CBSE Class 12 Maths Revision Notes Chapter 4

Class 12 Mathematics Chapter 4 Notes

In Class 12, students deal with high pressure and stress, mainly in Mathematics. The marks scored in the final board examination will be required to appear for future studies in their field of interest. Mathematics is considered to be a tricky subject. A strong understanding of all basic concepts and regular practice is very important to understanding complex topics. Students are advised to use the best study material, such as the Class 12 Mathematics chapter 4 notes, to help them get a clear understanding of determinants and concepts related to it. 

The concepts explained in this chapter are very useful to analyse and find the solution. In Class 12 Mathematics chapter 4 notes, students will gain knowledge of determinants up to order 3. With the help of determinants, we can find out the uniqueness of a solution. The chapter has a wide range of applications in the Engineering, Science, etc., sectors. Also, concepts based on the properties of the matrix, cofactors, inverse, adjoint, and area of a triangle are covered in the Class 12 Mathematics chapter 4 notes. 

Extramarks, an online learning platform, provides professional help and gives students the best academic notes to ace their exams. With the help of the class 12 chapter 4 mathematics notes, students can prepare for both the board exams and the competitive exams. Our notes are provided with detailed information, graphical representation, and easy language for students’ overall development.  

Key Topics Covered In Class 12 Mathematics Chapter 4 Notes

The main topics covered in the Class 12 Mathematics chapter 4 notes are:

  • Introduction
  • Determinant Definition
  • Properties
  • Area of a Triangle
  • Minors and Cofactors
  • Adjoint of a Matrix
  • The inverse of a Matrix
  • Applications
  • Solution of a system of linear equations

A brief of the key topics covered in class 12 Mathematics chapter 4 notes is as under.

INTRODUCTION: 

The Class 12 Mathematics chapter 4 notes include various complex concepts related to determinants and their applications. To ace this chapter, students must learn concepts introduced in Class 11 as well as chapter 3 of the NCERT books. 

RECALL:

If the system of equations is given as 

a1x + b1y=c1 and a2x + b2y=c2, then the matric representation will be

a1 b1 x =  c1
a2 b2 y c2

The solution of this system is all the values of variables x and y, which satisfy the linear equations in the system.

DEFINITION OF DETERMINANT: 

The determinant of any square matrix is a scalar value that is calculated from all elements of the matrix, which implies, a for every square matrix A of order n is associated with a number called the determinant of that matric A and is denoted by A (mod A), det (A) or Δ . 

Consider A = 

a11 b12
a21 b22

The scalar value for matrix A is a1b2–b1a2

A= a11b22–b12a21

a11 b12
a21 b22

NOTE: Only square matrices (number of rows and columns are equal) have determinants.

TYPES OF DETERMINANTS:

  1. First Order Determinant: The determinant of a matrix, say A of order one, is known as the First-order determinant. The value of the determinant is said to be the element of matrix A.  

Consider A = [2] 

Therefore determinant = 2= 2

  1. Second Order Determinant – The determinant of a matrix, say A of order two, is known as the Second-order determinant.

Consider A = 

4 2
5 3

Therefore, A= 

4 2
5 3

(4 x 3) – (5 x 2) = 12 – 10 = 2

  1. Third Order Determinant – The determinant of a matrix, say A of order three, is known as the third-order determinant.

Consider A = 

a1 b1 c1
a2 b2 c2
a3 b3 c3

Then A = 

a1 b1 c1
a2 b2 c2
a3 b3 c3

as a1(b2c3– b3c2) – b1(a2c3– a3c2) + c1(a2b3– a3b2)

NOTE: 

Consider two square matrices A and B having order n and A=kB, then |A|=kn|B|, n N

PROPERTIES OF DETERMINANTS:

Let us glance through some properties of the Determinants, which are discussed in Class 12 Mathematics Chapter 4 Notes.

Property 1– the value of the determinant remains the same even after interchanging the rows and columns. Symbolic representation Ci Ri

Consider  A= 

4 2
5 3

(4 x 3) – (5 x 2) = 12 – 10 = 2

And  A= 

4 5
2 3

(4 x 3) – (2 x 5) = 12 – 10 = 2

This implies that:

If A is a square matrix, then det (A) = det (A’), where A’ is the transpose of matrix A.

Property 2– From the matrix, if the elements of any two columns (or rows) of the determinants are interchanged, then the sign of determinants will change. Symbolic representation Ci Cj or Ri Rj

Property 3– If the matrix has any two rows or columns equal or identical,  then we can say that the value of the determinant is equal to 0.

