A.kaw Matrix Algebra Preliminary study Note 3. Binary Matrix Operations

"Matrix Algebra Preliminary" (Introduction to Matrix ALGEBRA) course by Prof. A.k.kaw (University of South Florida) is designed and taught.
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Summary

• addition of matrices
  The matrices $[a]$ and $[b]$ can is added only if they is the same size. The addition is then shown as $$[c]=[a]+[b]$$ where $ $c _{ij}=a_{ij} + b_{ij}$$ for example $$\begin{bmatrix}5& 2& 3\\ 1& 2& 7\end{bmatrix} + \begin{bmatrix}6& 7& -2\\ 3& 5& 19\end{bmatrix}$$ $$= \begin{bmatrix}5 + 6& 2 + 7& 3-2\\ 1 + 3& 2 + 5& 7 + 19\end{bmatrix} = \begin{bmatrix}11& 9& 1\\ 4& 7& 26 \end{bmatrix}$$
• Subtraction of matrices The matrices $[a]$ and $[b]$ can is subtracted only if they is the same size. The subtraction is then given by $$[d] = [a]-[b]$$ where $ $d _{ij}=a_{ij}-b_{ij}$$ for example $$\begin{bmatrix}5& 2& Amp 3\\ 1& 2& 7\end{bmatrix}-\begin{bmatrix}6& 7& -2\\ 3& 5& 19\end{bmatrix}$$ $$= \begin{bmatrix}5 -6& 2-7& 3-( -2) \ 1-3& 2-5& 7-19\end{bmatrix} = \begin{bmatrix}-1& -5& 5\\ -2& -3&am P -12\end{bmatrix}$$
• Multiplication of matrices
  Matrices $[a]$ and $[b]$ can be multiplied only if the number of columns of $[a]$ are equal to the number of rows of $[ b]$ to give $$[c]_{m\times n}=[a]_{m\times p}\cdot[b]_{p\times n}$$ where $ $c _{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_ {ip}b_{pj}= \sum_{k=1}^{p}a_{ik}b_{kj}$$ for each $i =1, \cdots, m$ and $j =1, \cdots, n$. That is, the $i $-th row and the $j $-th column of $[c]$ are calculated by multiplying the $i $-th row of $[a]$ by the $j $-th Column of $[b]$: $ $c _{ij}=\begin{bmatrix}a_{i1} & a_{i2}& \cdots & A_{ip}\end{bmatrix}\cdot \begin{bmatrix} b_{1j}\\ b_{2j}\\ \vdots\\ b_{pj}\end{bmatrix}$$ For example, $ $A =\begin{bmatrix}5& 2& 3\\ 1& 2& 7\END{BM atrix},\ b=\begin{bmatrix}3& -2\\ 5& -8\\ 9& -10\end{bmatrix},\ C=a\cdot B$$ We have $ $c _{11}=\begin{bmatrix} 5& 2& 3\end{bmatrix}\cdot \begin{bmatrix}3\\ 5\\ 9\end{bmatrix}=15+10+27=52,\ c_{12}=\begin{bmatrix}5& 2 & 3\end{bmatrix}\cdot \begin{bmatrix}-2\\ -8\\ -10\end{bmatrix}=-10-16-30 = -56,$$ $ $c _{21}=\begin{bmatrix}1& 2& 7\end{bmatrix}\cdot \begin{bmatrix}3\\ 5\\ 9\end{bmatrix}=3+10+63 =76,\ c_{22}=\begin{bmatrix}1& 2& 7\end{bmatrix}\cdot \begin{bmatrix}-2\\ -8\\ -10\end{bmatrix}=-2-16-70=-88 , $$ $$\rightarrow c=\begin{bmatrix}52& -56\\ 76& -88\end{bmatrix}.$$
• Scalar product of matrices
  If $[a]$ is a $m \times n$ matrix and $k $ are a real number, then the multiplication $[a]$ by a scalar $k $ is another $m \tim Es n$ Matrix $[b]$, where $b _{ij}=ka_{ij}$ for all $i $, $j $. For example, $$2\cdot\begin{bmatrix}5& 2& 3\\ 1& 2& 7\end{bmatrix} = \begin{bmatrix}10& 4& 6\\ 2&A mp 4& 14\end{bmatrix}$$
• Linear combination of matrices
  If $[a_1]$, $[a_2]$, $\cdots$, $[a_{p}]$ is matrices of the same size and $k _1$, $k _2$, $\cdots$, $k _p$ is scalars, then $ $k _1a_1 + k_2a_2 + \cdots + k_pa_p$$ is called a linear combination of $[a_1]$, $[a_2]$, $\cdots$, $[a_{p}]$.
• Rules of binary matrix operation
  • Commutative Law of addition
    If $[a ]$ and $[b]$ is $m \times n$ matrices, then $$[a]+[b] = [B] + [a]$$
  • associative law of Addition
    If $[a]$, $[b]$, and $[c]$ is all $m \times n$ matrices, then $$[a]+ ([B] + [C]) = ([A] + [B]) + [c]$$
  • associate Law of multiplication
    If $[a]$, $[b]$, and $[c]$ is $m \times n$, $n \times p$, and $p \times R $ Size matrices. respectively. Then $$[a]\cdot ([b]\cdot[c]) = ([a]\cdot[b]) \cdot[c]$$ and the resulting matrix size in both sides of the equation is $m \ t IMEs R $.
  • Distributive Law
    If $[a]$ and $[b]$ is $m \times n$ size matrices, and $[c]$ and $[d]$ are $n \tim Es p$ size matrices, then $$[a]\cdot ([C] + [D]) = [A]\cdot[c] + [a]\cdot[d]$$ $$ ([A] + [B]) \cdot[c] = [A]\cdot[c] + [B]\CD ot[c]$$ and the resulting matrix size on both sides of the equation are $m \times p$.

