## Rank of a matrix

In linear algebra, the rank of a matrix is the dimension of its row space or column space. It is important to note that the row space and column space of a matrix have equal dimensions.

### Definition of rank of a matrix

A nimber r is said to be a rank of a non-zero matrix A if

1. There exist at least one minor of order r of A which does not vanish and

2. Every minor of higher order than r is zero.

The rank of a matrix A is denoted by  $\rho&space;(A)$.
We have $\rho&space;(A)=r$

#### Another definition of rank of a matrix

The rank of a non-zero matrix is largest order of any non-vanishing minor of the matrix.

Remarks

From the above definition of rank of a matrix, we observe that

1. The rank of zero matrix is zero i.e.,  $\rho&space;(O)=0$ where O is a zero matrix,

2. The rank of a non-singular matrix of order n is n,

3.  $\rho&space;(A)\leq&space;r$, if every minor of order r+1 vanishes,

4.  $\rho&space;(A)\geq&space;r$, if there is a minor of order r which does not vanish.

### ☆Finding rank of matrix

1. Minor method

2. Normal form

3. Echelon form of matrix

#### 1. Example on minor method

Q1. Determine the rank of matrix
$\begin{bmatrix}&space;4&space;&&space;2&space;&&space;3\\&space;8&&space;5&space;&&space;2\\&space;12&-4&&space;5&space;\end{bmatrix}$
Ans

$A=\begin{bmatrix}&space;4&space;&&space;2&space;&&space;3\\&space;8&&space;5&space;&&space;2\\&space;12&-4&&space;5&space;\end{bmatrix}$
$|A|=\begin{vmatrix}&space;4&space;&&space;2&space;&&space;3\\&space;8&space;&&space;5&space;&&space;2\\&space;12&&space;-4&space;&5&space;\end{vmatrix}$
$=4\begin{vmatrix}&space;5&space;&&space;2\\&space;-4&space;&&space;5&space;\end{vmatrix}-2\begin{vmatrix}&space;8&space;&&space;2\\&space;12&&space;5&space;\end{vmatrix}+3\begin{vmatrix}&space;8&space;&&space;5\\&space;12&&space;-4&space;\end{vmatrix}$
$=4(25+8)-2(40-24)+3(-32-60)$
$=4(33)-2(16)+3(-92)$
$=132-32-276\neq&space;0$
$\therefore&space;\,&space;\:&space;\:&space;\:&space;\;&space;\;&space;\;&space;|A|\neq&space;0$
A is non singular
$\rho&space;(A)=3$

Q2. Determine the rank of matrix
$\begin{bmatrix}&space;2&space;&&space;3&space;&3&space;\\&space;3&&space;6&space;&12&space;\\&space;2&space;&&space;4&space;&&space;8&space;\end{bmatrix}$
Ans.
$Let\:&space;\:&space;A=\begin{bmatrix}&space;2&space;&&space;3&space;&3&space;\\&space;3&&space;6&space;&12&space;\\&space;2&space;&&space;4&space;&&space;8&space;\end{bmatrix}$
$|A|=\begin{vmatrix}&space;2&space;&&space;3&space;&3&space;\\&space;3&&space;6&space;&&space;12\\&space;2&&space;4&space;&&space;8&space;\end{vmatrix}$
$=2\begin{vmatrix}&space;6&space;&&space;12\\&space;4&space;&&space;8&space;\end{vmatrix}-3\begin{vmatrix}&space;3&space;&&space;12\\&space;2&space;&&space;8&space;\end{vmatrix}+3\begin{vmatrix}&space;3&space;&&space;6\\&space;2&space;&&space;4&space;\end{vmatrix}$
$=2(48-48)-3(24-24)+3(12-12)$
$=0-0+0=0$
$|A|=0$ i.e, only minor of order 3 of A vanishes.
Now we consider any minor of order 2.

Consider   $\begin{vmatrix}&space;2&space;&&space;3\\&space;3&space;&&space;6&space;\end{vmatrix}=12-9=3\neq&space;0$
There is a minor of order 2 of A which does not vanish

$\rho&space;(A)=2$

#### 2. Normal form of a matrix

$\begin{bmatrix}&space;I_{r}&space;&O&space;\\&space;O&space;&&space;O&space;\end{bmatrix},\:&space;\begin{bmatrix}&space;I_{r}&space;&&space;O&space;\end{bmatrix},\:&space;\begin{bmatrix}&space;I_{r}&space;\\&space;O&space;\end{bmatrix}$ are called the normal formes of matrix

