Lesson 8: Concept of Matrices
Lesson Objective
Dear learner,
at the end of this lesson, you will be able to:
- Define basic concepts about matrices
- Differentiate types of a matrix
- Establish matrix concept to different real life scenarios
- Perform operation on matrices
- Determine the sum and difference of two matrices of the same order
- Multiply a matrix by a scalar
- Compute the product of two matrices
Key Terms
- Size of a matrix
- Equality of matrices
- Column matrix
- Row matrix
- Square matrix
- Diagonal matrix
- Identity matrix
Brainstorming Activity
Three students Selam, Hagos and Chaltu have a number of 10, 50 and 25 cents coin in their pocket. The following table shows what they have:
Find
a. The number of cents each student have?
b. Determine number of cents in each row?
Solution:
you can describe each questions follows:
a) Selam can have =$\begin{pmatrix}
2 \\
3 \\
4
\end{pmatrix}$ cents which were put according to their order and Hagos can have =$\begin{pmatrix}
6 \\
2 \\
0
\end{pmatrix}$ coins while Chaltu has =$\begin{pmatrix}
4 \\
0 \\
5
\end{pmatrix}$ coins
b. in each row the coins are to be ordered like:
first row: $\begin{pmatrix}
2 \ 6 \ 4 \end{pmatrix}$ 10 cent coins only attained by the three students, but in:
the second row =$\begin{pmatrix}
3 \ 2 \ 0 \end{pmatrix}$ 25 cent coins while,
in the third row we can get = $\begin{pmatrix}
4 \ 0 \ 5 \end{pmatrix}$ 50 cent coins
1.1. Matrix
Definition: Let m and n be positive integers. A rectangular array of numbers in $\mathbb{R}$of the form:
$A= \begin{pmatrix}
a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\dots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \ \end{pmatrix}$ is an $m by n$ matrix, where $m by n$ or $m \times n$ is the size of the matrix. such that m: shows the horizontal rows and n: shows the vertical columns
Note:
i. the $i^{th}$ row of A = $\begin{pmatrix}
a_{i1} & a_{i2} & a_{i3} & … &a_{in}
\end{pmatrix}$
ii. the $J^{th}$ column of A = $\begin{pmatrix}
a_{1j} \\
a_{2j}\\
a_{3j} \\
.\\
.\\
.\\
a_{mj}
\end{pmatrix}$ is a column matrix.
iii. $a_{ij}$ is the (i, j)- entry of or element of A.
iv. the Matrix A is written as $A = (a_{ij})_mn$.
v. $m \times n$ is the size or order of matrix A.
Example 1 : Let Matrix $A = \begin{pmatrix}
1& 5 & 6 & 3 \\
2& 8 & -1 & 0 \\
4 & -9 & 12 & 10
\end{pmatrix}$ then find:
a. number of rows and columns
b. $a_{23}$
c. $a_{33}$
d. size or order of the matrix
e. entry (3,4)
Solution:
a). Matrix A has 3 rows and 4- columns.
b). In $a_{23}$ : 2 stands for second row and 3 – stands for third column and the entry is -1.
c). In $a_{33}$ the intersection for third row and third column is 12.
d). size of A = 3 X 4
e). entry(3,4) = $a_{34}$ = 10
Types of Matrices:
There are different types of matrices:
- Column matrix(column vector): A matrix having only one column:
$\begin{pmatrix}
a_{1} \\
a_{2}\\
a_{3} \\
.\\
.\\
.\\
a_{m}
\end{pmatrix}$ is a column matrix.
2. Row matrix (row vector): A matrix having only one row
$A = \begin{pmatrix}
a_{1} & a_{2} & a_{2} & … &a_{n}
\end{pmatrix}$
3. Zero matrix: A matrix with all zero entries
$A = \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
4. Square matrix: A matrix in which the number of rows is equal to number of columns.
$A = \begin{pmatrix}
1 & 2 & 3\\
4 & 7 & 2 \\
8 & 1 & 0
\end{pmatrix}$ is a 3 by 3 matrix.
5. Diagonal matrix: A square matrix with all zero entries except diagonal entries.
$A = \begin{pmatrix}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 2
\end{pmatrix}$ is a 3 by 3 Diagonal matrix.