Selected problems

1. For the following matrices $ $A =\begin{bmatrix}3& 0\\ -1& 2\\ 1& 1\end{bmatrix},\ b=\begin{bmatrix}4&-1 \ 0& 2\end{bmatrix},\ c=\begin{bmatrix}5& 2\\ 3& 5\\ 6& 7\end{bmatrix}.$$ Find where possible $4[A] + 5[C ]$, $[a]\cdot[b]$, $[a]-2[c]$.
Solution:
$$4[a] + 5[c] = \begin{bmatrix}12& 0\\ -4& 8\\ 4& 4\end{bmatrix} + \begin{bmatrix}25& 10\\ 15& 25\\ 30 & 35\end{bmatrix} = \begin{bmatrix}37& 10\\ 11& 33\\ 34& 39\end{bmatrix}$$ $$[A] \cdot [B] = \begin{bmatri x}12& -3\\ -4& 5\\ 4& 1\end{bmatrix}$$ $$[a]-2[c] = \begin{bmatrix}3& 0\\ -1& 2\\ 1& 1\end{bmatrix }-\begin{bmatrix}10& 4\\ 6& 10\\ 12& 14\end{bmatrix} = \begin{bmatrix}-7& -4\\ -7& -8\\ -11&-13 \end{bmatrix}$$

2. Food orders is taken from the engineering departments for a takeout. The order is tabulated in Table 1.

However they has a choice of buying this food from three different restaurants. Their prices for the three food items is tabulated in Table 2.

Show how much each department would pay for their order at each restaurant. Which restaurant would is more economical to order from for each department?
Solution:
Denote the food order and price matrices as $$[a]=\begin{bmatrix}25& 35& 25\\ 21& 20& 21\end{bmatrix},\ [B ]=\begin{bmatrix}2.42 & 2.38 & 2.46\\ 0.93 & 0.90 & 0.89\\ 0.95 & 1.03 & 1.13 \end{bmatrix}$$ the Total Fees Matrix $[c]$ is the product of $[a]$ and $[b]$: $$[c]=[a]\cdot [B] = \begin{bmatrix}116.8 & 116.75 & 12 0.9\\ 89.37 & 89.61 & 93.19\end{bmatrix}$$ Covert It using tabular is shown in Table 3.