Property 4– If every element of a row or a column in a matrix is multiplied by k, then to obtain the value of the determinant, the original determinant is multiplied by k, where k is a constant. 

Property 5– If the elements of any row or a column of a determinant are expressed as the sum of two or more terms, then we can say that the original determinant can also be expressed as the sum of two or more determinants. 

Property 6– If for each element of a row or a column of the given determinant, the equimultiples of the analogous elements of other row or column respectively are added, then we can say that the value of the original determinant remains the same. 

The symbolic representation of the operation performed is Ri Ri+ kRj or Ci Ci+ kCj. 

Property 7– If every element in a row or column of the given determinant is zero, then the value of the determinant is zero. 

Property 8- Consider an upper triangular matrix or low triangular matrix, i.e., all elements on one side of the diagonal of the matrix are zeroes. The value of the determinant of such a matrix can be calculated by multiplying all diagonal elements. 

In simple words, the determinant of an upper (lower) triangular matrix is equal to the product of all the elements in the diagonal.

Visit the Extramarks platform to access the class 12 mathematics chapter 4 notes. Gain in-depth information and understand each property in detail. 

AREA OF A TRIANGLE:

Let the vertices of a triangle be (x1, y1),(x2, y2) and (x3, y3). Then the area of the triangle is defined as 

A=12[x1(y2−y3) + x3(y3−y1) + x3(y1−y3)].

Representation of area in the form of determinant = = 12 x determinant 

Determinant is given as

x1 y1 1
x2 y2 1
x3 y3 1

Remember:

  1. The area is a positive quantity. Therefore determinant will be positive 
  2. In case the area is already given, then use both the positive and negative values of the determinant.  
  3. The area of a triangle with three collinear points is equal to zero. 

The class 12 mathematics notes chapter 4 includes several problems based on the area of the triangle for students to practice and master this concept. 

Minors of the Determinant:  

Representation: Mij

Minor of any element is calculated by deleting the ith row and jth column in which the element lies. For a21 we delete the 2nd row and first column. 

Minor of any element of the given determinant of order n, where n ≥ 2 is a determinant of order n – 1.

Cofactor of the Determinant:  

Representation: Aij

Cofactor of an element aij is calculated by multiplying the minor of the element with (-1)i+j

Cofactor of element aij (Aij) = (-1)i+jMij

Adjoint of a Matrix:

The adjoint of a matrix is the matrix obtained by the transpose of a cofactor matrix. 

Singular Matrices 

A singular matrix is defined as the square matrix whose determinant is zero.

Non-Singular Matrices 

A Non-singular matrix is defined as the square matrix whose determinant is a non-zero value. 

THEOREMS:

Theorem 1 – Let A be any square matrix of order n, then A (adj A) = (adj A) A =|A| I, where I is the identity matrix of the same order.

Theorem 2 – Let A and B be the non-singular matrices of order n, then their product, i.e., AB and BA, will also be non-singular matrices of the same order n.

Theorem 3 – Let A and B be two square matrices of order n. The determinant of the product of matrix AB is given as the product of the respective determinants. It is written as AB=|A||B|. 

Theorem 4 – A square matrix is invertible (inverse exists)  iff the matrix is non-singular. So, for any matrix A, the inverse of the matrix is A-1= 1A(adj A).

Refer to the class 12 chapter 4 mathematics notes to access unlimited problems for extra practice. 

APPLICATIONS: 

  • Matrices and determinants are used to solve systems of linear equations with respect to two or three variables. They are also used to check the consistency of the given system of linear equations.
  • A Consistent system is said to be a system of equations whose solution can be found or exists.
  • An Inconsistent system is said to be a system of equations whose solution cannot be found or does not exist.
  • The value of the determinant is a number that is capable of determining the uniqueness of the solution of the given system of linear equations.

SOLUTION TO THE SYSTEM OF LINEAR EQUATIONS: 

Consider a system, 

a1x + b1y=c1 and a2x + b2y=c2, then the matric representation will be

a1 b1 x = c1
a2 b2 y c2

It is represented as AX = B ….. (1)

Case 1: A  is a non-singular matrix

In this case, we premultiply A-1to both sides of equation 1

Therefore, we get (A-1) AX = (A-1).B

Using associativity, (A-1A) X = (A-1).B

I X = (A-1).B

We get, X = (A-1).B

The value of X provides a unique solution. This method is commonly known as the matrix method.

