1. Matrices
For example
$\begin{pmatrix}2 & 6 & 3 & -5\\15 & 8 & 1 & 14\\2 & 2 & 2 & -8\end{pmatrix},$ |
$\begin{pmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{pmatrix},$ |
$\begin{pmatrix}e & 3e\\0 & e\\ \frac{1}{2}e^5-\frac{1}{2}e & e^5 \end{pmatrix},$ |
$\begin{pmatrix}x^2\\y^2\\z^2\\w^2\end{pmatrix}.$ |
What is the size of the following matrices:
$\begin{pmatrix}1 & 3 & 7\end{pmatrix}$ | $\begin{pmatrix}a & b \\c & d \\e & f\end{pmatrix}$ | $\begin{pmatrix}\overline{1} & \overline{7} & \overline{8} \\ \overline{2} & \overline{9} & \overline{0}\end{pmatrix}$ | $\begin{pmatrix}7+i \\8-3i \\10+\sqrt{5}i\end{pmatrix}$ | $\begin{pmatrix}7\end{pmatrix}$ | $\begin{pmatrix}x^2 & 1 & 3 \\6 & x^2 & 5 \\9 & 7 & x^2\end{pmatrix}$ |
Answer | Answer | Answer | Answer | Answer | Answer |
Which of the following matrices are vectors? If a matrix below is not a vector, then name all vectors the matrix consists of.
$\begin{pmatrix}1\\2\\0\\1\end{pmatrix}$ | $\begin{pmatrix}x & y \\x^2 & y^2 \\v^3 & y^3\end{pmatrix}$ | $\begin{pmatrix}-\cos\alpha&\sin\alpha&\tan\alpha\end{pmatrix}$ | $\begin{pmatrix}f(2) & g(1) & 3 \\g(6) & f(5) & 2\end{pmatrix},$ $f(x)=x-4$ and $g(x)=\pi x$ | $\begin{pmatrix}7+i \\8-3i \\10+\sqrt{5}i\end{pmatrix}$ |
Answer | Answer | Answer | Answer | Answer |
A general notation of an $m \times n$ matrix $A$ is
$A=\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\a_{21} & a_{22} & \dots & a_{2n}\\a_{31} & a_{32} & \dots & a_{3n}\\ \vdots & \vdots & & \vdots\\a_{m1} & a_{m2} & \dots & a_{mn}\end{pmatrix}\hspace{2cm} \text{or}\hspace{2cm} A=(a_{ij}),$
where $i=1,2,\dots , m$ is the row index and $j=1,2,\dots , n$ is the column index. Usually, a capital letter is used for denoting a matrix. Row and column vectors are often denoted by a (noncapital) bold letter, e.g.,
$\mathbf{x}=\begin{pmatrix}x_1 & x_2 & \dots & x_n\end{pmatrix}\hspace{2cm} \text{or}\hspace{2cm} \mathbf{b}=\begin{pmatrix}b_1\\b_2\\\vdots\\b_m\end{pmatrix}.$
What is the $ij^{th}$ element of the matrix \begin{pmatrix}0 & 1 & 2 & 3 & 4 \\5 & 6 & 7 & 8 & 9\\10 & 11 & 12 & 13 & 14\end{pmatrix} when
Let $a_{14}=8$, $a_{27}=0$, $a_{13}=2$, $a_{21}=-4$, $a_{41}=-5$, $a_{12}=1$, $a_{31}=5$, $a_{23}=12$, $a_{25}=-1$, $a_{03}=10$, $a_{26}=7$, $a_{34}=4.$ Complete the matrix by entering the missing values (drag and trop the blue number below)
Matrices can be added, multiplied by a scalar, and multiplied by other matrices. Formulas for the operations are as follows
Operation | Formula | Example |
---|---|---|
Sum | $(a_{ij})+(b_{ij})=(a_{ij}+b_{ij})$ | $\begin{pmatrix}\color{blue}{1} & \color{blue}2\\\color{blue}0 &\color{blue}{-2}\end{pmatrix}+\begin{pmatrix}\color{red}{0} &\color{red}{-1}\\\color{red}1 & \color{red}2\end{pmatrix}=\begin{pmatrix}\color{blue}1+\color{red}0 & \color{blue}2+\color{red}{(-1)}\\\color{blue}0+\color{red}1 & \color{blue}{-2}+\color{red}2\end{pmatrix}=\begin{pmatrix}1 & 1\\1 & 0 \end{pmatrix}$ |
