Dear Student, you can find here the Matrix Theory and Exercises (basics) as per lecture planning.
A matrix that has $m$ rows and $n$ columns is called an $m \times n$ matrix (read: m by n). In the tag, "$m \times n$" the number of rows is always specified first, and the number of columns second. The tag $m \times n$ is called the order of the matrix.
The typical element of matrix $A={{\left( \begin{matrix} {{a}_{11}} & \ldots & {{a}_{1n}} \\ \vdots & \ddots & \vdots \\ {{a}_{m1}} & \cdots & {{a}_{mn}} \\ \end{matrix} \right)}_{m\times n}}$ are denoted by \[{{a}_{ij}},\text{ }i=1,2,\ldots ,m\text{ and }j=1,2,\ldots ,n\text{ }\] The $i^{th}$ row of matrix $A$ is ${{A}_{i}}={{\left( {{a}_{i1}},{{a}_{i2}},\ldots ,{{a}_{in}} \right)}_{1\times n}}$, the $j^{th}$ column of matrix $A$ is ${{B}_{j}}={{\left( \begin{matrix} {{a}_{1j}} \\ {{a}_{2j}} \\ {{a}_{3j}} \\ \vdots \\ {{a}_{mj}} \\ \end{matrix} \right)}_{m\times 1}}$
The DIAGONAL Matrix:
These are square matrices in which the elements that are not on principal diagonal are all zero. That is $A = (a_{ij})$, $a_{ij} = 0$, $i \ne j$, $i=1,2\ldots n$, and $j=1,2,\ldots n$. We can write $A = \text{diag} \begin{pmatrix} a_{11}&a_{22}&\ldots a_{nn}\end{pmatrix}$
The UNIT Matrix:
This is a square matrix with 1's on principal diagonal and 0's elewhere. It is designated by I. We have
$I = \begin{pmatrix}1&0&\ldots 0\\0&1&\ldots 0\\ \vdots&\ddots&\vdots \\0&0&\ldots 1\end{pmatrix}$
That is $I = (a_{ij})$ then $a_{ij}=1$, if $i=j$, $a_{ij}=0$, if $i \ne j$, $i=1,2\ldots n$, and $j=1,2,\ldots n$
The TRIANGULAR Matrix:
An upper triangular matrix is a square matrix all of whose nonzero elements are located on or above the principal diagonal. A lower triangular matrix is a square matrix all of whose nonzero elements are located on or below the principal diagonal. In other words, an upper triangular matrix has all zeros below its main diagonal, where as a lower triangular matrix has all zeros above its main diagonal.
That is $A = (a_{ij})$,
$a_{ij}=0$, if $i>j$,$i=1,2,\ldots, n$, $j=1,2,\ldots,n$, then matrix $A$ is upper triangular
Definition: Two $m \times n$ matrices $A=(a_{ij})$ and $B=(b_{ij})$, $i=1,2,\ldots, m$, $j=1,2,\ldots, n$ are added by adding corresponding elements. The expressions $A+B$ will designate the $m \times n$ matrix whose elements are $a_{ij}+b_{ij}$. Once again
$(a_{ij}+b_{ij})=(a_{ij})+b_{ij})$
Theorem:
If $A=(a_{ij})$ is an $m \times n$ matrix, then its negative, designated by $-A$, is the $m \times n$ matrix whose elements are $-a_{ij}$. By the difference $A - B$, we shall mean the matrix $A+(-B)$, that is to say, $A=(a_{ij})$ and $B=(b_{ij})$, $i=1,2,\ldots, m$, $j=1,2,\ldots, n$ then
$A-B=(a_{ij}-b_{ij})$, $i=1,2,\ldots, m$, $j=1,2,\ldots, n$
Theorem: For any matrix $A$
Definition:
If $A=(a_{ij})$ is an $m \times n$ matrix $i=1,2,\ldots, m$, $j=1,2,\ldots, n$, then $cA$ will designate a scalar product of $A$ by $c$. It is $m \times n$ matrix $i=1,2,\ldots, m$, $j=1,2,\ldots, n$, whose elements are $c \times a_{ij}$. In other words, $cA = (c \times a_{ij})$
Two matrices are conformable for multiplication when the number of columns of the first factor equals the number of rows of the second factor. The orders of two matrices $A$ and $B$ that are conformable for multiplication can be written as $ m \times n$ and $n \times p$ respectively.
A matrix that has $m$ rows and $n$ columns is called an $m \times n$ matrix (read: m by n). In the tag, "$m \times n$" the number of rows is always specified first, and the number of columns second. The tag $m \times n$ is called the order of the matrix.