Case 2: A  is a singular matrix

In this case, firstly, we calculate (adj A) B

Depending on the value of (adj A) B, we get our result.

(adj A)B is a non-zero matrix: Solution does not exist. The system of equations is said to be inconsistent with no solution.

(adj A)B is a zero matrix: Solution exists. The system of equations is said to be either consistent with many solutions or inconsistent.

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Chapter 4 Mathematics Class 12 Notes: Exercises & Answer Solutions

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Q.1 Find the area of a triangle whose vertices are given by (x1, y1), (x2, y2) and (x3, y3).

Ans

Area of Δ = 12 x1 y1 1x2 y2 1x3 y3 1

Q.2

Evaluate1     2    4 –1   3    0 4     1    0

Ans

1     2    4 –1   3     0 4     1    0 = 1 3    01    0 −2 −1    04      0+4−1    34      1  = 10−20+4−1−12 =   4−13 = − 52

Q.3

Show that a b ca+2x b+2y  c+2zx        y          z = 0  by using properties of determinants.

Ans

a b ca+2x b+2y  c+2zx        y          z = a b ca b cx   y     z+a b c2x 2y   2zx   y      zAll elements of R2 are expressed as sum of two terms. Therefore, the determinant can be expressed as sum of two determinant.      = 0 +2 a b ca b cx   y     z 2 is taken out common from R2.If any two rows or columns of a determinant are identical then the value of determinant is zero.     =  0 + 0 = 0

Q.4 If P, Q, R are three non-null square matrices of the same order, write the condition on P such that PQ = PR ⇒ Q = R.

Ans

P must be invertible or |P| ≠ 0.

Q.5

If A = 1 24 2, then show that 2A = 4A

Ans

   A = 1 24 2⇒ 2A = A = 2 48 4 ∴ A = 1 24 2 = 2−8 =−6   2A = 2 48 4 = 8−32 =−24⇒ 2A = -24 = 4×–6 = 4A

Q.6

Examine the consistency of the system of equations:5x–y+4z= 52x+3y+5z = 25x-2y+6z = -1

Ans

The system of equations can be written as AX = Bwhere A = 5 -1 42 3 55 -2 6 , X = xyz and B ​= 52−1Here, A = 518+10 +12−25 +4−4−15              = 528+−13+4−19              = 140−13−76              = 51≠0Since A ≠ 0, the system of equations is consistent.

Q.7

If A = 1 -1 20 2 -33 -2 4 and B –2 0 19 2 -36 1 -2​ then find AB.Using AB, solve the folowing system of equations: x−y+2z =1     2y−3z =13x−2y+4z = 2

Ans

Here, AB = 1 -1 20 2 -33 -2 4  –2 0 19 2 -36 1 -2              = −2−9 +1 2    0−2+2 1+3−4 0 +18−18  0+4−3 0−6+6−6−18+24 0−4+4     3 +6−8              = 1 0 00 1 00 0 1   Then AB = I   Thus A−1 = B   1 -1 20 2 -33 -2 4−1=  –2 0 19 2 -36 1 -2   Now given equations can be written as A1X = B1  where, A1 = 1 -1 20 2 -33 -2 4 X = xyz and B1​= 112Now  A1 = A     ∴ A1−1 = A−1 ⇒ A1−1 = –2 0 19 2 -36 1 -2  as X = A1−1 B1 ∴ X = –2 0 19 2 -36 1 -2112  X = –2 + 0 + 19 +2 -66 +1 -4053 Hence, x = 0, y = 5 and z = 3

Q.8 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it we get 11. By adding first and third numbers we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Ans

Let the first, second and third numbers be x, y and z.∴ x+y+z = 6  y+3z = 11   x + z = 2 y or x -2y+z = 0This system of equations can be written as AX = B, where A =  1 1 10 1 31 -2 1, X = xyz​  and B = 6110Here, A is non singular matrix and so its inverse exists.Now we find adjAc11 = 7        c12 = 3     c13 = −1c21 = −3      c22 = 0     c23 = 3c31 = 2        c32 = −3     c33 = 1Hence, adj A = A =  7 -3 23 0 -3–1 3 1Thus A−1= adjAA= 19 7 -3 23 0 -3–1 3 1Since  X = A−1BTherefore​ X = 19 7 -3 23 0 -3–1 3 16110           xyz = 19 42 -33 018 +0 -0–6 + 3 +0 = 19 9183= 123Thus     x =​1,  y = 2 and z = 3