#### Example

Q3. Prove that the matrix  $\begin{bmatrix}&space;1&space;&&space;2&space;&&space;3\\&space;2&space;&&space;3&space;&&space;0\\&space;0&space;&&space;1&space;&&space;2&space;\end{bmatrix}$ is equivalent to  $I_{3&space;}$.
Ans
$Let\:&space;\:&space;\:&space;\:&space;A=\begin{bmatrix}&space;1&space;&&space;2&space;&&space;3\\&space;2&space;&&space;3&space;&&space;0\\&space;0&space;&&space;1&space;&&space;2&space;\end{bmatrix}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;2&space;&&space;3\\&space;0&space;&&space;-1&space;&&space;-6\\&space;0&&space;1&space;&&space;2&space;\end{bmatrix},&space;by\:&space;\:&space;R_{2}\rightarrow&space;R_{2}-2R_{1}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;-1&space;&&space;-6\\&space;0&&space;1&space;&&space;2&space;\end{bmatrix},&space;by\:&space;\:&space;C_{2}\rightarrow&space;C_{2}-2C_{1},\:&space;C_{3}\rightarrow&space;C_{3}-3C_{1}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;6\\&space;0&&space;1&space;&&space;2&space;\end{bmatrix},&space;by\:&space;\:&space;R_{2}\rightarrow-&space;R_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;6\\&space;0&&space;0&space;&&space;-4&space;\end{bmatrix},&space;by\:&space;\:&space;R_{3}\rightarrow&space;R_{3}-R_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;0\\&space;0&&space;0&space;&&space;-4&space;\end{bmatrix},&space;by\:&space;\:&space;C_{3}\rightarrow&space;C_{3}-6C_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;0&space;&&space;0\\&space;0&space;&&space;1&space;&&space;0\\&space;0&&space;0&space;&&space;1&space;\end{bmatrix},&space;by\:&space;\:&space;R_{3}\rightarrow-\frac{1}{4}&space;R_{3}$
$A\sim&space;I_{3}$
Given matrix is equivalent to  $I_{3&space;}$.

#### 3. Echelon form

A matrix  $A=\left&space;[&space;a_{ij}&space;\right&space;]$ is said to be in echelon form if

1. The zero rows of A occur below all the non-zero rows of A

2. The number of rows before the first non-zero element in a row is less than the number of such zero in the next row.

#### Example

Q4. Reduce to row echelon form the matrix
$A=\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;2&space;&&space;-4&space;&&space;3&space;&5&space;\\&space;-1&space;&&space;2&space;&&space;6&space;&&space;-7&space;\end{bmatrix}$
Ans.
$A=\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;2&space;&&space;-4&space;&&space;3&space;&5&space;\\&space;-1&space;&&space;2&space;&&space;6&space;&&space;-7&space;\end{bmatrix}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;0&space;&&space;0&space;&&space;5&space;&-3\\&space;0&space;&&space;0&space;&&space;5&space;&&space;-3&space;\end{bmatrix},&space;\:&space;by&space;\:&space;R_{2}\rightarrow&space;R_{2}-2R_{1},\:&space;R_{3}\rightarrow&space;R_{3}+R_{1}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;0&space;&&space;0&space;&&space;1&space;&-\frac{3}{5}\\&space;0&space;&&space;0&space;&&space;5&space;&&space;-3&space;\end{bmatrix},&space;\:&space;by&space;\:&space;R_{2}\rightarrow&space;\frac{1}{5}R_{2}$
$\sim&space;\begin{bmatrix}&space;1&space;&&space;-2&space;&&space;-1&space;&&space;4\\&space;0&space;&&space;0&space;&&space;1&space;&-\frac{3}{5}\\&space;0&space;&&space;0&space;&&space;0&space;&&space;0&space;\end{bmatrix},&space;\:&space;by&space;\:&space;R_{3}\rightarrow&space;R_{3}-5R_{2}$

Which is in row echelon form.
Since there are two non-zero rows in row echelon form.

$\rho&space;(A)=2$

## Exact Differential Equations

The differential equation of type
$M\,dx&space;+N&space;\,&space;dy=0$ (where M and N are function of x and y) is called an exact differential equation when
$M\,dx&space;+N&space;\,&space;dy=du$ ( where u is a function of x and y).

Examples :

1.  $x\,dy&space;+y&space;\,&space;dx=0$ is an exact differential equation as  $x\,dy&space;+y&space;\,&space;dx=d(x,y)$

2.  $\frac{&space;x\,dy&space;-y&space;\,&space;dx&space;}{x^{2}}=0$ is an exact differential equation as $\frac{&space;x\,dy&space;-y&space;\,&space;dx&space;}{x^{2}}=d\left&space;(&space;\frac{y}{x}&space;\right&space;)$

3. The differential equation
$\sin&space;x\,&space;\cos&space;y\:&space;dy+\cos&space;x\:&space;\sin&space;y\:&space;dx=0$ is an exact differential equations as $\sin&space;x\,&space;\cos&space;y\:&space;dy+\cos&space;x\:&space;\sin&space;y\:&space;dx=d(\sin&space;x\:&space;\sin&space;y)$

#### Necessary and sufficient condition

Article. Find the necessary and sufficient condition that the equation  $M\,dx&space;+N&space;\,&space;dy=0$ ( where M and N are function of x and y with the condition that  $M,N,\frac{\partial&space;M}{\partial&space;y},\frac{\partial&space;N}{\partial&space;x}$ are continuous function of x and y) may be exact.