6. Scalar matrix : A diagonal matrix all diagonal entries are equal.
$ A = \begin{pmatrix}
2 & 0 & 0 \\
0 &2 & 0 \\
0 & 0 & 2
\end{pmatrix}$ is a 3 by 3 scalar matrix.
7. Identity matrix: A scalar or diagonal matrix with all the diagonal entries are equal to 1.
$A = \begin{pmatrix}
1 & 0 & 0 \\
0 &1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ is a 3 by 3 identity matrix
8. Triangular matrix:
a). Upper Triangular Matrix: A square matrix whose entries below the main diagonal are all zero.
$A = \begin{pmatrix}
1 & 2 & 2 \\
0 &1 & 3\\
0 & 0 & 1
\end{pmatrix}$ is a 3 by 3 upper triangular matrix.
b) Lower triangular matrix: A square matrix whose entriesabove the main diagonal are all zero.
$A = \begin{pmatrix}
1 & 0 & 0 \\
2 &1 & 0\\
3 & 8 & 1
\end{pmatrix}$ is a 3 by 3 lower triangular matrix
Equality of Matrices
Two matrices $ A = (a_{i,j})_{mn}$ and $B = (b_{i,j})_{pq}$ are said to be equal if and only if:
- they have the same size i.e. $m = p$ and $n = q$
- the corresponding entries are equal i.e. $a_{i,j} = b_{i,j}$
Example 2:
Find the values of x, y,a and b if:
a). $ \begin{pmatrix}
3x + 4y & 6 & x – 2y \\
a + b & 2a – b & -3\\
\end{pmatrix}$ = $\begin{pmatrix}
2 & 6 & 4 \\
5 &-5 & 3\\
\end{pmatrix}$
b). $ \begin{pmatrix}
x^{2} – 1 & 1 & 2 \\
-1 & a+b^{2} & -1\\
\end{pmatrix}$ = $\begin{pmatrix}
2x – 2 & 1 & 2 \\
-1 & -3a + b^{2} + 12 & -1\\
\end{pmatrix}$
Solution:
a) $ \begin{pmatrix}
3x + 4y & 6 & x – 2y \\
a + b & 2a – b & -3\\
\end{pmatrix}$ = $\begin{pmatrix}
2 & 6 & 4 \\
5 &-5 & 3\\
\end{pmatrix}$ if and only if 3x + 4y = 2 and x – 2y = 4 implies x = 2y + 4 and substituting x in to the first equation 3(2y + 4) +4y = 2
implies 10y = 2-12 then y = -1 where as x = 2(-1)+4 = 2.
therefore x = 2 and y = -1
by the same principle a + b = 5 and 2a -b = -5 and re-writing the first equation a = 5 – b. substituting it in to the second equation 2(5 -b) – b = -5
implies b = 5 while a = 5 – 5 = 0.
therefore a = 0 and b = 5
b). $ \begin{pmatrix}
x^{2} – 1 & 1 & 2 \\
-1 & a+b^{2} & -1\\
\end{pmatrix}$ = $\begin{pmatrix}
2x – 2 & 1 & 2 \\
-1 & -3a + b^{2} + 12 & -1\\
\end{pmatrix}$ implies $x^{2} – 1 = 2x -2$ $implies$ $x^{2} – 2x +2 = 0$ then $(x – 1)^{2} = 0$ therefore x = 1
similarly $a + b^{2} = -3a + b^{2} + 12$ $implies$ a = 3 since $b^{2}$ is cancelled.
the final result gives a = 3 and $b \in \mathbb{R}$.
1.2. Operations on Matrices
i). Addition and Subtraction of Matrices
Let $A = (a_ij)_mn$ and $B = (b_ij)_mn$ be two matrices of the same order, then their sum or difference denoted by $A \pm B = (a_ij)mn \pm (b_ij)mn = (a_ij \pm b_ij)_mn$. where $m, n \in Z^{+}$ and $a,b \in \mathbb{R}$
Example 3
- Let $A = \begin{pmatrix}
3 & 5 & 2.1 \\
4 & 2 & 6\\
0 & -1 & 5
\end{pmatrix}$ and $B = \begin{pmatrix}
7 & 6 & 5 \\
10 &2 & 2.15\\
9 & 8 & -6.5
\end{pmatrix}$ be matrices, then find $A + B$? - Let $A = \begin{pmatrix}
3 & 0 & 8\\
4 & 4 & 6\\
0 & -1 & 5
\end{pmatrix}$ and $B = \begin{pmatrix}
7 & 6 & 5 \\
10 &-8 & 9\\
9 & 12 & -11
\end{pmatrix}$ be matrices, then find A – B?