Q.9

Solve the following system of equations by matrix method.  3x-2y+3z = 82x + y–z    = 14x-3y+2z =4

Ans

The system of equations can be written as AX = B, whereA =  3   -2    32    1   -14   -3   2 x = xyz and B = 814Here, A = 32−3+24+4+3−6−4              = −17≠ 0 Hence, A is nonsingular matrix and so its inverse exists.Now cofactors of A arec11 = −1      c12  = −8    c13 = −10c12 = −1      c22  = −8    c23 = −1c31 = −1      c32  = −8    c33 = −7adjoint of A = CT= −1    −5   −1−8    −6     9−10     1     7∴  A−1 = adjAA = −17=  −1    −5   −1−8    −6     9−10     1     7So X = A−1B−117​ =  −1    −5   −1−8    −6     9−10     1     7 814  xyz = −117−17−34−51 = 123Hence, x = 1, y = 2 and z = 3.

Q.10

If a, b, care in A.P then without expanding show thatx+2 x+3 x+2ax+3 x+4 x+2bx+4 x+5 x+2c = 0

Ans

Here, Δ = x+2 x+3 x+2ax+3 x+4 x+2bx+4 x+5 x+2c             = x+2 x+3 x+2ax+3 x+4 x+2b 1    1 2c−b by applying R3→R3−R2            = x+2 x+3 x+2a 1 1 2 b−a 1    1 2c−b by applying R2→R2−R1 Now because a, b, c are in A.P., we haveb−a = c −bHence in Δ R2= R3If any two rows or columns of a determinant are identicalthen the value of determinant is zero.⇒ Δ = 0

Q.11

Without expanding, prove thatb+c q+r   y+zc+a   r+p   z+xa+b   p+q  x+y = 2 a q   xb   q yc r z

Ans

Here, LHS = b+c q+r   y+zc+a   r+p   z+xa+b   p+q  x+y  By applying R1→R1−R2+R3,  we getLHS = −2a −2p  −2xc+a   r+p   z+xa+b   p+q  x+y         = 2 −a −p     −xc+a   r+p   z+xa+b   p+q  x+y Taking 2 common from R1        = 2 −a −p   −xc       r       zb      q       y By applying R2→R2+R1 and R3→R3+R1        = 2 a  p   xc   r    zb  q    y  Taking – 1 common from R1        = 2 a  p   xb   q    yc   r    z  By applying R2↔R3        = RHS

Q.12

Using properties of determinants, show that 1+a     1     11     1+b     11     1     1+c = abc 1+1a+1b+1c = abc +bc + ca + ab.

Ans

Let Δ = 1+a     1     11     1+b     11     1     1+cBy applying c1→c2 and c3→c3−c2, we geta       1          0−b     1+b   −b0       1           cExpanding along R1, we obtain Δ = a 1+b   −b1         c−1 −b   −b0       c+ 0    = a c+bc+b−−bc    = ac+abc+ab+bc     = abc+bc+ca+ab     = abc a+1a+1b+1c

Q.13

If x, y, z are different and Let Δ = x     x2     1+x3y     y2     1+y3z      z2     1+z3 = 0,then show that 1+xyz is a factor of Δ.

Ans

Let Δ = x     x2     1+x3y     y2     1+y3z      z2     1+z3 = 0,All elements of c3 are expressed as sum of two terms.Therefore, the determinant can be expressed as sum oftwo determinants.         = x     x2     1y     y2     1z      z2     1 + x     x2     x3y     y2     y3z      z2     z3Using c3↔ c2 and then c1↔ c2 in first determinant and taking out x, y and z common from R1, R2 and R3 respectivelyin second determinant, we get = −12 1     x     x21     y     y21     z     z2 +xyz 1     x     x21     y     y21      z     z2Δ = 1     x     x21     y     y21     z     z2 1+xyzHence, 1+xyz​ is factor of Δ.

Q.14

Using properties of determinants, prove thata     a+b        a+b+c 2a     3a+2b   4a+3b+2c3a      6a+3b  10a+6b+3c = a3

Ans

By applying R2→R2−2R1 and R3→R3−3R1Δ = a a+b   a+b+c 0     a       2a+b0   3a      7a+3bNow applying R3→R3−R2, we getΔ = a a+b   a+b+c 0      a       2a+b0      0          aExpanding along c1 we obtain Δ = a  a 2a+b0      a  + 0+0    = aa2−0 = aa2 = a3

Q.15

 If A = 2   53   6, verify that A adjA = A I.

