.
\]
and $A_{ij}$ be the algebraic cofactors of $A$. Compute $A_{11}+A_{12}+A_{13}+A_{14}.$ | \[A_{11}+A_{12}+A_{13}+A_{14}=\left|\begin{array}{cccc}
1 &1 & 1 &1\\
1 & 0 &1 & 3\\
2 & 1 & -1 & 4\\
0 & 1 & 1& 5
\end{array}\right|=21.\] | 21. |
|
Find the solution $[x_1,x_2,x_3]$ to the following equations
\[
\left\{\begin{array}{c}
x_1+3x_2+3x_3=16, \\
3x_1+x_2+3x_3=14, \\
3x_1+3x_2+x_3=12. \\
\end{array}\right.
\] | The second equation subtracts the first one, leading to
\[x_1-x_2=-1.\]
The third equation subtracts the second one, leading to
\[x_2-x_3=-1.\]
Thus
\[x_2=x_1+1,\quad x_3=x_1+2.\]
Inserting $x_2,x_3$ into the first one, we deduce that
\[7x_1+9=16.\]
Then $x_1=1, x_2=2,x_3=3$ is the solution. | [1,2,3]. |
|
Find the positively definite matrix $A\in \mathbb{R}^{3\times 3}$ such that
\[
A^2 = \begin{pmatrix}
11&7 & 7 \\
7 &11 &7\\
7 &7 & 11\\
\end{pmatrix}.
\]
In your answer, present the matrix in the form of $[a_{11},a_{12},a_{13}; a_{21},a_{22},a_{23}; a_{31},a_{32},a_{33} ]$ | Let
\[B=\begin{pmatrix}
11&7 & 7 \\
7 &11 &7\\
7 &7 & 11\\
\end{pmatrix}.\]
The characteristic polynomial of B is
\[\left|\begin{array}{ccc}
\lambda-11&-7 & -7 \\
-7 & \lambda-11 &-7\\
-7 &-7 & \lambda-11\\
\end{array}\right|=(\lambda-25)(\lambda-4)^2.\]
Thus, the eigenvalues of A are $25,4,4$. The corresponding eigenvectors are $\left(
\begin{array}{c}
1 \\
1 \\
1 \\
\end{array}
\right),
$ $\left(
\begin{array}{c}
1 \\
-2 \\
1 \\
\end{array}
\right),
$
$\left(
\begin{array}{c}
1 \\
0 \\
-1 \\
\end{array}
\right).
$
Set
\[U=\begin{pmatrix}
\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} &\frac{-2}{\sqrt{6}} &0\\
\frac{1}{\sqrt{3}} &\frac{1}{\sqrt{6}} & \frac{-1}{\sqrt{2}}\\
\end{pmatrix},\]
then
\[B=U\begin{pmatrix}
25&0 & 0 \\
0 &4 &0\\
0 &0 & 4\\
\end{pmatrix}U^{-1}.\]
Thus
\[A=U\begin{pmatrix}
5&0 & 0 \\
0 &2 &0\\
0 &0 & 2\\
\end{pmatrix}U^{-1}=\begin{pmatrix}
3&1 & 1 \\
1 &3 &1\\
1 &1 & 3\\
\end{pmatrix}.\] | [3, 1, 1; 1, 3, 1; 1, 1, 3] |
|
Compute the volume of the triangular pyramid
generated by four points $(1,1,1),(2,5,5), $ $(5,2,5) $, and $(5,5,2)$ in $\mathbb{R}^3.$ | Using the geometry meaning of the determinant, we know the volume of the triangular pyramid can be represented as
\[\dfrac{1}{6}
\left|
\begin{array}{ccc}
1 &4 & 4 \\
4 & 1 & 4 \\
4 & 4& 1 \\
\end{array}
\right|=\dfrac{27}{2}.