Solution:
- A + B = $\begin{pmatrix}
3 & 5 & 2.1 \\
4 & 2 & 6\\
0 & -1 & 5\end{pmatrix}$ + $\begin{pmatrix}
7 & 6 & 5 \\
10 &2 & 2.15\\
9 & 8 & -6.5\end{pmatrix}$ = $\begin{pmatrix}
3+7 & 5+6 & 2.1+5 \\
4+10 &2+2 & 6+2.15\\
0+9 & -1+8 & 5-6.5 \end{pmatrix}$ = $\begin{pmatrix}
10 & 11 & 7.1 \\
14 & 4 & 8.15\\
9 & 7 & -1.5
\end{pmatrix}$ - A – B = $\begin{pmatrix}
3 & 0 & 8\\
4 & 4 & 6\\
0 & -1 & 5 \end{pmatrix}$ – $\begin{pmatrix}
7 & 6 & 5 \\
10 &-8 & 9\\
9 & 12 & -11 \end{pmatrix}$ = $\begin{pmatrix}
3 -7 & 0-6 & 8-5 \\
4-10 &4-(-8) & 6-9\\
0-9 & -1-12 & 5-(-11) \end{pmatrix}$ = $\begin{pmatrix}
-4 & -6 & 3 \\
-6 & 12 & -3\\
-9 & -13 & 16
\end{pmatrix}$
Properties of Addition of Matrices
Let A, B and C be matrices of real numbers of the same order, then the following properties are satisfied
A + B = B + A ………………………. commutative property
A + (B +C) = (A + B) + C ………. Associative propert
A + O = A = A + O ………………… Existence of additive identity
A + ( -A) = O ………….Existence of additive inverse
Example 4
- Let $A = \begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$, $B = \begin{pmatrix}-3& 6\\ 1& -2 \end{pmatrix}$. $C = \begin{pmatrix} -3&7\\5&2\end{pmatrix}$ and $D = \begin{pmatrix} 0& 0\\0&0\end{pmatrix}$. Find
a) A +B and B + A
b). (A + B) + c and A + (B +C)
c). A + D
d). A + (-A)
Solution:
a. A + B = $\begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$ + $\begin{pmatrix}-3& 6\\ 1& -2 \end{pmatrix}$ = $\begin{pmatrix} -1 & 7\\ 5& 1 \end{pmatrix}$ = B + A
b). A +(B + C) = $\begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$ + [$\begin{pmatrix}-3& 6\\ 1& -2 \end{pmatrix}$ + $\begin{pmatrix} -3 & 7\\5 & 2\end{pmatrix}$] = $\begin{pmatrix} -4 & 14\\10 & 3\end{pmatrix}$
c). A + D = $\begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$ + $\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$ = $\begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$
d). A + (-A) = $\begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$ – $\begin{pmatrix} 2 & 1\\ 4& 3 \end{pmatrix}$= $\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$
ii).Multiplication of matrices
a. Scalar Multiplication
Let $A = (a_{i,j})_{mn}$ be a matrix in the set of real numbers and $r \in \mathbb{R}$, then $rA = r (a_{i,j})_{mn} = (ra_{i,j})_{mn}$
Example 5:
- Let $A = \begin{pmatrix}
3 & 0 & 8 \\
4 & 4 & 6\\
0 & -1 & 5
\end{pmatrix}$ be 3 by 3 matrix. Then find
a). 6A
b). $\frac{-1}{2} A$
Solution:
a). $6A = 6 \begin{pmatrix}
3 & 0 & 8 \\
4 & 4 & 6\\