Ans

We have  A = 4−6 = −2A I= -2 1   00   1       = −2   00   −2          .…..1Cofactors of Ac11 = 4      c12 = −2c21 = −3    c22 = 1⇒ adjA = cT            = 4   −2−3    1Now A adjA = 1   23   44   −2−3    1                      = 1×4+2×−3     1×−2+2×13×4+4×−3     3×−2+4×1                      = 4−6     −2+212−12  −6+4                      = −2   20   −2        .… 2 by 1 and 2A adjA = AI

Q.16 Find the area of the triangle whose vertices are (3, 8), (-4, 2) and (5, 1).

Ans

 The area of triangle is given by Δ = 12 3      8     1 −4    2    1 5      1     1    = 12 32−1−8−4−5+1−4−10    = 12 3+72−14    = 612 Sq. units

Q.17

Without expanding, evaluate Δ =  1    a    bc 2    b    ca5    c    ab.

Ans

Applying R2→ R2−R1 and R3→ R3 −R1,​ We get Δ =  1     a           bc 0    b−c    ca−b0    c−a     ba−cTaking factors b−a and c−a common from R2 and R3, respectively we get Δ = b−ac−a 1    a     bc 0    1     −c0    1     −b    = b−ac−a−b+c    = a−bb−cc−a

Q.18

If A = 1    1    −2 2    1    −35    4    −9,  find A.

Ans

Expanding with R1A = 1 1   −34   −9 −1 2   −35   −9 −22    15    4     = −9+12−1−18+15−28−5     = 3−−3−23     = 3+3−6     = 0

Q.19

Find values of x for which 3   xx   1 = 3   24   1.

Ans

 We have, 3   xx   1 = 3   24   1.⇒            3−x2 = 3−8⇒​                x2 = 8⇒                 x = ±22

Q.20

Find values of x for which 1   xx   1 = 4   14   1

Ans

 We have, 1   xx   1 = 4   14   1⇒            1−x2 = 4−4⇒​            1−x2 = 0⇒                 x2 = 1⇒                 x = ±1

Q.21

Evaluate:  x      x+1 x -1       x.

Ans

x      x+1 x -1       x = xx−x−1x−1 = x2−x2+1= 1

Q.22 The value of determinant will _______ if its rows and columns are interchanged.

Ans

remain unchanged

Q.23 The value of determinant will be _____, if its two rows or two columns are interchanged with each other.

Ans

negative

Q.24 If A is an invertivle square matrix, then write the matrix adj (AT) – (adj A)T.

Ans

If A is an invertible square matrix, then adj (AT) = (adj A)T

∴ adj (AT) – (adj A)T is a null matrix.

Q.25

Without expanding, prove that x         y         z  y + z   z + x   x + y 1          1          1 = 0

Ans

By applying R2→R2+R1, we getx           y         z  y + z   z + x   x + y 1          1          1 = x            y         z  x + y   x + y   x + y + z 1          1           1                             = x    y    z  1     1     11     1     1 Taking x+y+z common from R2.                             = x+y+z × 0                             = 0

Q.26 Let A be a 3 × 3 square matrix such that A(adj A) = 2I, where I is the identity matrix. Write the value of |adj A|.

Ans

∵ A (adj (A)) = |A|I

∴ A (adj A) = 2I ⇒ |A| = 2

Now, |adj A| = |A|n-1 ⇒ |adj A| = 23-1 = 4

Q.27 If A is a square matrix of order 3 such that adj (2A) = kadj (A), then write the value of k.

Ans

Since, adj (kA) = kn – 1adj(A), where n is the order
∴ adj (2A) = 23 – 1adj A
⇒ adj (2A) = 4 adj A
⇒ k = 4

Q.28

Without expanding evaluate the determinant41    1    579    7   929    5    3

Ans

Let Δ = 41    1    579    7   929    5    3.Then, applying c1→c2+−8 c3, we get Δ = 1    1    57    7   95    5    3 = 0 ∵ c1 and c2 are identical

Q.29

If A = 1   23   4, verify A adjA = A I.

