“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
PDF格式学习笔记下载（Academia.edu）
第6章课程讲义下载（PDF）
Summary
- Gaussian elimination consists of two steps:
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Forward Elimination of Unknowns
In this step, the unknown is eliminated in each equation starting with the first equation. This way, the equations are reduced to one equation and one unknown in each equation. -
Back Substitution
In this step, starting from the last equation, each of the unknowns is found.
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Forward Elimination of Unknowns
- More about determinant
- Let $[A]$ be a $n\times n$ matrix. Then if $[B]$ is a $n\times n$ matrix that results from adding or subtracting a multiple of one row (column) to another row (column), then $\det(A) = \det(B)$.
- Let $[A]$ be a $n\times n$ matrix that is upper triangular, lower triangular or diagonal, then $$\det(A) = a_{11}\times a_{22}\times\cdots\times a_{nn} = \prod_{i=1}^{n}a_{ii}$$ This implies that if we apply the forward elimination steps of Gaussian elimination method, the determinant of the matrix stays the same according to the previous result. Then since at the end of the forward elimination steps, the resulting matrix is upper triangular, the determinant will be given by the above result.
Selected Problems
1. Using Gaussian elimination to solve $$\begin{cases}4x_1+x_2-x_3=-2\\ 5x_1+x_2+2x_3=4\\ 6x_1+x_2+x_3=6\end{cases}$$
Solution:
Forward elimination: $$\begin{bmatrix}4& 1& -1& -2\\ 5& 1& 2& 4\\ 6& 1& 1& 6\end{bmatrix}\Rightarrow \begin{cases} R_2-{5\over4}R_1\\ R_3-{3\over2}R_1\end{cases}\begin{bmatrix}4& 1& -1& -2\\ 0& -{1\over4}& {13\over4}& {13\over2}\\ 0& -{1\over2}& {5\over2}& 9\end{bmatrix}$$ $$\Rightarrow R_3-2R_2\begin{bmatrix}4& 1& -1& -2\\ 0& -{1\over4}& {13\over4}& {13\over2}\\ 0& 0& -4& -4\end{bmatrix}$$ Back substitution: $$\begin{cases}-4x_3=-4\\ -{1\over4}x_2+{13\over4}x_3={13\over2}\\ 4x_1+x_2-x_3=-2\end{cases} \Rightarrow \begin{cases}x_3=1\\ -{1\over4}x_2+{13\over4}={13\over2}\\ 4x_1+x_2-1=-2\end{cases}$$ $$\Rightarrow \begin{cases}x_3=1\\ x_2 = -13\\ 4x_1-13=-1 \end{cases}\Rightarrow \begin{cases}x_1 = 3\\ x_2=-13\\ x_3=1 \end{cases}$$
2. Find the determinant of $$[A] = \begin{bmatrix}25& 5& 1\\ 64& 8& 1\\ 144& 12& 1\end{bmatrix}$$
Solution:
Forward elimination $$[A] = \begin{bmatrix}25& 5& 1\\ 64& 8& 1\\ 144& 12& 1\end{bmatrix}\Rightarrow\begin{cases}R_2 - {64\over25}R_1\\ R_3-{144\over25}R_1\end{cases} \begin{bmatrix}25& 5& 1\\ 0& -{24\over5}& -{39\over25}\\ 0& -{84\over5}& -{119\over25} \end{bmatrix}$$ $$\Rightarrow R_3-{7\over2}R_2 \begin{bmatrix}25& 5& 1\\ 0& -{24\over5}& -{39\over25}\\ 0& 0 & {7\over10} \end{bmatrix}$$ This is an upper triangular matrix and its determinant is the product of the diagonal elements $$\det(A) = 25\times(-{24\over5})\times{7\over10}=-84 $$
3. Find the determinant of $$[A] = \begin{bmatrix}10& -7& 0\\ -3& 2.099& 6\\ 5& -1& 5\end{bmatrix}$$
Solution:
Forward elimination $$[A] = \begin{bmatrix}10& -7& 0\\ -3& 2.099& 6\\ 5& -1& 5 \end{bmatrix}\Rightarrow\begin{cases}R_2 + {3\over 10}R_1\\ R_3-{1\over2}R_1\end{cases} \begin{bmatrix}10& -7& 0\\ 0& -{1\over1000}& 6\\ 0& {5\over2}& 5 \end{bmatrix}$$ $$\Rightarrow R_3+2500R_2 \begin{bmatrix}10& -7& 0\\ 0& -{1\over1000}& 6\\ 0& 0 & 15005 \end{bmatrix}$$ This is an upper triangular matrix and its determinant is the product of the diagonal elements $$\det(A) = 10 \times(-{1\over1000})\times15005=-150.05$$
4. Using Gaussian elimination to solve $$\begin{cases}3x_1-x_2 - 5x_3 = 9\\ x_2-10x_3=0\\ -2x_1+x_2=-6\end{cases}$$
Solution:
Forward elimination: $$\begin{bmatrix}3& -1& -5& 9\\ 0& 1& -10& 0\\ -2& 1& 0& -6\end{bmatrix}\Rightarrow R_3+{2\over3}R_1 \begin{bmatrix}3& -1& -5& 9\\ 0& 1& -10& 0\\ 0& {1\over3}& -{10\over3}& 0\end{bmatrix}$$ $$\Rightarrow R_3-{1\over3}R_2 \begin{bmatrix}3& -1& -5& 9\\ 0& 1& -10& 0\\ 0& 0 & 0 & 0\end{bmatrix}$$ Back substitution: $$\begin{cases}x_2-10x_3=0\\ 3x_1-x_2-5x_3=9\end{cases} \Rightarrow \begin{cases}x_2 = 10x_3\\ 3x_1-15x_3 = 9\end{cases} \Rightarrow \begin{cases}x_1 = 5x_3+3\\ x_2 = 10x_3\end{cases}$$ where $x_3$ is arbitrary.