### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### Determinants Class XII Chapter 4 : Formulas & Basics

Class XII : Chapter : 4 : Determinants
Formulas and basic concepts based on the determinants chapter 4 class XII, properties of determinants, method of finding the solutions of linear equations by matrix method.

Determinant:
To every square matrix A = [aijof order n, we can associate a number (real or complex) called determinant of the square matrix  A. Determinant of matrix [A] is denoted by |A|

For Example: Let a matrix A of order 2 x 2 is given by $A=\; \left [ \begin{matrix} a &b \\ c & d \end{matrix} \right ]$$Then \; \; \left |A \right | =\left | \begin{matrix} a &b \\ c & d \end{matrix} \right |= ad-cb$Let a matrix of order  3 x 3 is given by : $A=\begin{bmatrix} 3 &4 &5 \\ -2& 5 & -4\\ 3 & -1 & 5 \end{bmatrix}$
Determinant of matrix A is denoted by |A| and is given by $|A|=\begin{vmatrix} 3 &4 &5 \\ -2& 5 & -4\\ 3 & -1 & 5\end{vmatrix}$Expand it along the first row  $|A|= 3\begin{vmatrix} 5 &-4 \\ -1&5 \end{vmatrix}-4\begin{vmatrix} -2 &-4 \\ 3&5 \end{vmatrix}+5\begin{vmatrix} -2 &5 \\ 3 &-1 \end{vmatrix}$ $|A|=3(25-4)-4(-10+12)+5(2-15)$ $|A|=3\times 21-4\times 2+5\times -13$ $|A|=63-8-65=-8-2=-10$
Note:
1. For matrix A, |A| is read as determinant of A and not modulus of A.

2. Only square matrix have determinant.

3. For easier calculation to find the determinants of matrix of order 2 x 2, we shall expand the determinant along that row or column which contains maximum number of zeroes.
4. Rule to write the sign (+ or - ) of the position of the element while finding the determinant.

Position of a11 is  (-1)1+1 = (-1)2 = +ve

Position of a12 is  (-1)1+2 = (-1)3 = - ve

Position of a13 is  (-1)1+3 = (-1)4 = +ve

Position of a21 is  (-1)2+1 = (-1)3 = - ve

Position of a22 is  (-1)2+2 = (-1)4 = +ve

……………….. and so on

For 2 x 2 Matrix we can take sigh as follows $\begin{bmatrix} + &- \\ -&+ \end{bmatrix}$
For 3 x 3 Matrix we can take sign as follows $\begin{bmatrix} + & - &+ \\ -& + & -\\ +& - &+ \end{bmatrix}$

Important Properties of determinants :

Property 1: The value of the determinant remains unchanged if its rows and columns are interchanged.      |A| = |A'|

Property 2: If any two rows (or columns) of a determinant are interchanged then the sign of the determinant is changes.

Property 3 : If in a matrix any two rows (or columns) are identical then the determinant of the matrix is zero. $|A|=\begin{vmatrix} a &b &c \\ a &b & c\\ 3& 5 & 7 \end{vmatrix}=0$

Property 4: If each element of a row (or column) of a determinant is multiplied by a constant k, then the value of the determinant gets multiplied by k. or multiplying a determinant by any constant  k(say) means, multiply each element of only one row (or column) by k.

Property 5: If some or all elements of a row or column of a determinant are expressed as sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.
$If\; A=\left | \begin{matrix} a+b &c+d &e+f \\ 1&2 &3 \\ 4&5 &6 \end{matrix} \right | \; then\; A\; can\; be\; written\; as$$A=\left | \begin{matrix} a & c&e \\ 1& 2 & 3\\ 4&5 & 6 \end{matrix} \right |+\left | \begin{matrix} b &d &f \\ 1 &2 &3 \\ 4 &5 &6 \end{matrix} \right |$Same property can be applied in columns.

Property  6: If each element of a row (or column) are multiplied by the same number and is added to the corresponding elements of the other row (or column) , then the value of the determinants remain unchanged.
This means that the value of determinant remain unchanged if we apply the following operations
$R_{i}\rightarrow R_{i}+kR_{j}\; \;Or \; \; R_{i}\rightarrow R_{i}-kR_{j}\; \; Or$$C_{i}\rightarrow C_{i}+kC_{j}\; \;Or \; \; C_{i}\rightarrow C_{i}-kC_{j}\; \;$
Property 7: If A is a matrix of order n x n then  |kA| = kn|A|
Similarly   If A is a matrix of order 3 x 3 then  |kA| = k3|A|   or
If A = kB, where A and B are two square matrices of order n,
Then   |A| = kn |B|
Property 8: If all elements of a row or a column of a matrix are  zero then the value of that determinant is also become zero. $|A|=\begin{vmatrix} 0 &0 &0 \\ a &b & c\\ 3& 5 & 7 \end{vmatrix}=0$

Method of finding the area of triangle with the help of Determinants

If A is a matrix and A' is the transpose of the matrix then |A'| = |A|

Area of Triangle ABC with vertices (x1, y1),  (x2, y2) ,  (x3, y3)
$=\frac{1}{2}\left | \begin{matrix} x_{1} &y_{1} &1 \\ x_{2}& y_{2} &1 \\ x_{3}&y_{3} &1 \end{matrix} \right |$
$=\frac{1}{2}\left | x_{1}\left ( y_{2}-y_{3}\right )+x_{2}\left ( y_{3}-y_{1} \right )+x_{3}\left ( y_{1}-y_{2} \right ) \right |$

Note :
1. Area is a positive quantity, so we always take the absolute value of the determinant for finding area.