The typical element of matrix $A={{\left( \begin{matrix} {{a}_{11}} & \ldots & {{a}_{1n}} \\ \vdots & \ddots & \vdots \\ {{a}_{m1}} & \cdots & {{a}_{mn}} \\ \end{matrix} \right)}_{m\times n}}$ are denoted by \[{{a}_{ij}},\text{ }i=1,2,\ldots ,m\text{ and }j=1,2,\ldots ,n\text{ }\] The $i^{th}$ row of matrix $A$ is ${{A}_{i}}={{\left( {{a}_{i1}},{{a}_{i2}},\ldots ,{{a}_{in}} \right)}_{1\times n}}$, the $j^{th}$ column of matrix $A$ is ${{B}_{j}}={{\left( \begin{matrix} {{a}_{1j}} \\ {{a}_{2j}} \\ {{a}_{3j}} \\ \vdots \\ {{a}_{mj}} \\ \end{matrix} \right)}_{m\times 1}}$
Exercise:
Construct an $m \times n$ matrix whose
elements ${{a}_{ij}}$are given by
- $a_{ij}=1$,$i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 2, n = 2$
- $a_{ij}=0$, if $i=j$, $a_{ij}=1$ if $i \ne j$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 3, n = 3$
- $a_{ij}=i+j$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 3, n = 3$
- $a_{ij}=i \times j$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 3, n = 3$
- $a_{ij}= $ minimum of $i$ and $j$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 3, n = 3$
- $a_{ij}=(i-j)^{2}$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 2, n = 2$
- $a_{ij}=i^{2} \times j+ j^{2} \times i$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $ for $m = 2, n = 2$
Equality of
two matrices:
Definition:
If $A$ is an $m \times n$ matrix with elements \[{{a}_{ij}},\text{ }i=1,2,\ldots
,m\text{ and }j=1,2,\ldots ,n\text{ }\] and $B$ is a $p \times q$ matrix with elements \[{{b}_{ij}},\text{
}i=1,2,\ldots ,p\text{ and }j=1,2,\ldots ,q\text{ }\]then we shall write $A = B$ if and only if
$m = p, n = q$ and ${{a}_{ij}}={{b}_{ij}},$\[i=1,2,\ldots
,m\text{ and }j=1,2,\ldots ,n\text{ }\]
The equality sign" = " when written
between two matrices is essentially a new symbol, and one should note that it
behaves in the same way as does the ordinary equality relating ordinary
numbers. That is to say, matrix equality has these three properties:
1. It is reflexive. This means that for any matrix $A, A = A$.
2. It is symmetric. This means that if $A = B$, then $B = A$.
3. It is transitive. This means that if $A, B$ and $C$ are matrices, then $A = B$, and $B = C$ together imply that $A = C$.
Exercise
Which of the following statements are true? (1-3)
- $(0,0)=(0,0,0)$
- $(0-0,0)=(0-0,0-0)$
- $\begin{pmatrix}1 - 1 & 2 - 2\\ x -x & 1 - 1\end{pmatrix}=(0)$
- If $ \begin{pmatrix} x & 2x+y \\ x +z & 2y+ w \end{pmatrix} = \begin{pmatrix} 6 & 2\\ 4 & 3\end{pmatrix}$ find values of $x,y,z$ and $w$
The
transpose of matrix:
If $A$ is an $m \times n$ matrix, the transpose of matrix $A$ denoted by $A^{\prime}$ is an $n \times m$ matrix.
If $A = (a_{ij})$, $i=1,2,\ldots ,m$ and $j=1,2,\ldots ,n $
If $A^{\prime} = A$ (i.e. $b_{ij}=a_{ij}$, $i=1,2,\ldots ,n$ and $j=1,2,\ldots ,n $) the matrix $A$ is called symmetric matrix.
If $A^{\prime} = -A$ (i.e. $m = n$ and $b_{ij}=-a_{ij}$, $i=1,2,\ldots ,n$ and $j=1,2,\ldots ,n $) the matrix $A$ is called skew -symmetric matrix.
Exercise:
Exercise:
- If $A = \begin{pmatrix}2&1\\3&4\\5&6\end{pmatrix}$, construct $A^{\prime}$
- If $A = \begin{pmatrix}2&1&6&9&4\end{pmatrix}$, construct $A^{\prime}$
- If $A = \begin{pmatrix}1&2&x\\y&3&6\\4&z&0\end{pmatrix}$, construct $A^{\prime}$
- Which of following Matrices are symmetric?
