A.kaw Matrix Algebra Preliminary study notes: 1. Introduction

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"Matrix Algebra Preliminary" (Introduction to Matrix ALGEBRA) course by Prof. A.k.kaw (University of South Florida) is designed and taught.
1th. Download the course handout (PDF)

Summary

  • Matrix
    A matrix is a rectangular array of elements. Matrix $A $ is denoted by $ $A = \begin{bmatrix}a_{11} & \cdots & a_{1n}\\ \vdots&\vdots&\vdots \ A_{m1}&am P \cdots & A_{mn} \end{bmatrix}$$
  • Vector
    A vector is a matrix of that have only one row or one column. For example, $[1, 2, 3]$ is a row vector of Dimension 3, and $\begin{bmatrix}1 \ 2 \ 3 \end{bmatrix}$ is a column vector of Dimension 3.
  • Equal matrices
    Matrices $[a]$ and $[b]$ are equal if the size of $[a]$ and $[b]$ is the same, which is, the number of rows and columns of $[a]$ is same as that of $[b]$. and $a _{ij}=b_{ij}$ for all $i $ and $j $.
  • Zero Matrix
    A matrix whose all entries was zero is called a zero matrix, which is, the $a _{ij}=0$ for all $i $ and $j $. For example, $ $A = \begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$
  • Submatrix
    If Some rows or/and columns of a matrix $[a]$ is deleted (no rows or columns could be deleted), the remaining matrix is cal Led a submatrix of $[a]$. For example, some of the Submatrix of $\begin{bmatrix}1 & 2 \ 3 & 4\\ 5 & 6 \end{bmatrix}$ is $$[1],\ [1, 2] , \ \begin{bmatrix}1\\3\\5\end{bmatrix},\ \begin{bmatrix} 1 & 2\\3 & 4 \end{bmatrix},\ \begin{bmatrix} 1 & 2\\5 & 6 \end{bmatrix},\ \begin{bmatrix}1 & 2 \ 3 & 4\\ 5 & 6 \end{bmatrix}.$$
  • Square Matrix
    If the number of rows of a matrix is equal to the number of columns of a matrix, then the matrix is called a square matrix . For example, $ $A = \begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}$$
  • Diagonal Matrix
    A square matrix with all non-diagonal elements equal to zero are called a diagonal matrix, that's, only the diagonal Entri Es of the square matrix can is Non-zero, $a _{ij} = 0$ for $i \neq j$. For example, $ $A =\begin{bmatrix}1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5 \end{bmatrix}$$
  • Identity Matrix
    A diagonal matrix with all diagonal elements equal to 1 are called an identity matrix, that's, $a _{ij}=0$, $i \neq j$ for a ll $i $, $j $ and $a _{ii}=1$ for all $i $. For example, $ $A = \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$
  • Upper Triangular matrix
    A $n \times n$ matrix for which $a _{ij} = 0$, $i > j$ for all $i $, $j $ was called an upper triangular matrix. That's, the elements below the diagonal entries is zeros. For example, $ $A = \begin{bmatrix}1 & 0 & 3\\ 0 & 5 & 6\\ 0 & 0 & 9 \end{bmatrix}$$
  • Lower Triangular matrix
    A $n \times n$ matrix for which $a _{ij} = 0$, $j > i$ for all $i $, $j $ is called a lower triangular matrix. That's, the elements above the diagonal entries is zeros. For example, $ $A = \begin{bmatrix}1 & 0 & 0\\ 4 & 5 & 0\\ 0 & 8 & 9 \end{bmatrix}$$
  • tridiagonal Matrix
    A tridiagonal matrix is a square matrix in which all elements not on the Follo Wing is zero:the major diagonal, the diagonal above the major diagonal, and the diagonal below the major diagonal. For example, $ $A = \begin{bmatrix}1 & 2 & 0 & 0\\ 4 & 5 & 6 & 0\\ 0 & 0 & 7 & 8\\ 0&am P 0& -1& 2 \end{bmatrix}$$ Note that a non-square The matrix also has diagonal entries. For an $m \times n$ matrix, the diagonal entries is $a _{11}$, $\cdots$, $a _{kk}$ where $k =\min\{m, n\}$. For example, $ $A = \begin{bmatrix}1& 2\\ 3& 4 \ \ 5& 6\end{bmatrix}$$ The diagonal elements is $a _{11}=1$ and $a _{22}=4$.
  • diagonally dominant matrix
    an $n \times n$ square matrix $[a]$ is A diagonal dominant matrix of $$ |a_{ii}|\geq \sum_{j=1,\ i\neq j}^{n}|a_{ij}|$$ for $i = 1, \cdots, n$ and $$|a_{ii}| > \sum_{j=1,\ i\neq j}^{n}|a_{ij}|$$ for at least one $i $. That's, for each row, the absolute value of the diagonal element was greater than or equal to the sum of the absolute valu Es of the rest of the elements of that row, and that the inequality are strictly greater than for at least one row. For example, $ $A = \begin{bmatrix}15& 6& 7\\ 2& -4& -2\\ 3& 2& 6 \end{bmatrix}$$ is A diagonal dom Inant Matrix since $$\begin{cases}|a_{11}| = \geq |a_{12}| + |a_{13}| =13\\ |a_{22}|= 4 \geq |a_{21}| + |a_{23}| = 4\\ |a_{33}| = 6 \geq |a_{31}| + |a_{32}| = 5 \end{cases}$$ and for at least one row, which is row 1 and row 3 in this case, the inequality is a strictly greater tha N Inequality.