\] | 13.5 |
|
Find the values of $[a,b]$ such that $(1,2,1)^{\top}$ is an eigenvector of the matrix
$\left(
\begin{array}{ccc}
1 & 2 &1\\
3 & a & b \\
a & 0 & b \\
\end{array}
\right)
$. Present the answer as $[a,b]$. | Let $\lambda$ be the eigenvalue corresponding to eigenvector $(1,2,1)^{\top}$, then
\[\left(
\begin{array}{ccc}
1 & 2 &1\\
3 & a & b \\
a & 0 & b \\
\end{array}
\right)\left(
\begin{array}{c}
1\\
2 \\
1 \\
\end{array}
\right)=\lambda \left(
\begin{array}{c}
1\\
2 \\
1 \\
\end{array}
\right).\]
From this, we deduce that
\[6=\lambda, 3+2a+b=2\lambda, a+b=\lambda.\]
Solving the equation, we have $a=b=3.$ | [3,3] |
|
Find the matrix $A$ whose eigenvalues are 2,3,6 and corresponding eigenvectors are
$\begin{pmatrix} 1\\0 \\ -1 \end{pmatrix}, \begin{pmatrix}1\\1\\1 \end{pmatrix}, \begin{pmatrix}1\\-2\\1 \end{pmatrix}$ respectively.\\
In your answer, present the matrix in the form of $[a_{11}, a_{12}, a_{13}; a_{21}, a_{22}, a_{23}; a_{31}, a_{32}, a_{33} ]$. | Let
\[U=\left(
\begin{array}{ccc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\
0 & \frac{1}{\sqrt{3}} &\frac{-2}{\sqrt{6}}\\
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\
\end{array}
\right)\]
then
\[AU=U\left(
\begin{array}{ccc}
2 & 0 &0 \\
0 & 3 &0 \\
0 & 0 & 6 \\
\end{array}
\right).\]
Note that $U$ is an orthogonal matrix, then $U^{-1}=U^{\top}$. Thus
\[A=U\left(
\begin{array}{ccc}
2 & 0 &0 \\
0 & 3 &0 \\
0 & 0 & 6 \\
\end{array}
\right)U^{\top}=\left(
\begin{array}{ccc}
3 & -1 & 1 \\
-1 & 5 & -1 \\
1 & -1 & 3 \\
\end{array}
\right).\] | [3, -1, 1; -1, 5, -1; 1, -1, 3]. |
|
Compute the rank of the matrix
\[\left(
\begin{array}{ccc}
1 & 1 & 1 \\
2 & 0 & 3 \\
3 & 1 & 4 \\
\end{array}
\right)
\] | By elementary transformation of matrix, we have
\[\left(
\begin{array}{ccc}
1 & 1 & 1 \\
2 & 0 & 3 \\
3 & 1 & 4 \\
\end{array}
\right)\to \left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & -1 &1 \\
3 & 1 & 4 \\
\end{array}
\right)\to \left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & -2 &1 \\
0 & -2 & 1 \\
\end{array}
\right) \to \left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & -2 &1 \\
0 & 0 & 0 \\
\end{array}
\right).\]
It implies that the rank is 2. | 2 |
|
[Rank of a matrix]
Compute the dimension of the linear subspace generated by the following vectors
\[\left(\begin{array}{c}
1 \\
1 \\
1 \\
1
\end{array}\right),\left(\begin{array}{c}
1 \\
2 \\
1 \\
0
\end{array}\right), \left(\begin{array}{c}
0 \\
-1 \\
3 \\
4
\end{array}\right),\left(\begin{array}{c}
2 \\
2 \\
5 \\
5
\end{array}\right).
\] | Let
\[A=\left(
\begin{array}{cccc}
1 & 1 & 0 & 2 \\
1 & 2 & -1 & 2 \\
1 & 1 & 3 & 5 \\
1 & 0 & 4 & 5 \\
\end{array}
\right)
\]
Then
the dimension of the linear subspace generated by the column vectors of matrix A is $\text{rank}(A).$
By elementary transformation of matrix, we have
\[A\to \left(
\begin{array}{cccc}
1 & 1 & 0 & 2 \\
0 & 1 & -1 & 0 \\
0 & 0 & 3 & 3 \\
0 & -1 & 4 & 3 \\
\end{array}
\right)\to \left(
\begin{array}{cccc}
1 & 1 & 0 & 2 \\
0 & 1 & -1 & 0 \\
0 & 0 & 3 & 3 \\
0 & 0 & 3 & 3 \\
\end{array}
\right)\to \left(
\begin{array}{cccc}
1 & 1 & 0 & 2 \\
0 & 1 & -1 & 0 \\
0 & 0 & 3 & 3 \\
0 & 0 & 0 & 0 \\
\end{array}
\right).\]
It shows that the rank of $A$ is 3. Thus, the dimension is 3. | 3 |
|
Let the matrix
$A=\left(
\begin{array}{ccc}
2 & -2 & 1 \\
4 & -4 & 2 \\
6 & -6 & 3 \\
\end{array}
\right)
$.
Compute the product matrix $A^{2024}.$
In your answer, present the matrix in the form of $[a_{11}, a_{12}, a_{13}; a_{21}, a_{22}, a_{23}; a_{31}, a_{32}, a_{33} ]$. | Note that $A^2=A,$ so $A^{2024}=A=\left(
\begin{array}{ccc}
2 & -2 & 1 \\
4 & -4 & 2 \\
6 & -6 & 3 \\
\end{array}
\right).$ | [2, -2, 1; 4, -4, 2; 6, -6, 3]. |
|
Compute $|A^{-1}|$ for $A=\left(
\begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & 0 & 1 \\
\end{array}
\right).
$ | Since $AA^{-1}=I_3$ and $|AA^{-1}|=|A||A^{-1}|,$ we then obtain
\[|A^{-1}|=\dfrac{1}{|A|}=1.\] | 1. |