Thus, Burcholestrol is the cheapest for the Mechanocal department, which is 116.75. And Macfat is the cheapest for the civil department, which is 89.37.

3. Given $$[a] = \begin{bmatrix}2& 3& 5\\ 6& 7& 9\\ 2& 1& 3\end{bmatrix},\ [B]= \begin{bmatrix} 3& 5\\ 2& 9\\ 1& 6\end{bmatrix} ,\ [C]= \begin{bmatrix}5& 2\\ 3& 9\\ 7& 6\end{bmatrix}.$$ Illustra Te the distributive law of Binary matrix Operations: $$[a]\cdot ([b]+[c]) = [A]\cdot[b] + [a]\cdot[c]$$
solution: $$[b]+[c] = \begin{bmatrix}8& 7\\ 5& 18\\ 8& 12\end{bmatrix},\ [A]\cdot ( [b]+[c]) = \begin{bmatrix}71& 128\\ 155& 276\\ 45& 68\end{bmatrix}$$ $$[a]\cdot [B]=\begin{bmatrix}17& 67\\ 41& 147\\ 11& 37\end{bmatrix},\ [A]\cdot [C] = \begin{bmatrix}54& 61\\ 114& 129\\ 34& 31\end{bmat Rix}\, [A]\cdot [B]+[a]\cdot [c]= \begin{bmatrix}71& 128\\ 155& 276\\ 45& 68\end{bmatrix}$$ Thus $$[A]\cdot ([ B]+[C]) = [A]\cdot[b] + [a]\cdot[c]$$

4. Let $[i]$ is a $n \times n$ identity matrix. Show that $[a]\cdot[i] = [i]\cdot[a]=[a]$ for every $n \times n$ matrix $[a]$.
solution:
Let $[c]_{n\times n}=[a]_{n\times n}\cdot[i]_ {N\times n}$. So we have $ $c _{ij}=a_{i1}i_{1j} + \cdots + A_{i, j-1}i_{j-1, J} + A_{IJ}I_{JJ} + a_{i, j+1}i_{j+1, J}+\cdots + a_{in}i_{n J} = \sum_{p=1}^{n} a_{ip}i_{pj}$$ for each of the $i =1, \cdots, n$ and $j =1, \cdots, n$. Since $ $i _{ij}=\begin{cases}0 & i\neq j \ \ 1 & i=j\end{cases}$$ Thus $ $c _{ij} = \sum_{p=1}^{n} A_{IP}I_{PJ} = A_{i J}I_{JJ} = a_{ij}$$ is, $[a]\cdot[i] = [a]$. Similarly, denote $[d]_{n\times n}=[i]_{n\times N}\cdot [A]_{n\times n}$, and $ $d _{ij}=i_{i1}a_{1j} + \cdots + i_{i, i-1}a _{i-1, J} + I_{ii}a_{ij} + i_{i, i+1}a_{i+1, J}+\cdots + i_{in}a_{nj} = \sum_{p=1}^{n} i_{ip}a_{pj}$$ Because $i _{ij}=1$ W Hen $i =j$, otherwise $i _{ij}=0$. Thus, $ $d _{ij}= \sum_{p=1}^{n} I_{IP}A_{PJ} = a_{ij}$$ is, $[i]\cdot[a]=[a]$.