0 & -1 & 5\end{pmatrix}$ = $\begin{pmatrix}
6 \times 3 & 6 \times 0 & 6 \times 8 \\
6 \times 4 & 6 \times 4 & 6 \times 6\\
6\times 0 & 6 \times (-1) & 6 \times 5\end{pmatrix}$= $\begin{pmatrix}
18 & 0 & 48 \\
24 & 24 & 36\\
0 & -6 & 30\end{pmatrix}$
b). $\frac{-1}{2} A$ = $\frac{-1}{2} \times \begin{pmatrix}
3 & 0 & 8 \\
4 & 4 & 6\\
0 & -1 & 5
\end{pmatrix}$ = $\begin{pmatrix}
3/2 & 0 & 4 \\
2 & 2 & 3\\
0 & -1/2 & 5/2\end{pmatrix}$
Properties of Scalar Multiplication of Matrices
Let A and B be matrices of real numbers of the same order, then the following properties hold:
- r(A+B) = rA + rB ……. scalar multiplication is distributive over addition of matrices
- (r+s)A = rA + sA …….. scalar multiplications are distributive over matrix addition
- (rs)A = r(sA) = s(rA) and IA = AI = A
https://www.geogebra.org/m/hzkcxsex
b. Multiplication of Matrices
Let $ A = (a_{i,j})_{mn}$ and $B = (b_{i,j})_{pq}$ be matrices, then the product AB exists If and only if: The number of column of A is equal to number of rows of B. i.e. n = p. order of AB is $m \times q$
Example 6:
Identify which pair of matrices are comfortible for multiplication
a). $A = \begin{pmatrix}
4 & 0 & -8 \\
4 & 8 & 12 \end{pmatrix}$ and $B = \begin{pmatrix}
42 & 0 & 2 \\
0 & 8 & 5 \end{pmatrix}$
b). $A = \begin{pmatrix}
2 & 0 & -8 \\
3 & 5 & 1\end{pmatrix}$ and $B = \begin{pmatrix}
2 & 3 \\
0 & 5 \\
8 & 1 \end{pmatrix}$
Solution:
a). A and B are not multiplicative cmfortible therefore AB doesn’t exist since the number f columns of A is different from number of rows of B.
b). A and B are multiplicative comfortible since the number of columns of A is equal to the number of rows of B.
Multiplication Rules:
Let $A=\begin{pmatrix} a & b \\c & d \end{pmatrix}$ and $B= \begin{pmatrix}w& x & r \\h & y & s \end{pmatrix}$ be matrices where a,b,c,d, w, x, y, r, s$ \in \mathbb{R}$, then AB is defined using the technique that uses rows of A and columns of B.
define for matrix A $R_{1} = (ab)$, $R_{2} = (cd)$ and
for matrix B $C^{1} = \begin{pmatrix}w \\ h\end{pmatrix}$,
$C^{2} = \begin{pmatrix}x \\ y \end{pmatrix}$,
$C^{3} = \begin{pmatrix} r\\s \end{pmatrix}$.
$AB = \begin{pmatrix}R_{1}C^{1} & R_{1}C^{2} &R_{1}C^{3}\\ R_{2}C^{1}& R_{2}C^{2} &R_{2}C^{3}\end{pmatrix}$
= $\begin{pmatrix} (ab)\begin{pmatrix}w \\ h\end{pmatrix} & (ab) \begin{pmatrix}x \\ y \end{pmatrix}& (ab)\begin{pmatrix} r\\s \end{pmatrix} \\ (cd) \begin{pmatrix}w \\ h\end{pmatrix} & (cd) \begin{pmatrix}x \\ y \end{pmatrix} & (cd)\begin{pmatrix} r\\s \end{pmatrix} \end{pmatrix}$