Ans

We have AI = −2  1   00   1                    = −2    00    −2        .….. 1Cofactors of Ac11 = 4       c12 = −3c21 = −2    c22 = 1⇒ adj A = CT              = 4    −2−3    1 Now A adjA = 1     23    ​ 4 4   −2−3   1                      = 1×4+2×−3     1×−2+2×1 3×4+4×−3     3×−2+4×1                       = 4−6      −2+2 12−12   −6+4                      = −2     0 0     −2        .……2 by 1 and 2A adjA = AI

Q.30

Using properties of determinants, prove thatb2+c2     ab       ac   ba   c2+a2    bcca         cb    a2+b2 = 4a2b2c2.

Ans

Let  Δ b2+c2     ab       ac   ba   c2+a2    bcca         cb    a2+b2  Multiplying R1,R2  and R3 by a, b and c respectively, we getΔ = 1abc ab2+c2     a2b       a2c b2a    bc2+a2    b2cc2a         c2b    ca2+b2⇒ Δ = abcabc b2+c2     a2        a2b2    c2+a2      b2c2       c2    a2+b2 Taking a, b anc c common from c1, c2 and c3 respectively.⇒      Δ =  2b2+c2     2a2+c2   2 a2+b2      b2            c2+a2              b2      c2            c2               a2+b2 Applying R1, R2 and R3⇒     Δ = 2  b2+c2     a2+c2    a2+b2     b2        c2+a2              b2     c2        c2              a2+b2 Taking 2 common from R1⇒     Δ = 2  b2+c2     a2+c2    a2+b2   −c2            0              −a2   −b2         −a2               0 Applying R2→R2−R1 and R3→R3−R1⇒     Δ = 2  0       c2     b2−c2    0   −a2−b2   −a2   0 Applying R1→R1+R2+R3⇒  Δ = 2−c2  −c2   −a2−b2     0+b2 −c2      0−b2   −a2 ⇒  Δ = 2a2b2c2+a2b2c2 = 4a2b2c2

Q.31

If fx = ax2+bx+c isaquadratic function such that f1 = 8,f2 = 11 and f−3 = 6,  find fx by using determinants. Also, find f0.

Ans

We have fx = ax2+bx+c ∴          f1 = 8 ⇒ a+b+c =​ 8             f2 = 11 ⇒ 4a+2b+c =​ 11 and       f−3 = 6 ⇒ 9a−3b+c =​ 6 Thus, we obtain the following system of equations  a+b+c = 8 4a+2b+c = 11 9a  -3b+c = ​6 Here,  D = 1 1 14 2 19 -3 1 = 12+3−4−9+112−18     = 5+5−30 = −20  D1 = 8 1 111 2 16 -3 1 = 82+3−111−6+133−12     = 40−5−45 = −10  D2 = 1 8 14 11 19 6 1 = 111−6−84−9+124−99     = 5+40−75 = −30  D3 = 1 1 84 2 119 -3 6 = 112+33−124−99+8−12−18     = 45+75−240 = −120a = D1D = −10−20 = 12b = D2D = −30−20 = 32c = D3D = −120−20 = 6Hence, fx = 12 x2+32x+6⇒ f0 = 6.

Q.32

If m ∈ N and m ≥ 2, without expanding, prove that1 1 1mc1   m+1c1 m+2c1mc2   m+1c2 m+2c2   = 1

Ans

Let  Δ = 1 1 1mc1   m+1c1 m+2c1mc2   m+1c2 m+2c2 . Then        Δ = 1 1 1mc1   m+1c1 m+1c0+m+1c1mc2   m+1c2 m+1c1+m+1c2   ∵ nct+nct−1= n+1cr     Δ = 1 1 1mc1   m+1c1 m+1c0  mc2   m+1c2 m+1c1   Applying c3→ c3−c2⇒   Δ = 1 1 1mc1  m+1c1 +mc0+m+1c0mc2  mc1+ mc2       m+1c1   ∵ ncr+ncr−1= n+1cr⇒   Δ = 1 0 0mc1    m+1c0 m+1c0mc2    mc1     m+1c1   Applying c2→ c2−c1⇒   Δ = mc0    m+1c0 mc1    m+1c1   Expanding along R1⇒   Δ = 1      1 m  m+1 = m+1−m = 1

Q.33

Solve the following system of equations by cramer‘s rule: 2x+3y+10z = 4, 4x+6y+5z = 1, and 6x+9y+20z = 2.