5. Consider there is only the computer companies in a country. The companies is named Dude and Imac. Each year, company 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 2002, Dude have ${1/6}^{th}$ of the market and Imac have ${5/6}^{th}$ of the market.
(A) What is the distribution of the customers between the and the companies in 2003? Write the answer first as multiplication of the matrices.
(B) What would is distribution when the market becomes stable?
Solution:
(A) denote $D _n$ and $M _n$ as the market share of Dude and Imac in the $n $-th year, respectively. $$\begin{bmatrix}d_n \ m_n \end{bmatrix} = \begin{bmatrix}{1\over5} & {2\over3} \ \ {4\over5} & {1\OVER3}\END{BMA Trix}\cdot \begin{bmatrix}d_{n-1} \ M_{n-1} \end{bmatrix}$$ Thus $$\begin{bmatrix}d_{2003} \ \ M_{2003} \end{bmatrix} = \ BEGIN{BMATRIX}{1\OVER5} & {2\over3} \ \ {4\over5} & {1\over3}\end{bmatrix}\cdot \begin{bmatrix}d_{2002} \ \ M_{ 2002} \end{bmatrix} = \begin{bmatrix}{1\over5} & {2\over3} \ \ {4\over5} & {1\over3}\end{bmatrix}\cdot \begin{ BMATRIX}{1\OVER6} \ \ {5\OVER6} \end{bmatrix}= \begin{bmatrix}{53\over90} \ \ {37\over90} \end{bmatrix}$$
(B) The stable system means the market share won't be changed from year to year, which is, $$\begin{cases}d = {1\over5}d +{2\over3}m \ m = {4\over5}d + {1\over3}m\end{cases}\rightarrow {4\over5}d-{2\over3}m=0$$ on the other hand, $D + M =1$, Thus we have $$\begin{cases}{4\over5}d-{2\over3}m=0\\ D + M =1\end{cases}\rightarrow\begin{cases}d={5\over11}\\ M={6\ over11}\end{cases}$$ Hence The stable market share of Dude and Imac are ${5\over11}$ and ${6\over11}$, respectively.

6. Given $$[a]=\begin{bmatrix}12.3& -12.3& 10.3\\ 11.3& -10.3& -11.3\\ 10.3& -11.3& -12.3\end{ bmatrix},\ [b]=\begin{bmatrix}2& 4\\ -5& 6\\ 11& -20\end{bmatrix}$$ if $[c] = [A]\cdot[B]$, then $c _{3 1}$?
Solution:
$ $c _{31} = \begin{bmatrix}10.3& -11.3& -12.3\end{bmatrix} \cdot \begin{bmatrix}2\\-5\\11\end{bmatrix} = 10.3\ TIMES2+11.3\TIMES5-12.3\TIMES11 = -58.2$$

7. $[a]$ and $[b]$ are square matrices of $n \times n$ order. Then $ ([a]-[b]) ([a]-[b]) $ are equal to ().
Solution:
$$ ([a]-[b]) ([a]-[b]) = [A] ([a]-[b])-[b] ([a]-[b]) =[a]^2-[a][b]-[b][a]+[b]^2$$ Note that $[a][b]$ and $[b][a]$ are not Equal to all other unless $[a][b]=[b][a]$.

8. Given $[a]$ is A rectangular matrix and $c [a]=0$, then-what's the values of $c $ and $[a]$?
Solution:
$c [A]=0\rightarrow c=0$ or $[a]=[0]$.

9. You sell Jupiter and fickers Candy bars. The sales in January is Jupiter and Fickers, respectively. In February, the sales is Jupiter and Fickers, respectively. If a Jupiter bar costs 2 dollars and a fickers bar costs 7 dollars, then what's the total sales amount in each month?
Solution:
$$\begin{bmatrix}25& 30\\ 75& 35\end{bmatrix}\cdot\begin{bmatrix} 2 \ 7 \end{bmatrix} =\begin{bmatrix} 260 \ 39 5 \end{bmatrix} $$ Thus, the total sales amount in January and February is 260 dollars and 395 dollars, respectively.

A.kaw Matrix Algebra Preliminary study Note 3. Binary Matrix Operations