Mathematics Algebra of Matrices For CBSE-NCERT
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### Topic Covered

star Addition/Subtraction of matrices
star Multiplication of a matrix by a scalar
star Multiplication of matrices

### Operations on Matrices

color{green} ✍️ we shall study certain operations on matrices, namely, addition of matrices, multiplication of a matrix by a scalar, difference and multiplication of matrices.

color{green} ✍️  The sum of two matrices is a matrix obtained by adding the corresponding elements of the given matrices. Furthermore, the two matrices have to be of the same order.

color{green}{A = [a_(ij)]_(mxxn) \ \ \ \ B = [b_(ij)]_(mxxn)}
color{green}{A+B = [a_(ij) + b_(ij)]_(mxxn)}

color{green} ✍️  Thus, if A = [(a_(11),a_(12),a_(13)),(a_(21),a_(2,2),a_(2,3))] and B = [(b_(11),b_(12),b_(13)),(b_(21),b_(2,2),b_(2,3))] A+B =[(a_(11) + b_(11) ,a_(12)+b_12,a_(13)+b_13),(a_(21)+b_21,a_(22)+b_22,a_(23)+b_23)]

color{red}{"Key Concept :"} Subtraction(Difference) peform same way as addition in matrix.

color{green}{A = [a_(ij)]_(mxxn) ;\ \ \ \ B = [b_(ij)]_(mxxn)}

color{green}{A-B = [a_(ij) - b_(ij)]_(mxxn)}

Point to remember : The addition And subtraction is not defined for different order matrices.
Q 3184578457

Given A = [ ( sqrt (3) ,1, -1 ), ( 2,3, 0) ]  and B = [ ( 2 , sqrt (5) ,1 ), ( -2,3, 1/2) ] , find A+B

Class 12 Chapter 3 Example 6
Solution:

Since A, B are of the same order 2 × 3. Therefore, addition of A and B is defined
and is given by

A+B = [ ( 2+sqrt (3) , 1+ sqrt (5) , 1-1 ), ( 2-2 , 3+3 , 0+ 1/2 ) ] = [ ( 2+ sqrt (3) , 1+ sqrt (5) , 0 ), ( 0, 6 , 1/2) ]
Q 3164678555

Find X and Y , if X+ Y = [ ( 5,2),( 0,9 ) ] and X - Y = [ (3,6), ( 0,-1) ]

Class 12 Chapter 3 Example 9
Solution:

We have  (X+ Y) + (X-Y) = [( 5,2), ( 0, 9 ) ] + [ ( 3,6), ( 0,-1) ]

or  (X+X ) + (Y-Y) = [ (8,8 ), (0,8) ] => 2X = [ (8,8), (0,8) ]

or X= 1/2 [ (8,8), ( 0,8) ] = [ (4,4),(0,4) ]

Also  (X+ Y ) - (X- Y ) = [ (5,2), (0,9) ] - [ (3,6),( 0,-1) ]

or  (X-X) + (Y+Y) = [ (5-3 , 2-6 ) , (0, 9+1) ] = 2Y = [ ( 2,-4), (0,10) ]

OR y=1/2 [ (2,-4), ( 0,10) ] = [ (1,-2), (0,5) ]

(i) color{red}{"Commutative Law :"} If A = [a_(ij)], B = [b_(ij)],C = [c_(ij)] are matrices of the same order, say m × n,

Then color{orange}{A + B = B + A}

Now A + B = [a_(ij)] + [b_(ij)] = [a_(ij) + b_(ij)] = [b_(ij) + a_(ij)]
(addition of numbers is commutative) = ([b_(ij)] + [a_(ij)]) = B + A

(ii) color{red}{"Associative Law :"}

 color{orange}{A + B) + C = A + (B + C}

Now (A + B) + C = ([a_(ij)] + [b_(ij)]) + [c_(ij)]
= [a_(ij) + b_(ij)] + [c_(ij)]
= [a_(ij) + (b_(ij) + c_(ij))]
= [a_(ij)] + ([b_(ij)] + [c_(ij)]) = A + (B + C)

(iii) color{red}{"Existence of additive identity :"}
Let A = [a_(ij)] be an m xx n matrix and O be an m × n zero matrix,
Then color{orange}{A + O = O + A = A.}

(iv) color{red}{"The existence of additive inverse :"}
A = [a_(ij)]_(m xx n) be any matrix, then we have another matrix as – A = [– a_(ij)]_(m xx n) such

That color{orange}{A + (– A) = (– A) + A= O.} So – A is the additive inverse of A or negative of A

