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这是伦敦政经学校线性代数课程的代写成功案例。
MA222课程简介
Teacher responsible
Dr James Ward
Availability
This course is compulsory on the BSc in Data Science. This course is available as an outside option to students on other programmes where regulations permit. This course is available with permission to General Course students.
Pre-requisites
Students should ideally have taken the course Mathematical Methods (MA100) or equivalent, entailing intermediate-level knowledge of linear algebra, linear independence, eigenvalues and diagonalisation.
Prerequisites
This course develops ideas first presented in MA100. It consists of the linear algebra part of MA212, covering the following topics: Vector spaces and dimension. Linear transformations, kernel and image. Real inner products. Orthogonal matrices, and the transformations they represent. Complex matrices, diagonalisation, special types of matrix and their properties. Jordan normal form, with applications to the solutions of differential and difference equations. Singular values, and the singular values decomposition. Direct sums, orthogonal projections, least square approximations, Fourier series. Right and left inverses and generalized inverses.
MA222 Linear algebra HELP（EXAM HELP， ONLINE TUTOR）
Determine if the following system is consistent:
$$
\begin{aligned}
x_1-2 x_2+x_3 & =0 \
2 x_2-8 x_3 & =8 \
5 x_1-5 x_3 & =10
\end{aligned}
$$
SOLUTION This is the system from Example 1. Suppose that we have performed the row operations necessary to obtain the triangular form
$$
\begin{aligned}
x_1-2 x_2+x_3 & =0 \
x_2-4 x_3 & =4 \
x_3 & =-1
\end{aligned} \quad\left[\begin{array}{rrrr}
1 & -2 & 1 & 0 \
0 & 1 & -4 & 4 \
0 & 0 & 1 & -1
\end{array}\right]
$$
At this point, we know $x_3$. Were we to substitute the value of $x_3$ into equation 2 , we could compute $x_2$ and hence could determine $x_1$ from equation 1 . So a solution exists; the system is consistent. In fact, $x_2$ is uniquely determined by equation 2 since $x_3$ has only one possible value, and $x_1$ is therefore uniquely determined by equation 1 . So the solution is unique.
Determine if the following system is consistent:
$$
\begin{aligned}
x_2-4 x_3 & =8 \
2 x_1-3 x_2+2 x_3 & =1 \
4 x_1-8 x_2+12 x_3 & =1
\end{aligned}
$$
SOLUTION The augmented matrix is
$$
\left[\begin{array}{rrrr}
0 & 1 & -4 & 8 \
2 & -3 & 2 & 1 \
4 & -8 & 12 & 1
\end{array}\right]
$$
To obtain an $x_1$ in the first equation, interchange rows 1 and 2:
$$
\left[\begin{array}{rrrr}
2 & -3 & 2 & 1 \
0 & 1 & -4 & 8 \
4 & -8 & 12 & 1
\end{array}\right]
$$
To eliminate the $4 x_1$ term in the third equation, add -2 times row 1 to row 3 :
$$
\left[\begin{array}{rrrr}
2 & -3 & 2 & 1 \
0 & 1 & -4 & 8 \
0 & -2 & 8 & -1
\end{array}\right]
$$
Next, use the $x_2$ term in the second equation to eliminate the $-2 x_2$ term from the third equation. Add 2 times row 2 to row 3 :
$$
\left[\begin{array}{rrrr}
2 & -3 & 2 & 1 \
0 & 1 & -4 & 8 \
0 & 0 & 0 & 15
\end{array}\right]
$$
The augmented matrix is now in triangular form. To interpret it correctly, go back to equation notation:
$$
\begin{aligned}
2 x_1-3 x_2+2 x_3 & =1 \
x_2-4 x_3 & =8 \
0 & =15
\end{aligned}
$$
The equation $0=15$ is a short form of $0 x_1+0 x_2+0 x_3=15$. This system in triangular form obviously has a built-in contradiction. There are no values of $x_1, x_2, x_3$ that satisfy (8) because the equation $0=15$ is never true. Since (8) and (5) have the same solution set, the original system is inconsistent (i.e., has no solution).
Row reduce the matrix $A$ below to echelon form, and locate the pivot columns of $A$.
$$
A=\left[\begin{array}{rrrrr}
0 & -3 & -6 & 4 & 9 \
-1 & -2 & -1 & 3 & 1 \
-2 & -3 & 0 & 3 & -1 \
1 & 4 & 5 & -9 & -7
\end{array}\right]
$$
SOLUTION Use the same basic strategy as in Section 1.1. The top of the leftmost nonzero column is the first pivot position. A nonzero entry, or pivot, must be placed in this position. A good choice is to interchange rows 1 and 4 (because the mental computations in the next step will not involve fractions).
$$
\left[\begin{array}{rrrrr}
1 & 4 & \text { Pivot } \
-1 & -2 & -1 & 3 & 1 \
-2 & -3 & 0 & 3 & -1 \
0 & -3 & -6 & 4 & 9
\end{array}\right]
$$
Create zeros below the pivot, 1 , by adding multiples of the first row to the rows below, and obtain matrix (1) below. The pivot position in the second row must be as far left as possible-namely, in the second column. Choose the 2 in this position as the next pivot.
$$
\left[\begin{array}{rrrrr}
1 & 4 & 5 & -9 & -7 \
0 & 2 & 4 & -6 & -6 \
0 & 5 & 10 & -15 & -15 \
0 & -3 & -6 & 4 & 9
\end{array}\right]
$$
Add $-5 / 2$ times row 2 to row 3 , and add $3 / 2$ times row 2 to row 4 .
$$
\left[\begin{array}{rrrrr}
1 & 4 & 5 & -9 & -7 \
0 & 2 & 4 & -6 & -6 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & -5 & 0
\end{array}\right]
$$
The matrix in (2) is different from any encountered in Section 1.1. There is no way to create a leading entry in column 3! (We can’t use row 1 or 2 because doing so would destroy the echelon arrangement of the leading entries already produced.) However, if we interchange rows 3 and 4 , we can produce a leading entry in column 4 .
The matrix is in echelon form and thus reveals that columns 1,2 , and 4 of $A$ are pivot columns.
$$
A=\left[\begin{array}{rrrr|r}
0 & -3 & -6 & 4 & 9 \
-1 & -2 & -1 & 3 & 1 \
-2 & -3 & 0 & 3 & -1 \
1 & 4 & 5 & -9 & -7
\end{array}\right] \text { Pivot positions }
$$
MY-ASSIGNMENTEXPERT™可以为您提供LSE.AC.UK MA222 LINEAR ALGEBRA线性代数的代写代考和辅导服务！
