Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 51ACE Linear Algebra, Multivariable Calculus, and Modern Applications线性代数代写代考和辅导服务！
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Multivariable calculus is a branch of mathematics that deals with functions of several variables, also known as vector calculus. It extends the concepts of differential calculus and integral calculus to higher dimensions. In multivariable calculus, we study functions of two or more independent variables and their derivatives, integrals, and applications.
The key concepts in multivariable calculus include partial derivatives, gradients, directional derivatives, the chain rule, double and triple integrals, line integrals, surface integrals, and the divergence theorem and Stokes’ theorem.
Multivariable calculus has a wide range of applications in science, engineering, economics, and other fields. It is used to model physical phenomena, such as fluid flow and electromagnetism, to optimize systems, such as in finance and management, and to understand complex data, such as in machine learning and data science.
Prove or disprove: A system of linear equations is homogeneous if and only if the
system has the zero vector as a solution.
This is a true statement. A proof is:
( $\Rightarrow$ ) Suppose we have a homogeneous system $\mathcal{C S}(A, 0)$. Then by substituting the scalar zero for each variable, we arrive at true statements for each equation. So the zero vector is a solution. This is the content of Theorem HSC.
( $\Leftarrow$ ) Suppose now that we have a generic (i.e. not necessarily homogeneous) system of equations, $\mathcal{L} \mathcal{S}(A, \mathbf{b})$ that has the zero vector as a solution. Upon substituting this solution into the system, we discover that each component of $\mathbf{b}$ must also be zero. So $\mathbf{b}=\mathbf{0}$.
Find $\alpha$ and $\beta$ that solve the vector equation.
$$
\alpha\left[\begin{array}{l}
2 \
1
\end{array}\right]+\beta\left[\begin{array}{l}
1 \
3
\end{array}\right]=\left[\begin{array}{l}
5 \
0
\end{array}\right]
$$
Performing the indicated operations (Definition CVA, Definition CVSM), we obtain the vector equations
$$
\left[\begin{array}{l}
5 \
0
\end{array}\right]=\alpha\left[\begin{array}{l}
2 \
1
\end{array}\right]+\beta\left[\begin{array}{l}
1 \
3
\end{array}\right]=\left[\begin{array}{l}
2 \alpha+\beta \
\alpha+3 \beta
\end{array}\right]
$$
Since the entries of the vectors must be equal by Definition CVE, we obtain the system of equations
$$
\begin{aligned}
& 2 \alpha+\beta=5 \
& \alpha+3 \beta=0
\end{aligned}
$$
which we can solve by row-reducing the augmented matrix of the system,
$$
\left[\begin{array}{ccc}
2 & 1 & 5 \
1 & 3 & 0
\end{array}\right] \stackrel{\text { RREF }}{\longrightarrow}\left[\begin{array}{ccc}
1 & 0 & 3 \
0 & 1 & -1
\end{array}\right]
$$
Thus, the only solution is $\alpha=3, \beta=-1$.
Suppose that $S=\left{\left[\begin{array}{c}-1 \ 2 \ 1\end{array}\right],\left[\begin{array}{l}3 \ 1 \ 2\end{array}\right],\left[\begin{array}{l}1 \ 5 \ 4\end{array}\right],\left[\begin{array}{c}-6 \ 5 \ 1\end{array}\right]\right}$. Let $W=\langle S\rangle$ and let $\mathbf{y}=\left[\begin{array}{c}-5 \ 3 \ 0\end{array}\right]$. Is $\mathbf{y} \in W$ ? If so, provide an explicit linear combination that demonstrates this.
Form a linear combination, with unknown scalars, of $S$ that equals $\mathbf{y}$,
$$
a_1\left[\begin{array}{c}
-1 \
2 \
1
\end{array}\right]+a_2\left[\begin{array}{l}
3 \
1 \
2
\end{array}\right]+a_3\left[\begin{array}{l}
1 \
5 \
4
\end{array}\right]+a_4\left[\begin{array}{c}
-6 \
5 \
1
\end{array}\right]=\left[\begin{array}{c}
-5 \
3 \
0
\end{array}\right]
$$
We want to know if there are values for the scalars that make the vector equation true since that is the definition of membership in $\langle S\rangle$. By Theorem SLSLC any such values will also be solutions to the linear system represented by the augmented matrix,
$$
\left[\begin{array}{ccccc}
-1 & 3 & 1 & -6 & -5 \
2 & 1 & 5 & 5 & 3 \
1 & 2 & 4 & 1 & 0
\end{array}\right]
$$
Row-reducing the matrix yields,
$$
\left[\begin{array}{lcccc}
{[1} & 0 & 2 & 3 & 2 \
0 & 1 & 1 & -1 & -1 \
0 & 0 & 0 & 0 & 0
\end{array}\right]
$$
From this we see that the system of equations is consistent (Theorem RCLS), and has a infinitely many solutions. Any solution will provide a linear combination of the vectors in $R$ that equals $\mathbf{y}$. So $\mathbf{y} \in S$, for example,
$$
(-10)\left[\begin{array}{c}
-1 \
2 \
1
\end{array}\right]+(-2)\left[\begin{array}{l}
3 \
1 \
2
\end{array}\right]+(3)\left[\begin{array}{l}
1 \
5 \
4
\end{array}\right]+(2)\left[\begin{array}{c}
-6 \
5 \
1
\end{array}\right]=\left[\begin{array}{c}
-5 \
3 \
0
\end{array}\right]
$$