### Multiplication of a matrix by a scalar

color{green} ✍️if A = [a_(ij)]_(mxxn) is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by the scalar k.

color{green} ✍️ In other words, kA = k[a_(ij)]_(mxxn) = [k(a_(ij))_(mxxn))]  that is, (i, j)^(th) element of kA is ka_(ij) for all possible values of i and j.

color{green}{"Negative of a matrix : The negative of a matrix is denoted by" – A."} We define –A = (– 1) A.
Q 3114678550

If A = [ (1,2,3), ( 2,3,1) ]  and B= [ ( 3,-1,3 ), ( -1,0,2 ) ]  , then find 2A – B .
Class 12 Chapter 3 Example 7
Solution:

We have

2A- B = 2 [( 1,2,3), ( 2,3,1) ] - [ ( 3,-1,3),( -1,0,2) ]

= [ ( 2,4,6), ( 4,6,2) ] + [ (-3,1,-3), ( 1,0,-2) ]

= [ (2-3 ,4+1, 6-3), ( 4+1 , 6+0 , 2-2 ) ] = [ ( -1,5,3 ), ( 5,6,0) ]
Q 3124778651

Find the values of x and y from the following equation:

2 [ ( x,5), ( 7,y-3) ] + [ (3,-4), ( 1,2 ) ] = [ ( 7,6), ( 15,14) ]
Class 12 Chapter 3 Example 10
Solution:

We have

2 [ (x,5), ( 7, y-3) ] + [ ( 3,-4),( 1,2) ] = [ (7,6), ( 15,14) ] => [ (2x ,10 ), ( 14, 2y-6) ] + [ ( 3,-4), ( 1,2) ] = [ (7,6), ( 15,14) ]

or  [ (2x+3, 10-4), ( 14+1 , 2y -6 +2 ) ] = [ (7,6), ( 15,14) ] => [ (2x+3, 6), ( 15, 2y -4) ] = [ (7,6), ( 15,14) ]

or 2x +3 = 7 and 2y -4 =14 (Why?)

or 2x = 7 – 3 and 2y = 18

or x= 4/2 and y= 18/2

i.e., x= 2 and  y = 9

### Properties of scalar multiplication of a matrix

color{green} ✍️ If A = [a_(ij)] and B = [b_(ij)] be two matrices of the same order, say m × n, and k and I are scalars, then

(i)color{orange}{ k(A +B) = k A + kB},

(ii) color{orange}{(k + I)A = k A + I A = kA +A}
Q 3144678553

If A = [ (8,0), ( 4,-2), ( 3,6) ]  and B= [ ( 2,-2),(4,2),(-5,1) ] , then find the matrix X, such that

2A + 3X = 5B.
Class 12 Chapter 3 Example 8
Solution:

We have 2A + 3X = 5B
or 2A + 3X – 2A = 5B – 2A
or 2A – 2A + 3X = 5B – 2A (Matrix addition is commutative)
or O + 3X = 5B – 2A (– 2A is the additive inverse of 2A)
or 3X = 5B – 2A (O is the additive identity)

or X = 1/3 (5B-2A)

or  X = 1/3 (5 [ (2,-2), (4,2), (-5,1) ] -2 [ (8,0), ( 4,-2), (3,6) ] ) = 1/3 ( [ (10 , -10), (20,10), (-25,5) ] + [ (-16,0), (-8,4), (-6,-12) ] )

= 1/3 [ (10-16 , -10+0 ), ( 20-8 , 10+4), ( -25-6, 5-12) ] = 1/3 [ (-6,-10),( 12,14),( -31,-7) ] = [ ( -2, (-10)/3), ( 4, 14/3) , ( (-31)/3 , (-7)/3) ]
Q 3164778655

Two farmers Ramkishan and Gurcharan Singh cultivates only three
varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these
varieties of rice by both the farmers in the month of September and October are given
by the following matrices A and B.

(i) Find the combined sales in September and October for each farmer in each
variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October.
Class 12 Chapter 3 Example 11
Solution:

(i) Combined sales in September and October for each farmer in each variety is
given by

(ii) Change in sales from September to October is given by

(iii) 2% of B = 2/100 xx B = 0.02 xx B

Thus, in October Ramkishan receives Rs 100, Rs 200 and Rs 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of Rs
400, Rs 200 and Rs 200 in the sale of each variety of rice, respectively.