Selected problems

1. Given $ $A =\begin{bmatrix}6& 2& 3& 9\\ 0& 1& 2& 3\\ 0& 0& 4& 5\\ 0& 0& 0& ; 6 \end{bmatrix}$$ then $[a]$ is A () matrix.

Solution:

This was an upper triangular matrix.

2. A square matrix $[a]$ is lower triangular if ().

Solution:

Lower triangular matrix: $a _{ij} = 0$ for $j > i$.

3. Given $ $A = \begin{bmatrix} 12.3& -12.3& 20.3\\ 11.3& -10.3& -11.3\\ 10.3& -11.3& -12.3\END{BMA Trix},\ B = \begin{bmatrix} 2& 4\\ -5& 6\\ 11& -20\end{bmatrix}$$ Then if $[c] = [a]\cdot[b]$, then $c _{31}= ( )$.

Solution:

$ $c _{31} = \begin{bmatrix}10.3 & -11.3 &-12.3\end{bmatrix}\cdot \begin{bmatrix}2\\ -5\\ one \end{bmatrix}$$ $$= 10 .3\times2 + ( -11.3) \times ( -5) + ( -12.3) \times11= -58.2$$

4. The following system of equations has () solutions. $$\begin{cases}x + y =2\\ 6x + 6y=12 \end{cases}$$

Solution:

$x =2-y$ where $y $ is arbitrary. Thus it has infinite solutions.

5. Consider there is only the computer companies in a country. The companies is named Dude and Imac. Each year, Dude keeps ${1/5}^{th}$ of it customers, while the rest switch to Imac. The Imac keeps ${1/3}^{rd}$ of its customers, and the rest switch to Dude. If in 2003, Dude had ${1/6}^{th}$ of the market and Imac had ${5/6}^{th}$ of the market, what would be is the share of Dude Co Mputers when is the market becomes stable?

Solution:

Since we want when the market is stable, the market share should does not change from year to year. Let $D $ and $M $ denote the market of Dude and Imac, respectively. Thus we have $$\begin{cases} d_n = {1\over5}d + {2\over3}m\\ m_n= {4\over5}d + {1\over3}m\end{cases}\rightarrow \BEGIN{BMA trix}d_n\\ m_n \end{bmatrix} = \begin{bmatrix}{1\over5} & {2\over3}\\ {4\over5}& {1\over3} \end{bmatrix}\cdot \ begin{bmatrix}d\\ m\end{bmatrix}$$ $D _n = d$ and $M _n=m$ eventually. That is $$\begin{cases}{4\over5}d-{2\over3} m=0 \ d+m=1\end{cases}\rightarrow\begin{cases}d = {5\over11}\\ M= {6\over1 1} \end{cases}$$ Hence The final market share of Dude would be $\displaystyle{5\over11}$.

6. Three Kids-jim, Corey and David receive an inheritance of \$2,253,453. The money was put in three trusts, but isn't divided equally to begin with. Corey ' s trust is the three times that of David's because Corey made an A in Dr Kaw's class. Each trust was put in an interest generating investment. The three trusts of Jim, Corey and David pays an interest of 6\%, 8\%, 11\%, respectively. The total interest of the three trusts combined at the end of the first year is \$190,740.57. The equations to find the trust money of Jim (J), Corey (C) and David (D) in a matrix form is ().

Solution:

From the given conditions, we have $$\begin{cases}j + C +d =2253453\\ c=3d\\ 0.06J + 0.08C + 0.11D = 190740.57\end{cases}$ $ $$\rightarrow \begin{cases}j + C +d =2253453\\ c-3d = 0\\ 0.06J + 0.08C + 0.11D = 190740.57\end{cases}$$ $$\rightarrow \ begin{bmatrix}1& 1& 1\\ 0& 1& -3\\ 0.06& 0.08& 0.11 \end{bmatrix}\cdot \begin{bmatrix}J\\ C\\ D \e Nd{bmatrix} = \begin{bmatrix}2253453\\ 0\\ 190740.57 \end{bmatrix}$$

7. Which of the following matrices is strictly diagonally dominant? $ $A = \begin{bmatrix}15 &6 &7\\ 2 &-4 &2\\ 3& 2 &6 \end{bmatrix},\ B = \begin{bmatrix}5 &6 &am p;7\\ 2 &-4 &2\\ 3& 2 &-5 \end{bmatrix},\ C = \begin{bmatrix}5&3 &2\\ 6 &-8 &2\\ 7&-5 &12 \end{bmatrix}. $$

Solution:

For $A $, $$\begin{cases}|a_{11}|=15 > |a_{12}| + |a_{13}| = 13\\ |a_{22}| = 4 = |a_{21}| + |a_{23}| = 4\\ |a_{33}| = 6 > |a_{31}| + |a_{32}| = 5 \end{cases}$$ So it is strictly diagonal dominant. For $B $, $$|b_{11}| = 5 < |b_{12}| + |b_{13}| = 13$$ So it was not strictly diagonal dominant. For $C $, $$\begin{cases}|c_{11}|=5 = |c_{12}| + |c_{13}| = 5\\ |c_{22}| = 8 = |c_{21}| + |c_{23}| = 8\\ |c_{33}| = |c_{31}| + |c_{32}| = \end{cases}$$ So it was not strictly diagonal dominant.

A.kaw Matrix Algebra Preliminary study notes: 1. Introduction

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