= $\begin{pmatrix} aw +bh &ax +by & ar + bs\\ cw + dh & cx +dy & cr +ds\end{pmatrix}$
Example 7 :
Let $A = \begin{pmatrix}2 & 0 & 8\\ 3 & 5& 1\end{pmatrix}_{2\times 3}$ and $B = \begin{pmatrix} 2& 3\\ 0& 5\\ 8& 1\end{pmatrix}_{3\times 2}$ be matrices. Find
a). AB
b). BA
Solution:
a). step 1: List rows of A and columns of B
Rows of A$R_{1}= \begin{pmatrix} 2&0 &8 \end {pmatrix}$ and $R_{2} =\begin{pmatrix} 3 & 5 &1 \end{pmatrix}$
while columns of B are $C^{1} = \begin{pmatrix}2 \\0\\ 8 \end{pmatrix}$ and $C^{2} = \begin{pmatrix} 3 \\ 5\\ 1\end{pmatrix}$
step 2: detrmine AB
$AB = \begin{pmatrix}R_{1}C^{1} & R_{1}C^{2} \\ R_{2}C^{1}& R_{2}C^{2} \end{pmatrix}$ = $\begin{pmatrix} 2\times 2 +0 \times 0 + 8\times 8 & 2\times 3 + 0\times 5 + 8 \times 5\\ 3 \times 2 +5 \times 0 + 1 \times 8 & 3 \times 3 + 5 \times 5 + 1 \times 1 \end{pmatrix} _{2\times 2}$ = $\begin{pmatrix}4 + 64 & 6 +40\\ 6 + 8 & 9 + 25 + 1\end{pmatrix}$ = $\begin{pmatrix} 70 & 46\\ 14 &35\end{pmatrix}_{2\times 2}$
b). List rows of B $R_{1} = (23)$, $R_{2} = (0 5)$, $R_{3} = (8 1)$
columns of A $C^{1} =\begin{pmatrix}2\\3\end{pmatrix}$, $C^{2} =\begin{pmatrix}0\\5\end{pmatrix}$, $C^{3} =\begin{pmatrix}8\\1\end{pmatrix}$
then BA = $\begin{pmatrix}(23)\begin{pmatrix}2\\3\end{pmatrix} & (23)\begin{pmatrix}0\\5\end{pmatrix} & (2 3) \begin{pmatrix}8\\1\end{pmatrix}\\(05)\begin{pmatrix}2\\3\end{pmatrix} & (05)\begin{pmatrix}0\\5\end{pmatrix} &(0 5)\begin{pmatrix}8\\1\end{pmatrix}\\ (81)\begin{pmatrix}2\\3\end{pmatrix} & (8 1)\begin{pmatrix}0\\5\end{pmatrix} & (8 1) \begin{pmatrix}8\\1\end{pmatrix} \end{pmatrix}_{3\times 3}$ = $\begin{pmatrix} 2\times2 +3\times 3 & 2\times 0+3\times 5&2 \times 8+3\times 1\\0\times2 +5\times 3 &0\times 0+5\times 5& 0\times 0 +5\times 1\\ 8 \times2 +1 \times3 & 8 \times 0+1 \times5 & 8\times8 +1\times1 \end{pmatrix}_{3\times3}$ = $\begin{pmatrix}13&15&19\\17&25&5\\19&5&65\end{pmatrix}|-{3\times 3}$
Properties of Matrix multiplication:
$\text{Let}~A,~B~\text{and}~C~\text{be matrices in real numbers and}~r\in\mathbb{R},~\text{then}$
A(B+C) = AB+AC ……………left distributive property
(A+B)C = AC+BC………..…right distributive property
A(BC) = (AB)C……………….associative property
r(AB) = (rA)B=A(rB)………..scalar multiplication
1.3. Transpose of Matrices and Its Property
Definition:
For any matrix A, the matrix obtained from A by interchanging its rows and columns is the transpose of A denoted by$A^{T}$.
i . e. if $A = (a_{i,j})_{mn}$, then $A^{T} = (a_{j,i})_{nm}$.