- $A = \begin{pmatrix}1&2&3\\2&4&6\\3&6&5\end{pmatrix}$
- $B = \begin{pmatrix}-1&3&-1\\2&4&2\\-1&2&0\end{pmatrix}$
- $C = \begin{pmatrix}-1&6&3\\2&-1&4\\3&2&1\end{pmatrix}$
- $D = \begin{pmatrix}0&5&2\\5&6&4\\2&4&-7\end{pmatrix}$
- Which of following Matrices are skew -symmetric?
- $A = \begin{pmatrix}0&6&7\\-6&0&8\\-7&-8&0\end{pmatrix}$
- $B = \begin{pmatrix}0&1&-1\\1&0&2\\1&2&0\end{pmatrix}$
- $C = \begin{pmatrix}0&3&4\\-3&1&-7\\-4&7&2\end{pmatrix}$
- $D = \begin{pmatrix}0&5&1\\-5&0&2\\-1&-2&0\end{pmatrix}$
This is a matrix whose elements are all zero. It is
designated by O
$O = \begin{pmatrix}0&0&\ldots 0\\\vdots&\ddots&\vdots \\0&0&\ldots 0\end{pmatrix}$
That is $O = (a_{ij})$ then $a_{ij}=0$, $i=1,2\ldots m$, and $j=1,2,\ldots n$
$O = \begin{pmatrix}0&0&\ldots 0\\\vdots&\ddots&\vdots \\0&0&\ldots 0\end{pmatrix}$
That is $O = (a_{ij})$ then $a_{ij}=0$, $i=1,2\ldots m$, and $j=1,2,\ldots n$
These are square matrices in which the elements that are not on principal diagonal are all zero. That is $A = (a_{ij})$, $a_{ij} = 0$, $i \ne j$, $i=1,2\ldots n$, and $j=1,2,\ldots n$. We can write $A = \text{diag} \begin{pmatrix} a_{11}&a_{22}&\ldots a_{nn}\end{pmatrix}$
The UNIT Matrix:
This is a square matrix with 1's on principal diagonal and 0's elewhere. It is designated by I. We have
$I = \begin{pmatrix}1&0&\ldots 0\\0&1&\ldots 0\\ \vdots&\ddots&\vdots \\0&0&\ldots 1\end{pmatrix}$
That is $I = (a_{ij})$ then $a_{ij}=1$, if $i=j$, $a_{ij}=0$, if $i \ne j$, $i=1,2\ldots n$, and $j=1,2,\ldots n$
An upper triangular matrix is a square matrix all of whose nonzero elements are located on or above the principal diagonal. A lower triangular matrix is a square matrix all of whose nonzero elements are located on or below the principal diagonal. In other words, an upper triangular matrix has all zeros below its main diagonal, where as a lower triangular matrix has all zeros above its main diagonal.
That is $A = (a_{ij})$,
Definition: Two $m \times n$ matrices $A=(a_{ij})$ and $B=(b_{ij})$, $i=1,2,\ldots, m$, $j=1,2,\ldots, n$ are added by adding corresponding elements. The expressions $A+B$ will designate the $m \times n$ matrix whose elements are $a_{ij}+b_{ij}$. Once again
$(a_{ij}+b_{ij})=(a_{ij})+b_{ij})$
Theorem:
- Matrix addition is commutative. That is for any two matrices $A$ and $B$ conformable for addition $A+B=B+A$
- Matrix addition is associative. That is for any three matrices $A,B$ and $C$ that are conformable for addition,$(A+B)+C=A+(B+C)$
- If $A$ is an $m \times n$ matrix, and $O$ is the $m \times n$ zero matrix,then $A+O=O+A$
If $A=(a_{ij})$ is an $m \times n$ matrix, then its negative, designated by $-A$, is the $m \times n$ matrix whose elements are $-a_{ij}$. By the difference $A - B$, we shall mean the matrix $A+(-B)$, that is to say, $A=(a_{ij})$ and $B=(b_{ij})$, $i=1,2,\ldots, m$, $j=1,2,\ldots, n$ then
$A-B=(a_{ij}-b_{ij})$, $i=1,2,\ldots, m$, $j=1,2,\ldots, n$
Theorem: For any matrix $A$
- $-O=O$
- $A-A=A+(-A)$
- $-(-A)=A$
Definition:
If $A=(a_{ij})$ is an $m \times n$ matrix $i=1,2,\ldots, m$, $j=1,2,\ldots, n$, then $cA$ will designate a scalar product of $A$ by $c$. It is $m \times n$ matrix $i=1,2,\ldots, m$, $j=1,2,\ldots, n$, whose elements are $c \times a_{ij}$. In other words, $cA = (c \times a_{ij})$