Scalar multiplication | $c\cdot (a_{ij})=(c\cdot a_{ij})$ | $\color{blue}{-1}\cdot\begin{pmatrix}1 & 8 \\5 &-2 \end{pmatrix}=$ $\begin{pmatrix}\color{blue}{-1}\cdot 1 & \color{blue}{-1}\cdot 8 \\\color{blue}{-1}\cdot 5 &\color{blue}{-1}\cdot (-2)\end{pmatrix}=\begin{pmatrix}-1 & -8 \\-5 & 2 \end{pmatrix}$ |
Matrix multiplication | Let $AB=(c_{ij})$, then $c_{ij}=\sum^m_{r=1}a_{ir}b_{rj}$, where $A=(a_{ij})$ is $n\times m$ matrix and $B=(b_{ij})$ is $m\times p$ matrix. | $\begin{pmatrix}\color{blue}2 & \color{blue}3 & \color{blue}4\\ \color{red}1 & \color{red}0 & \color{red}0\end{pmatrix}\begin{pmatrix}\color{LimeGreen}6& \color{Cyan}{10}\\\color{LimeGreen}7 & \color{Cyan}5\\\color{LimeGreen}8 & \color{Cyan}1\end{pmatrix}=$ $\begin{pmatrix}\color{blue}2\cdot \color{LimeGreen}6+\color{blue}3\cdot \color{LimeGreen}7+\color{blue}4\cdot \color{LimeGreen}8 & \color{blue}2\cdot \color{Cyan}{10}+\color{blue}3\cdot \color{Cyan}5+\color{blue}4\cdot \color{Cyan}1\\\color{red}1\cdot \color{LimeGreen}6+\color{red}0\cdot \color{LimeGreen}7+\color{red}0\cdot \color{LimeGreen}8 & \color{red}1\cdot \color{Cyan}{10}+\color{red}0\cdot \color{Cyan}5+\color{red}0\cdot \color{Cyan}1\end{pmatrix}=$ $\begin{pmatrix}65& 39\\6 & 10\end{pmatrix}$ |
Note that the dimension of matrices is important in the operations of adding and multiplying matrices:
- two matrices can be added only when they have same dimensions;
- two matrices can be multiplied if and only if the number of row in first matrix equals to the number of columns in the second matrix, i.e.
$\underset{n\times \color{blue}m}{A}\cdot \underset{\color{blue}m\times p}B=\underset{n\times p}{AB}.$
Click on corresponding elements of the matrices for calculating the elements of the sum (over $\mathbb Z_6$).
Which of the following operations can be performed and what is the result:
$\begin{pmatrix}1 & -1 & 2\\-2 & 1 & 0\end{pmatrix}\!+\!\begin{pmatrix}-2 & 1\\2 & -1\end{pmatrix}$ | $\begin{pmatrix}2 \\-1 \\4\end{pmatrix}\begin{pmatrix}2 & -2 & 3\end{pmatrix}$ | $3\!\cdot\!\begin{pmatrix}1&0&4\\3&1&-2\\2&1&-1\end{pmatrix}$ |
Answer | Answer | Answer |
$(-2)\!\cdot\!\begin{pmatrix}10 & -3\\-8 & 5\\20 & 3\end{pmatrix}\!+\!\begin{pmatrix}-8 & 4\\6 & 3\\-15 & 17\end{pmatrix}$ | $\begin{pmatrix}1 & 2\\-1 & 3\\5 & -1\end{pmatrix}\begin{pmatrix}-1 & 2\end{pmatrix}$ | $\begin{pmatrix}8&-6\\1&-1\\3&2\end{pmatrix}\begin{pmatrix}0&3\\10&6\end{pmatrix}$ |
Answer | Answer | Answer |
The diagram to the right describes also matrix multiplication.
Here are multiplied matrices $A$ with dimension $4\times 2$ and $B$ with dimension $2\times 3$. The matrix $AB$ is with dimension $4\times 3$.
Let $AB=(c_{ij})$. Then
$c_{12}=a_{11}b_{12}+a_{12}b_{22}$
and
$c_{33}=a_{31}b_{13}+a_{32}b_{23}$.
What are the values of indexes $i$ and $j$ if $c_{ij}=a_{41}b_{11}+a_{42}b_{21}$? Answer
Use IPython for calculating
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Another important operation on matrices is that of taking the transpose. It is denoted by placing a $T$ as an exponent on the initial matrix.