Example 8:
1 find the transpse of the follwing matrices
a). $A = \begin{pmatrix}2&3&5\\8&9&2\\074&1\end{pmatrix}$
b). $B =(6 4 8 1)$
c). D = (9)
2. if $A = \begin{pmatrix}-1&2&0\\-1&-2&1\\3&1&4\end{pmatrix}$, then find matrixB such that$B – 2A^{t} = A$
Solution:
1a). $A^{t} = \begin{pmatrix}2&8&0\\3&9&4\\5&2&1\end{pmatrix}$
b). $B^{t} = \begin{pmatrix}6\\5\\8\\1 \end{pmatrix}$
c). $D^{t} = (9)$
2. $B = A +2A^{t}$ = $\begin{pmatrix}-1&2&0\\-1&-2&1\\3&1&4\end{pmatrix} + \begin{pmatrix}-2&-2&6\\4&-4&2\\0&2&8\end{pmatrix}$ = $\begin{pmatrix}-3&0&6\\3&-6&3\\3&3&12\end{pmatrix}$
Properties of Transpose of A Matrix
Let A and B be matrices, A+B, AB are defined and $K \in \mathbb{R}$ then the following properties are satisfied:
i $ (A^{T})^{T} = A$ iv. $(KA)^{T} = K A^{T}$
ii .$(A + B)^{T} = A^{T} + B^{T}$ v. $I_{n}^{T} = I_{n}$ , where I is an identity matrix
iii . $(AB)^{T} = B^{T} A^{T}$
Example 9:
i. If $
A = \begin{pmatrix}
1 & 2 & 3 \\
-1 & 0 & 1\\
2 & -1 &-3 \\
\end{pmatrix}
$, then verify that $(A^t)^t = A.$
ii. If $A = \begin{pmatrix}
4 & 0 & -8 \\
4 & 8 & 12 \end{pmatrix}$, $
B = \begin{pmatrix}
-1 & 2 \\
3 & 0 \\
2 & 5 \\
\end{pmatrix}$ and $C = \begin{pmatrix}
-1 & 3 & 6 \\
-3 & 2 & -2 \end{pmatrix}$, then verify that
a. $(AB)^t=B^tA^t$ $\quad$ b. $(A+C)^t= A^t + C^t$
Solution:
1. $A = \begin{pmatrix}
1 & 2 & 3\\
-1 & 0 & 1\\
2 & -1 & -3\\
\end{pmatrix}$ $\Rightarrow A^t = \begin{pmatrix}1 & -1 & 2\\
2 & 0 & -1\\
3 & 1 & -3\\
\end{pmatrix}$ and $(A^t)^t = \begin{pmatrix}
1 & 2 & 3\\
-1 & 0 & 1\\
2 & -1 & -3\\
\end{pmatrix}=A$.
2. $\text{a.}\ AB = \begin{pmatrix} 1 & 2 & 3\\
2 & 1 & 3\\
\end{pmatrix}$ $\begin{pmatrix}
-1 & 2 \\
3 & 0 \\
2 & 5 & \\
\end{pmatrix}$=$\begin{pmatrix}
11 & 17 \\
7 & 19 \\
\end{pmatrix}$ $\implies (AB)^t = \begin{pmatrix}11 & 7 \\
17 & 19 \\ \end{pmatrix}$
$B^t A^t = \begin{pmatrix}-1 & 2 \\
3 & 0 \\
2 & 5 & \\
\end{pmatrix}^t$ $ \begin{pmatrix}1 & 2 & 3\\
2 & 1 & 3\\
\end{pmatrix}^t=\begin{pmatrix}-1 & 3 & 2\\
2 & 0 & 5\\
\end{pmatrix}\begin{pmatrix}1 & 2 \\
2 & 1 \\
3 & 3 \\
\end{pmatrix}=\begin{pmatrix}11 & 7 \\
17 & 19 \\
\end{pmatrix}.$
Therefore, $(AB)^t=B^tA^t.$
b. $A+C)^t=A^t +C^t.$
$\therefore (A + C)^t = (\begin{pmatrix} 1 & 2 & 3\\
2 & 1 & 3\\
\end{pmatrix}+\begin{pmatrix}-1 & 3 & 6\\
-3 & 2 & -2\\
\end{pmatrix})^t$=$\begin{pmatrix} 0 & 5 & 9\\
-1 & 3 & 1\\
\end{pmatrix}^t=\begin{pmatrix}0 & -1 \\
5 & 3 \\
9 & 1 \\
\end{pmatrix}$.
$A^t+C^t=\begin{pmatrix}1 & 2 & 3\\
2 & 1 & 3\\
\end{pmatrix}^t+\begin{pmatrix}-1 & 3 & 6\\
-3 & 2 & -2\\
\end{pmatrix}^t=\begin{pmatrix}1 & 2 \\
2 & 1 \\
3 & 3 \\
\end{pmatrix}+\begin{pmatrix}-1 & -3 \\
3 & 2 \\
6 & -2 \\
\end{pmatrix}=\begin{pmatrix}0 & -1 \\
5 & 3 \\
9 & 1 \\
\end{pmatrix}.$