Proof  1. Necessary condition
$\frac{\partial&space;M}{\partial&space;y}=\frac{\partial&space;N}{\partial&space;x}$

2. Condition is sufficient

$\int_{y\:&space;constant&space;}M\:&space;dx+\int&space;(terms\:&space;in\:&space;N\:&space;not\:&space;containing\:&space;x)&space;dy=c$

### Integrating factor

An integrating factor (abbreviatef I.F) of a differential equation is such a factor such that if the equation is multiplied by it, the result equation is exact.

### Five rules for finding integrating factor

If  $M\,dx&space;+N&space;\,&space;dy=0$ is not exact and it is difficult to find integrating factor, then following five rules help us in finding integrating factor.

Rule 1. If the equation  $M\,dx&space;+N&space;\,&space;dy=0$ is homogenous in x and y i.e. if M and N are homogenous function of the same degree in x and y, then  $\frac{1}{Mx+Ny}$ is an I.F. provided $M\,dx&space;+N&space;\,&space;dy\neq&space;0.$

Rule 2. If the equation  $M\,dx&space;+N&space;\,&space;dy=0$ is of the form  $f_{1}(x,y)\:&space;y\:&space;dx&space;+&space;f_{2}(x,y)\:&space;x\:&space;dy=0$, then  $\frac{1}{Mx-Ny}$ is an I.F. provided  $M\,dx&space;+N&space;\,&space;dy\neq&space;0.$

Rule 3. If the equation  $M\,dx&space;+N&space;\,&space;dy=0$$\frac{\frac{\partial&space;M}{\partial&space;y}-\frac{\partial&space;N}{\partial&space;x}}{N}$ is a function of x only =f(x) then  $e^{\int&space;f(x)dx}$ is an I.F.

Rule 4. If the equation  $M\,dx&space;+N&space;\,&space;dy=0$$\frac{\frac{\partial&space;N}{\partial&space;x}-\frac{\partial&space;M}{\partial&space;y}}{M}$ is a function of y only =f(y) then  $e^{\int&space;f(y)dy}$ is an I.F.

Rule 5. If the equation is
$x^{a}y^{b}(my\:&space;dx&space;+nx\:&space;dy)+x^{a'}y^{b'}(m'y\:&space;dx+n'x\:&space;dy)=0$, then  $x^{h}y^{k}$ is an I.F. when  $\frac{a+h+1}{m}=\frac{b+k+1}{n},\frac{a'+h+1}{m'}=\frac{b'+k+1}{n'}$ .

## Example

Q1. Solve the differential equation
$\fn_phv&space;(\cos&space;x\cos&space;y-\cot&space;x)dx-(\sin&space;x\sin&space;y)dy=0$.
Sol. The given differential equation is
$\fn_phv&space;(\cos&space;x\cos&space;y-\cot&space;x)dx-(\sin&space;x\sin&space;y)dy=0$
Comparing it with $M\,dx&space;+N&space;\,&space;dy=0$, we get
$\fn_phv&space;M=\cos&space;x\cos&space;y-\cot&space;x,\:&space;N=-\sin&space;x\sin&space;y$
$\fn_phv&space;\frac{\partial&space;M}{\partial&space;y}=-\cos&space;x\sin&space;y\:&space;\:&space;and\:&space;\:&space;\frac{\partial&space;N}{\partial&space;x}=-\cos&space;x\sin&space;y$
$\frac{\partial&space;M}{\partial&space;y}=\frac{\partial&space;N}{\partial&space;x}$
Given equation is exact and its solution is

$\int_{y\:&space;constant&space;}M\:&space;dx+\int&space;(terms\:&space;in\:&space;N\:&space;not\:&space;containing\:&space;x)&space;dy=c$

$\fn_phv&space;\int_{y\:&space;constant&space;}(\cos&space;x\cos&space;y-\cot&space;x)\:&space;dx+0=c$

$\fn_phv&space;(\cos&space;y)\left&space;(&space;\int&space;\cos&space;x\:&space;dx&space;\right&space;)-\int&space;\cot&space;x\:&space;dx=c$

$\fn_phv&space;\cos&space;y\sin&space;x-\log&space;\left&space;|&space;\sin&space;x&space;\right&space;|=c.$