Operation | Formula | Example |
---|---|---|
Transpose | $(a_{ij})^T=(a_{ji})$ | $\begin{pmatrix}\color{blue}1 & \color{blue}2 & \color{blue}{-6}\\ \color{red}3 & \color{red}5 & \color{red}4\end{pmatrix}^T=\begin{pmatrix}\color{blue}1 & \color{red}3\\ \color{blue}2 & \color{red}5\\ \color{blue}{-6} & \color{red}4\end{pmatrix}$ |
Thus taking the transpose means reorganize matrix entries so that row of the initial matrix will be columns in the transpose. First row will be first column, second row will be second column and so on.
Which of the matrices
$(a)\quad\begin{pmatrix}2 & 6 & 3\\15 & 8 & 1\\2 & 2 & 2\\ 0 & 1 & 0\\-5 & 14 & -8\end{pmatrix}$ |
$(b)\quad\begin{pmatrix}3&6&2\\1&8&15\\2&2&2\\0&1&0\\-8&14&-5\end{pmatrix}$ |
$(c)\quad\begin{pmatrix}-8&14&-5\\0&1&1\\2&2&2\\1&8&15\\3&6&2\end{pmatrix}$ |
is the transpose of the matrix $\begin{pmatrix}3 & 1 & 2 & 0 & -8\\6 & 8 & 2 & 1 & 14\\2 & 15 & 2 & 0 & -5\end{pmatrix}$? Answer
Let
$A=\begin{pmatrix}1 & 5\\4 & 3\\5 & 4\end{pmatrix}$, |
$\mathbf{b}=\begin{pmatrix}1\\2\\3\end{pmatrix}$, |
$\mathbf{c}=\begin{pmatrix}1\\2\\3\end{pmatrix}^T$. |
Calculate $5\cdot AA^T+\mathbf{b}\mathbf{c}$ by using IPython .
Sum of matrices, scalar multiplication, matrix multiplication and the transpose are basic matrix operations. These operations have several useful properties. For example, knowing that $$c\cdot A+c\cdot B=c\cdot (A+B),$$ where $A$ and $B$ are matrices and $c$ is a constant, it is possible to reduce number of multiplications. Thus, instead of
$$-2\cdot\begin{pmatrix}10 & -3\\-8 & 5\\20 & 3\end{pmatrix}\!+\!(-2)\cdot\begin{pmatrix}-8 & 4\\6 & 3\\-15 & 17\end{pmatrix}=\begin{pmatrix}-2\color{red}\cdot10 & -2\color{red}\cdot(-3)\\-2\color{red}\cdot(-8) & -2\color{red}\cdot5\\-2\color{red}\cdot20 & -2\color{red}\cdot3\end{pmatrix}\!+\!\begin{pmatrix}-2\color{red}\cdot(-8) & -2\color{red}\cdot4\\-2\color{red}\cdot6 & -2\color{red}\cdot3\\-2\color{red}\cdot(-15) & -2\color{red}\cdot17\end{pmatrix}\!=\!\begin{pmatrix}-4 & -2\\4 & -16\\-10 & -40\end{pmatrix},$$ which is $12$ multiplications, we can manage only with $6$:
$$-2\cdot\!\begin{pmatrix}10 & -3\\-8 & 5\\20 & 3\end{pmatrix}\!+\!(-2)\cdot\!\begin{pmatrix}-8 & 4\\6 & 3\\-15 & 17\end{pmatrix}=-2\cdot\!\Bigg[\begin{pmatrix}10 & -3\\-8 & 5\\20 & 3\end{pmatrix}\!+\!\begin{pmatrix}-8 & 4\\6 & 3\\-15 & 17\end{pmatrix}\Bigg]=-2\cdot\!\begin{pmatrix}2 & 1\\-2 & 8\\5 & 20\end{pmatrix}\!=\!\begin{pmatrix}-4 & -2\\4 & -16\\-10 & -40\end{pmatrix}.$$
Let $A$, $B$ and $C$ be matrices such that all the operations below are defined, and let $k$ and $l$ denote some constants (for example real numbers) . Here are some additional useful properties concerning the matrix operations:
Property | Example |