Ans

Let 1x = μ, 1y = ν and1z = ω.  Then, the above system of equation can bewritten as     2μ+3ν+10ω = 4    4μ−6ν+5ω = 1    6μ+9ν−20ω = 2   Here, D = 2 3 104 -6 56 9 -20              = 2120−45−3−80−30+1036+36⇒    D = 150+330+720 = 1200       D1 = 4 3 101 -6 52 9 -20           = 4120−45−3−20−10+109+12 ⇒  D1 = 300+90+210 = 600       D2 = 2 4 104 1 56 2 -20            = 2−20−10−4−80−30+108+6⇒  D2 = −60+440+20 = 400       D3 = 2 3   44 -6 16 9 2            = 2−12−9−38−6+436+36⇒  D3 = −42−6+288= 240  ∴ μ = D1D = 6001200 = 12⇒ 1x = 12⇒ x = 2  ν = D2D= 4001200  = 13⇒ 1y = 13⇒ y= 3, ω  = D3D = 2401200 = 15⇒    1z = 15    Hence, x = 2, y = 3 and z = 5.

Q.34

A triangle has its theree sides equal to a, b and c, If the coordinatesof its vertices are A x1, y1, Bx2, y2 and C x3, y3, show that   x1 y1 2x2 y2 2x3 y3 22 = a+b+c)b+c−a a+b−c  by using determinants. 

Ans

Let Δ be the area of triangle ABC. Then,         Δ = 12x1 y1 1x2 y2 1x3 y3 1⇒        2Δ = x1 y1 1x2 y2 1x3 y3 1⇒       4Δ =2 x1 y1 1x2 y2 1x3 y3 1 Multiplying both sides by 2.⇒       4Δ = x1 y1 2x2 y2 2x3 y3 2 All the elements of c3 are multiplied by 2.⇒       16Δ2 = x1 y1 2x2 y2 2x3 y3 22 On squaring both sides.   .….i We also know that the area of triangle ABC is given by     Δ = ss−as−bs−c, where s = 12a+b+c But, s  = 12 a+b+c⇒ s−a  = 12 a+b+c−c              = 12 b+c−a Similarly, s –c = 12a+b−c.∴ Δ2 = 12 a+b+c.12 b+c−a.12c+a−b.12a+b−c.⇒ 16Δ2 = a+b+cb+c−ac+a−ba+b−c     .…. ii   From i andii we get         x1 y1 2x2 y2 2x3 y3 2  = a+b+cb+c−ac+a−ba+b−c

Q.35

Using cramer‘s rule, find values of a and b for whichthe system of equations2x+ay+6z = 8  a+2y+bz = 5  x+ y+ 3z  = 4has i  a unique solutionii infinitely many soutionsiii no solutions.

Ans

We have,  D = 2 6 61 2 b1 1 3⇒  D = 26−b−a3−b+61−2⇒  D = 12- 2b – 3b + ab – 6⇒  D = 6 – 3a – 2b + ab = b−3a−2  D = 8 a 65 2 b4 1 3   D1 = 86−b−a15−4b+65−8⇒D1 = 48−8b−15a+4ab−18⇒D1 = 30−15a−8b+4ab= a−24b−15   D2   2 8 61 5 b1 4 3     D2 = 215−4b−83−b+64−5⇒D2 =30−8b−24+8b−6= 0     and   D3   2 a 81 2 51 1 4     D3 =28−5−a4−50+81−2⇒ D3 = 6+a−8=a−2i For unique solution, we must have  D ≠ 0 ⇒ a−2 b−3 ≠ 0          ⇒ a≠2, or b ≠ 3.ii for infinitely many solutions, we must have     D = D1 = D2 = D3 = 0⇒ a−2 b−3 = 0, a−24b−15 = 0 and a −2 =0.⇒   a= 2 Putting a = 2 in hte given system of equations, we get  2x + 2y + 6z =8     x + 2y + 6z =5    x + y + 3z =4    x + y + z =4    x + 2y + bz =5 Putting z = k, we get   x + y = 4 –k x + 2y = 5 –bk Solving these two equations, we get  x = 3 − 3k + bk, y = 1−bk+k  Thus, the given system has infinitely many solutions given by  x = 3− 2k+bk, y = 1−bk+k,z= k, where k ∈ R. Hence, the system has infinitely many solutions for a = 2. iii  For no solution, we must have D = 0 and at least one of D1, D2 and is non zero. Clearly, for b = 3, we have  D = 0 and D3 ≠ 0.  Hence, the system has no solution for b = 3.

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