Theorem: Let $A$ and $B$ be two $m \times n$ matrices, and let $a$ and $b$ are ordinary numbers (scalar) Then,
- $a(bA)=(ab)A$
- $(a+b)A=aA+bB$
- $a(A+B)= aA+aB$
- $(1)A=A$
- $(-1)A=-A$
- $aO=O$
- $OA=O$
- If $A=\begin{pmatrix}1 & 2 \\ 3 & -4 \end{pmatrix}$, $B=\begin{pmatrix}2 & -7\\5 & 8 \end{pmatrix}$ and $C=\begin{pmatrix}1 & 5\\0 & 2\end{pmatrix}$ Find $A + B - C$
- If $A=\begin{pmatrix}2 & 3 \\ -1 & 4\\1 & 0 \end{pmatrix}$, $B=\begin{pmatrix}-1 & 2\\3 & 1\\5&4 \end{pmatrix}$ and $C=\begin{pmatrix}2 & -1\\2 & 3\\-4&1\end{pmatrix}$ Find $A$$ -B +2C$
Definition: Let $A$ and $B$ be $1 \times n$ or $n \times 1$ matrices. $A$ has elements $a_{1},a_{2},\ldots, n$ and B has elements $b_{1},b_{2},\ldots, n$ The inner product of A and B is given by the formula $A \cdot B=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}$
Exercise:
If $A = (1, 1, 1)$, $B = (3, 1, 4)$, $C = (0, 1, 2)$,
compute the quantities $A \cdot A$, $A \cdot B$, $A \cdot C$, $(A - B) \cdot (A + B)$, $(2A) \cdot (3B)$, $A\cdot (B
+ C)$,
$(A + B) \cdot C$, $(A - B) \cdot (C - B)$, $(A + 2B + C) \cdot (A + B
+ 2C)$.
Two matrices are conformable for multiplication when the number of columns of the first factor equals the number of rows of the second factor. The orders of two matrices $A$ and $B$ that are conformable for multiplication can be written as $ m \times n$ and $n \times p$ respectively.
Designate the $m$ rows of $A$ by $A_{1},A_{2},\ldots ,A_{m}$ (Consider them as row vectors.) Designate the $p$ columns
of $B$ by $B_{1},B_{2},\ldots,B_{p}$ (Consider them as
column vectors.) These vectors all have $n$ elements (because the number of
columns of $A$ is $n$ and the number of rows of $B$ is also $n$). For each integer $i$ between $1$ and $m$, inclusive, and for each integer $j$ between $1$ and $p$, inclusive,
the inner product $A_{i} \cdot B_{j}$ can therefore be formed. This
number will be the element $c_{ij}$in the $(i,j)^{th}$ position of
the product $C = AB$.
Definition: If $A$ is an $m \times n$ matrix with rows $A_{1},A_{2},\ldots ,A_{m}$ and $B$ is an $n \times p$ matrix with columns $B_{1},B_{2},\ldots,B_{p}$, then the product $AB$ is an $m \times p$ matrix $C$ whose elements $c_{ij}$ are given by the formula $c_{ij}=A_{i} \cdot B_{j}$
We may also write $C = AB =$ $\begin{pmatrix}A_{1}\cdot B_{1} &A_{1}\cdot B_{2}& \ldots A_{1}\cdot B_{p}\\ A_{2}\cdot B_{1} &A_{2}\cdot B_{2} & \ldots A_{2}\cdot B_{p}\\ \vdots &\vdots & \vdots\\ A_{m}\cdot B_{1} &A_{m}\cdot B_{2}& \ldots A_{m}\cdot B_{p} \end{pmatrix}$
Exercise:
- Find elements $c_{22},c_{32},c_{23},c_{33}$ in the product $C=AB$ where, $A = \begin{pmatrix} 2 & 3 & 4\\1 & 2 & 3\\-1&1&2\end{pmatrix}$, $B=\begin{pmatrix} 1&3 &0\\-1 & 2 & 1\\0 & 0 & 2\end{pmatrix}$
- Given $A = \begin{pmatrix} 1 & 1 & -1\\2 & -3 & 4\\3&-2&3\end{pmatrix}$, $B=\begin{pmatrix} -1&-2 &-1\\6 & 12 & 6\\5 & 10 & 4\end{pmatrix}$, $C = \begin{pmatrix}-1 & -1 & 1\\2 & 2 &-2\\-3&-3&3 \end{pmatrix}$ Find
- $AB$
- $BA$
- $AC$
- $CA$
- For the following pair of matrices compute $AB$ and $BA$ whichever exist.
