Ok so the kernel of ##T## is ##(x,y,z)## such that ##T(x,y,z)=0## & this only occurs when we have ## (1,1,1)## so I guess that is the basis for the kernel right???
Okay so I found the eigenvalues to be ##\lambda = 0,-1,2## with corresponding eigenvectors ##v =
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix},
\begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix},
\begin{pmatrix}
1 \\
1 \\
0
\end{pmatrix}
##.
Not sure what to do next. Thanks!!!
Show that ##\{\mathbf{v}_1\}## is linearly independent. Simple enough lets consider
$$c_1\mathbf{v}_1 = \mathbf{0}.$$
Our goal is to show that ##c_1 = 0##. By the definition of eigenvalues and eigenvectors we have ##A\mathbf{v}_1= \lambda_1\mathbf{v}_1##. Lets multiply the above equation and...
The solution is 3: It's just ##(g\circ f)'(6) = (-1,-2)\cdot (4,-3) = (-1\times 4)+((-2)\times (-3)) = -4+6 = 2## using the multi-variate chain rule and the dot product.
Is this correct and if not how do I go about doing it?
Thanks!
Okay (1) and (2) are done.
So for (3), assuming ##t > 0##, ##f\circ \mathbf{r} = \ln{(e^{\sin{(t)}})}^2+\ln{(e^{\sin{(t)}})}^2 = \sin^2{(t)}+\cos^2{(t)} = 1## so the derivative is ##0##.
Yeah that certainly doesn't make sense!!!
1. Suppose ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty.
2. Suppose ##x_1,x_2\in T##. We then have that ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##, i.e. ##x_1+x_2\in T##.
3. Suppose ##\mathbf{x}\in T##, with ##\lambda\in \mathbb{R}##...
Thanks!!!
1.Since ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty.
2. Suppose ##x_1,x_2\in T##. Then there exists vectors ##x_1,x_2\in S## such that we have ##A(x_1)## and ##A(x_2)##. We then have that ##x_1+x_2\in S## and ##A(x_1+x_2) = A(x_1)+A(x_2)##, i.e. ##x_1+x_2\in T##.
3...
1. We find the partial derivatives of ##f## with respect to ##x## and ##y## to get ##f_x = \frac{2\ln{(x)}}{x}## and ##f_y = \frac{2\ln{(y)}}{y}.## This makes the gradient vector
$$\nabla{f} = \begin{bmatrix}
f_x \\
f_y
\end{bmatrix} = \begin{bmatrix}
\frac{2\ln{(x)}}{x} \\
\frac{2\ln{(y)}}{y}...
1. Lets show the three conditions for a subspace are satisfied:
Since ##\mathbf{0}\in \mathbb{R}^n##, ##A\times \mathbf{0} = \mathbf{0}\in S##.
Suppose ##x_1, x_2\in \mathbb{R}^n##, then ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##.
Suppose ##x\in S## and ##\lambda\in \mathbb{R}##, then ##A(\lambda x) =...
There are sets of the form ##\left\{(x,y)\in \mathbb{R}^2: f(x,y) = \ln{\left(3+(x+y)^2\right)} = c\right\}## where ##c## is some fixed number ##> 1##. Lets see what happens for a few values of ##c##.
Suppose ##c = 2##, then ##\ln{\left(3+(x+y)^2\right)} = 2 \Longleftrightarrow (x+y)^2 =...
Hold up.
I can just calculate ##\int_{0.5}^{1} (2x-1)^4 \; dx=0.1##.
Then calculate ##\int_{7/8}^{1} 8x-7 \; dx=0.0625##.
Now take the difference to get ##\frac{3}{80}=0.0375##.
1. Homework Statement
A wire pattern is inserted into a ##10##cm square by making a horizontal line in the middle of the square (not all the way across and with length ##x##) and connecting the ends of this line to the closest two corners. What is the minimum value of ##x##?
2. Homework...
Ok I found the equation of the tangent curve in the standard way to get ##y=8x-7## which cuts the x axis at ##x=\frac{7}{8}##.
To find P I found the x coordinate Of the turning point which is ##x=0.5## so now we have our bounds of our integral so we just calculate ##\int_{0.5}^{7/8}...
I seem to read it fine once I open the attachment.
It is asking given we have ##f(x) = (2x-1)^4##. The curve meets the ##x##-axis at a point ##P## and the line on the graph is a tangent to the curve at the point ##Q(1,1)##.
Find the area of the region bounded by the curve, the ##x##-axis, and...
Member warned to type the problem statement, not just post an image with type that is too small to read
1. Homework Statement
See attached.
2. Homework Equations
3. The Attempt at a Solution
Ok so the first thing you wanna do is find the equation of the tangent line which is done in the...
1. Homework Statement
A curve is defined by the parametric equations ##x=t^3+1## and ##y=t^2+1##.
Show that ##\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^4}## is a constant.
2. Homework Equations
3. The Attempt at a Solution
So you differentiate both equations wrt ##t## then apply...
1. Homework Statement
If ##\text{arg}(w)=\frac{\pi}{4}## and ##|w\cdot \bar{w}|=20##, then what is ##w## of the form ##a+bi##.
2. Homework Equations
3. The Attempt at a Solution
The only way for the argument of ##w## to be ##\frac{\pi}{4}## is when ##a+bi## where ##a=b \in \mathbb{Z}##...
My solution is ##-\frac{106}{3}## which is clearly wrong as it has to be positive.
My limits of integration on ##y## is ##0## and ##-2## for ##R_1## and ##0## and ##2## for ##R_2##.