2. When area is given, then use both positive and negative values of the determinants for calculation.

3. The area of the triangle formed by joining three collinear points is always zero.

Minors and cofactors
Minor:
Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. Minor of element aij is denoted by Mij.
Co-factors :
Co-factors of an element aij, is denoted by Aij is defined by Aij = (-1)i+j Mij, where Mij is minor of aij.
Note:
1. Sum of the product of the elements of any row (or column) with their corresponding co-factors, gives determinant.
|A| = a11A11 + a12A12 + a13A13

2. If elements of any row (or column) are multiplied with the co-factors of any other row (or column) then their sum is zero.
a11A12 + a12A13 + a13A11 = 0

Adjoint of a matrix is defined as the transpose of the co-factor matrix.

If A is any square matrix then :  A(adj A) = (adj A) A = |A|I

A square matrix A is said to be singular if  |A| = 0

A square matrix A is said to be non- singular if |A| ≠ 0

If A is a square matrix of order n then |adj.A| = |A|n-1

If AB = BA = I then A and B are said to be the inverse of each other. or A-1 = B or B-1 =  A  and  (A-1)-1 = A

Inverse of the matrix:
Inverse of matrix A is calculated by using the formula :$A^{-1}=\frac{1}{\left | A \right |}\times AdjointA$

Method of finding Inverse of a Matrix
Step 1 :  Name the given matrix A (say)

Step 2 : Find determinant of matrix A or find |A|

Step 3 : If |A| = 0, then A is a singular matrix and  A-1 does not exist. If |A| ≠0, then A is a non singular matrix and A-1 exists.

Step 4 : If A-1 exists then find cofactor matrix of A.

Step 5 : From co-factor matrix find adjoint A.

Step 6 : Find the inverse of Matrix A by using the formula:  $A^{-1}=\frac{1}{\left | A \right |}\times AdjointA$

Applications of Determinants and Matrices

Consistent System:-
If given system of linear equations either have one or more solutions then the system is called consistent. If   |A| ≠ 0, then system of equations have unique solution, and the system is called consistent.
Inconsistent System:-
System of linear equations is said to be inconsistent if its solution does not exist.

If A is a singular matrix, then |A| = 0.  In this case , we calculate (adj. A)B and following cases arise.
Case I
1. If |A| = 0 and (adj. A)B ≠ O, then solution does not exist and the system of equations is called inconsistent.
Case II
2.  If |A| = 0 and (adj. A)B = O, then the system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution.
a) If system of equations  have many solution then the system is consistent.
b)  If system have no solution then it is inconsistent.

Flow Diagram to understand consistency or inconsistency

Solution of system of linear equation using matrix method

Let the given system of linear equations is

a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3

This system of equations can be written in matrix form as
$A=\left [\begin{matrix} a_{1} &b_{1} &c_{1} \\ a_{2}& b_{2} & c_{2}\\ a_{3}& b_{3} & c_{3} \end{matrix} \right ],\; \; X= \left [\begin{matrix} x\\y \\z \end{matrix} \right ],\; \; B= \left [\begin{matrix} d_{1}\\d_{2} \\d_{3} \end{matrix} \right ]$
In Matrix form system of equations can be written as
AX = B   ⇒ X =  A-1 B

Steps to follow for solving system of linear equations

Step 1 : Write the given equations in the form AX = B, where A is the coefficient matrix, X is the variable matrix and B is the matrix of constant terms.

Step 2 : Find |A| , If system of equations is non-singular i.e.  |A| ≠ 0, system of equations have unique solution.

Step 3 : If |A| ≠ 0, then find  A-1 by using the steps discussed above.

Step 4 : Find the value of the variables x, y, z by using the formula : X = A-1 B

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Cramer's Rule: (Only for Applied Maths Students)
Let system of linear equations given by
$a_{1}x+b_{1}y=d_{1}\\a_{2}x+b_{2}y=d_{2}$
Solution of these equations by using cramer's rule is given by
$x=\frac{\Delta _{1}}{\Delta },\;\; y=\frac{\Delta _{2}}{\Delta }$
$Where\; \Delta =\begin{vmatrix} a_{1}& b_{1}\\ a_{2}& b_{2} \end{vmatrix},\; \; \Delta _{1}=\begin{vmatrix} d_{1} & b_{1}\\ d_{2} &b_{2} \end{vmatrix},\; \; \Delta _{2}=\begin{vmatrix} a_{1} &d_{1} \\ a_{2} & d_{2} \end{vmatrix}$$Provided:\; \; \Delta \neq 0$
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Let system of equations are given by
$a_{1}x+b_{1}y+c_{1}z=d_{1}$$a_{2}x+b_{2}y+c_{2}z=d_{2}$$a_{3}x +b_{3}y +c_{3}z = d_{3}$
Solution of system of equations is given by
$x=\frac{\Delta _{1}}{\Delta },\;\; y=\frac{\Delta _{2}}{\Delta },\;\; z=\frac{\Delta _{3}}{\Delta }$
$Where\; \; \Delta =\begin{vmatrix} a_{1} &b_{1} & c_{1}\\ a_{2} & b_{2} &c_{2} \\ a_{3}& b_{3} & c_{3} \end{vmatrix},\; \; \Delta_{1} =\begin{vmatrix} d_{1} &b_{1} & c_{1}\\ d_{2} & b_{2} &c_{2} \\ d_{3}& b_{3} & c_{3} \end{vmatrix}$
$Where\; \; \Delta_{2} =\begin{vmatrix} a_{1} &d_{1} & c_{1}\\ a_{2} & d_{2} &c_{2} \\ a_{3}& d_{3} & c_{3} \end{vmatrix},\; \; \Delta_{3} =\begin{vmatrix} a_{1} &b_{1} & d_{1}\\ a_{2} & b_{2} &d_{2} \\ a_{3}& b_{3} & d_{3} \end{vmatrix}$$Provided:\; \; \Delta \neq 0$
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