Q2. Solve the following differential equation $\fn_phv&space;(y^{3}-2yx^{2})dx+(2xy^{2}-x^{3})dy&space;=0$
Sol.
The given differential equation is
$\fn_phv&space;(y^{3}-2yx^{2})dx+(2xy^{2}-x^{3})dy&space;=0$

Which is homogenous in x,y.
Comparing with  $M\,dx&space;+N&space;\,&space;dy=0$, we get

$\fn_phv&space;M=y^{3}-2yx^{2},\:&space;\:&space;\:&space;N=2xy^{2}-x^{3}$

$\fn_phv&space;I.F.=&space;\frac{1}{Mx+Ny}=\frac{1}{xy^{3}-2x^{3}y+2xy^{3}-x^{3}y}=\frac{1}{3xy(y^{2}-x^{2})}$
Multiple both side by  $\fn_phv&space;\frac{1}{3xy(y^{2}-x^{2})}$, we get,

$\fn_phv&space;\frac{y^{2}-2x^{2}}{3xy(y^{2}-x^{2})}dx+\frac{(2y^{2}-x^{2})}{&space;3xy(y^{2}-x^{2})&space;}dy=0$

Which is exact and its solution is

$\fn_phv&space;\frac{1}{3}\int_{y\:&space;constant&space;}\frac{y^{2}-2x^{2}}{x(y^{2}-x^{2})}\:&space;dx+\frac{1}{3}\int&space;\frac{1}{y}&space;dy=c$

$\fn_phv&space;\frac{1}{3}\int_{y\:&space;constant&space;}\left&space;(&space;\frac{1}{x}-\frac{x}{y^{2}-x^{2}}&space;\right&space;)\:&space;dx+\frac{1}{3}\int&space;\log&space;\left&space;|&space;y&space;\right&space;|&space;dy=c$

$\fn_phv&space;\frac{1}{3}\left&space;[&space;\log|x|&space;+\frac{1}{2}\log|y^{2}-x^{2}|+&space;\log|y|&space;\right&space;]=c$

$\fn_phv&space;\log&space;|(x^{2}y^{2})(y^{2}-x^{2})|=6c$

$\fn_phv&space;x^{2}y^{2}(y^{2}-x^{2})=|e^{6c}|=C$

Q3. Solve tge differential equation
$\fn_phv&space;(x^{3}y^{4}+x^{2}y^{3}+xy^{2}+y)dx+(x^{4}y^{3}-x^{3}y^{2}+x^{2}y+x)dy=0$
Sol.
The given differential equation is

$\fn_phv&space;(x^{3}y^{4}+x^{2}y^{3}+xy^{2}+y)dx+(x^{4}y^{3}-x^{3}y^{2}+x^{2}y+x)dy=0$

$\fn_phv&space;(x^{3}y^{3}+x^{2}y^{2}+xy+1)y\:&space;dx+(x^{3}y^{3}-x^{2}y^{2}+xy+1)x\:&space;dy=0$

Which is of form  $\fn_phv&space;f_{1}(x,y)y\:&space;dx+f_{2}(x,y)x\:&space;dy=0$

Comparing it with  $M\,dx&space;+N&space;\,&space;dy=0$, we get

$\fn_phv&space;M=y(x^{3}y^{3}+x^{2}y^{2}+xy+1),\:&space;N=x(x^{3}y^{3}-x^{2}y^{2}+xy+1)$

I.F.= $\frac{1}{Mx-Ny}$

=$\fn_phv&space;\frac{1}{&space;xy(x^{3}y^{3}+x^{2}y^{2}+xy+1)-\:&space;xy(x^{3}y^{3}-x^{2}y^{2}+xy+1)&space;}$

=$\fn_phv&space;\frac{1}{2x^{3}y^{3}}$
Multiple both side by $\fn_phv&space;\frac{1}{2x^{3}y^{3}}$, we get

$\fn_phv&space;\frac{&space;x^{3}y^{3}+x^{2}y^{2}+xy+1&space;}{2x^{3}y^{2}}dx+\frac{&space;x^{3}y^{3}-x^{2}y^{2}+xy+1&space;}{2x^{2}y^{3}}dy=0$

Which is exact and its solution is

$\fn_phv&space;\frac{1}{2y^{2}}\int&space;\left&space;(&space;y^{3}+\frac{y^{2}}{x}+\frac{y}{x^{2}}+\frac{1}{x^{3}}&space;\right&space;)dx+\frac{1}{2}\int&space;\left&space;(&space;-\frac{1}{y}&space;\right&space;)dy=c$

$\fn_phv&space;\frac{1}{2y^{2}}&space;\left&space;(&space;xy^{3}&space;+y^{2}\log&space;x-\frac{y}{x}-\frac{1}{2x^{2}}\right&space;)-\frac{1}{2}\log&space;y=c$

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