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$[A^T]^T=A$ | $\Bigg[\begin{pmatrix}-1&2\\4&-7\\3&5\end{pmatrix}^T\Bigg]^T=\Bigg[\begin{pmatrix}-1&4&3\\2&-7&5\end{pmatrix}\Bigg]^T=\begin{pmatrix}-1&2\\4&-7\\3&5\end{pmatrix}$ |
$A+B=B+A$ | $\begin{pmatrix}1\\4\\3\end{pmatrix}\!+\!\begin{pmatrix}0\\5\\6\end{pmatrix}=\begin{pmatrix}1+0\\4+5\\3+6\end{pmatrix}=\begin{pmatrix}0+1\\5+4\\6+3\end{pmatrix}=\begin{pmatrix}0\\5\\6\end{pmatrix}\!+\!\begin{pmatrix}1\\4\\3\end{pmatrix}$ |
$A+(-1)\cdot A=\mathbf{0}$, where $\mathbf{0}$ denotes matrix consisting of only zeros | $\begin{pmatrix}1&2&4\\3&4&6\end{pmatrix}\!+\!(-1)\!\cdot\!\begin{pmatrix}1&2&4\\3&4&6\end{pmatrix}=\begin{pmatrix}1-1&2-2&4-4\\3-3&4-4&6-6\end{pmatrix}=\begin{pmatrix}0&0&0\\0&0&0\end{pmatrix}$ |
$k\cdot A+l\cdot A=(k+l)\cdot A$ | $-4\!\cdot\!\begin{pmatrix}1\\5\\4\end{pmatrix}\!+\!5\!\cdot\!\begin{pmatrix}1\\5\\4\end{pmatrix}=\begin{pmatrix}-4\\-20\\-16\end{pmatrix}\!+\!\begin{pmatrix}5\\25\\20\end{pmatrix}=\begin{pmatrix}1\\5\\4\end{pmatrix}\quad\text{and}\quad (-4+5)\!\cdot\!\begin{pmatrix}1\\5\\4\end{pmatrix}=\begin{pmatrix}1\\5\\4\end{pmatrix}$ |
In general $AB\neq BA$ | $\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}2&1\\4&3\end{pmatrix}\quad\text{but}\quad \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}3&4\\1&2\end{pmatrix}$ |
$[AB]C=A[BC]$ | $\Bigg[\begin{pmatrix}1 & 2 & 3\end{pmatrix}\begin{pmatrix}1 & 3\\-1 & -2\\2 & 5\end{pmatrix}\Bigg]\begin{pmatrix}1 & 1\\1 & 2\end{pmatrix}=[\begin{pmatrix}5&14\end{pmatrix}]\begin{pmatrix}1 & 1\\1 & 2\end{pmatrix}=\begin{pmatrix}19&33\end{pmatrix}$ |
$\begin{pmatrix}1 & 2 & 3\end{pmatrix}\Bigg[\begin{pmatrix}1 & 3\\-1 & -2\\2 & 5\end{pmatrix}\begin{pmatrix}1 & 1\\1 & 2\end{pmatrix}\Bigg]=\begin{pmatrix}1 & 2 & 3\end{pmatrix}\Bigg[\begin{pmatrix}4&7\\-3&-5\\7&12\end{pmatrix}\Bigg]=\begin{pmatrix}19&33\end{pmatrix}$ | |
$[A+B]C=AC+BC$ | $\Bigg[\begin{pmatrix}1&2\\3&1\end{pmatrix}\!+\!\begin{pmatrix}-1&2\\2&-3\end{pmatrix}\Bigg]\begin{pmatrix}1\\3\end{pmatrix}=\Bigg[\begin{pmatrix}0&4\\5&-2\end{pmatrix}\Bigg]\begin{pmatrix}1\\3\end{pmatrix}=\begin{pmatrix}12\\-1\end{pmatrix}$ |
$\begin{pmatrix}1&2\\3&1\end{pmatrix}\begin{pmatrix}1\\3\end{pmatrix}\!+\!\begin{pmatrix}-1&2\\2&-3\end{pmatrix}\begin{pmatrix}1\\3\end{pmatrix}=\begin{pmatrix}7\\6\end{pmatrix}\!+\!\begin{pmatrix}5\\-7\end{pmatrix}=\begin{pmatrix}12\\-1\end{pmatrix}$ | |
$(AB)^T=B^TA^T$ | $\Bigg[\begin{pmatrix}1&-2\\1&-1\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}\Bigg]^T=\Bigg[\begin{pmatrix}0\\1\end{pmatrix}\Bigg]^T=\begin{pmatrix}0&1\end{pmatrix}$ |
$\begin{pmatrix}2\\1\end{pmatrix}^T\begin{pmatrix}1&-2\\1&-1\end{pmatrix}^T=\begin{pmatrix}2&1\end{pmatrix}\begin{pmatrix}1&1\\-2&-1\end{pmatrix}=\begin{pmatrix}0&1\end{pmatrix}$ |
Note that matrix multiplication is not commutative, i.e. even if both the matrices $AB$ and $BA$ do exist, in general
$$AB\neq BA.$$