- $A=\begin{pmatrix}1&2&3&4&5\\6&7&8&0&3\\6&-3&4&-2&1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 2\\3 & 4\\5 & 6\\6 & 5\\7 & 0\end{pmatrix}$
- $A=\begin{pmatrix} 2&3&4&5\end{pmatrix}$, $B=\begin{pmatrix} 1&2&3&4\end{pmatrix}$
- $A=\begin{pmatrix} 0&1&2\\1 &2&0\end{pmatrix}$,$B=\begin{pmatrix} 0&1\\-1 & 0\\0 & 1\end{pmatrix}$
- $A= \begin{pmatrix}1 & 2\\2 & 3\\3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 3 & -2 & 1\\0 & 1 & -2\\3& -4 & 5\end{pmatrix}$
- $A=\text{diag}\begin{pmatrix}\lambda_{1}&\lambda_{2}&\lambda_{3}&\lambda_{4} \end{pmatrix}$ $B =\text{diag}\begin{pmatrix}\mu_{1}&\mu_{2}&\mu_{3}&\mu_{4} \end{pmatrix}$
- Multiply
- $\begin{pmatrix} 1 & 6 & 1\\0& 1 & 2\\ 0& 0 & 1\end{pmatrix}$$\begin{pmatrix} x\\y\\z\end{pmatrix}$
- $\begin{pmatrix} 1 & 0 & 0\\0& 4 & 0\\ 0& 0 & 5\end{pmatrix}$$\begin{pmatrix} x\\y\\z\end{pmatrix}$
- $\begin{pmatrix}1 & 4 & 3 & 2 \end{pmatrix}$ $\begin{pmatrix} x\\y\\z\\w\end{pmatrix}$
There are two types of
matrix multiplication: in the product $AB$, $B$ is said to be premulliplied by $A$ or
to be multiplied by $A$ on the left, while in the product $BA$, $B$ is said to be
postmultiplied by $A$ or to be multiplied by $A$ on the right. Two matrices $A$ and $B$ for which $AB = BA$ are said to commute
with one another.
Exercise:
- Do the matrices $\begin{pmatrix} 1&2&1\\2&1&1\\2&-1&1\end{pmatrix}$ and $\begin{pmatrix}1&1&1\\1&-1&1\\-1&1&1 \end{pmatrix}$ commute?
- If the matrix $\begin{pmatrix} 1&2\\3&0\end{pmatrix}$ commutes with matrix $\begin{pmatrix} a&b\\c&d\end{pmatrix}$ what must be the relationship among the quantities $a$, $b$, $c$ and $d$?
Theorem: If $A$, $B$, and $C$, are matrices; if $I$ is the unit matrix and $O$ is the zero
matrix, then, with proper attention paid to conformability, we have,
Back to Basic Mathematics
- $(AB)C = A(BC)$ the associative law;
- $(A + B)C = AC + BC$ the distributive laws;
- $A(B + C) = AB + AC$ the distributive laws;
- $OA = AO = O$ multiplication by zero;
- $AI = IA = A$ multiplication by the unit.
Corollary: If a and b are scalars and if $A$ and $B$ are conformable
for matrix multiplication, then $(aA)(bB)=ab(AB)$
Exercise: Verify
the two distributive laws numerically with selection
- $A = \begin{pmatrix}1&2\\3 & 4 \end{pmatrix}$, $B = \begin{pmatrix}1&0\\2 & 3 \end{pmatrix}$ and $C = \begin{pmatrix}1&-1\\0 & 1 \end{pmatrix}$
- $A = \begin{pmatrix}1&-1\\2 & 0 \end{pmatrix}$, $B = \begin{pmatrix}0&4\\1 & 2 \end{pmatrix}$ and $C = \begin{pmatrix}1&1\\1& 1 \end{pmatrix}$
Theorem: If $A$ and $B$ are conformable for multiplication, then $(AB)^{\prime}=(B^{\prime}A^{\prime})$
Corollary: For matrices $A,B,C,D$ we can write $(ABCD)^{\prime}=(D^{\prime}C^{\prime}B^{\prime}A^{\prime})$. In words: the transpose of the product of any number of matrices is
equal to the product of the transposes of the individual matrices formed in the
reverse order.
Exercise: If $A = \begin{pmatrix} 1&1&1\\-1&1&-1\\1&1&-1\end{pmatrix}$ and
$(AB)^{\prime}=(B^{\prime}A^{\prime})$Back to Basic Mathematics