### Multiplication of matrices

color{green} ✍️ The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B.

color{green} ✍️ Let A = [a_(ij)] be an m xx n matrix and B = [b_(jk)] be an n × p matrix. Then the product of the matrices A and B is the matrix C of order color{green}{m × p.}

color{green} ✍️ To get product, we take the i^(th) row of A and k^(th) column of B, multiply them element wise and take the sum of all these products.

color{green} ✍️ In other words, if A = [a_(ij)]_(m xx n), B = [b_(jk)]_(n xx p), then the i^(th) row of A is [a_(i1) a_(i2) a_(i3).....a_(i n)] and the k^(t h) column of B is

color{green} ✍️ The matrix C = [c_(ik)]_(m × p) is the product of A and B.
Q 3144878753

Find AB, if A = [ ( 6,9), ( 2,3) ]  and B = [ (2,6,0 ), ( 7,9,8) ]
Class 12 Chapter 3 Example 12
Solution:

The matrix A has 2 columns which is equal to the number of rows of B.
Hence AB is defined. Now

AB = [ ( 6(2) +9 (7 ) , 6 (6) + 9 (9) , 6 (0) + 9 ( 8) ), ( 2(2) +3 (7) , 2 ( 6) + 3 (9) ,2 (0) +3(8 ) ) ]

= [ ( 12+63 , 36 +81, 0+72 ), ( 4+21 ,12+27 , 0+24) ] = [ ( 75, 117 ,72), ( 25 ,39 ,24) ]
Q 3114878759

If A= [ ( 1,-2,3), (-4,2,5) ] and B = [ (2,3), ( 4,5), (2,1) ] , then find AB, BA. Show that

AB ≠ BA.
Class 12 Chapter 3 Example 13
Solution:

Since A is a 2 × 3 matrix and B is 3 × 2 matrix. Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively. Note that

AB = [ (1,-2,3), (-4,2,5) ] [ (2,3),( 4,5),( 2,1) ] = [ (2-8+6, 3-10+3), (-8+8 +10 , -12+10 + 5 ) ] = [ (0,-4), (10,3) ]

and BA = [ (2,3), (4,5), (2,1) ] [ (1,-2,3), (-4,2,5) ] = [ ( 2-12, -4+6, 6+15 ), ( 4-20 , -8 +10 ,12+25), ( 2-4, -4+2,6+5) ] = [ ( -10,2,21), (-16,2,37 ), ( -2, -2, 11 ) ]

Clearly AB ≠ BA
In the above example both AB and BA are of different order and so AB ≠ BA. But
one may think that perhaps AB and BA could be the same if they were of the same
order. But it is not so, here we give an example to show that even if AB and BA are of
same order they may not be same.
Q 3144078853

If A = [ (1,0 ), ( 0,-1) ]  and B = [ ( 0,1), (1,0 ) ] , then AB = [ (0,1), ( -1, 0 ) ]

and BA = [ ( 0 ,-1), (1,0) ] Clearly AB ≠ BA.

Thus matrix multiplication is not commutative.
Class 12 Chapter 3 Example 14
Solution:

Zero matrix as the product of two non zero matrices
We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. This need
not be true for matrices, we will observe this through an example.
Q 3164078855

Find AB, if A = [ (0,-1), (0,2) ] and B = [ ( 3,5), ( 0, 0 ) ]
Class 12 Chapter 3 Example 15
Solution:

We have AB = [ (0,-1), (0,2) ] [ (3,5), (0,0) ] = [ (0,0), (0,0 ) ]

Thus, if the product of two matrices is a zero matrix, it is not necessary that one of
the matrices is a zero matrix.

### Properties of multiplication of matrices

1. The associative law For any three matrices A, B and C. We have color{orange}{(AB) C = A (BC),}
whenever both sides of the equality are defined.

2. The distributive law For three matrices A, B and C.

(i)color{orange}{ A (B+C) = AB + AC}
(ii)color{orange}{ (A+B) C = AC + BC,} whenever both sides of equality are defined.

3. The existence of multiplicative identity For every square matrix A, there exist an identity matrix of same order such that color{orange}{IA = AI = A} Now, we shall verify these properties by examples.
Q 3134178952

If A = [ ( 1,1,-1), (2,0,3), ( 3,-1,2) ] , B = [ (1,3), (0,2), (-1,4) ] and C= [ ( 1,2,3,-4), ( 2,0,-2,1) ] , find

A(BC), (AB)C and show that (AB)C = A(BC).
Class 12 Chapter 3 Example 16
Solution:

We have AB = [ (1,1,-1), (2,0,3 ), ( 3,-1,2) ] [ (1,3), ( 0,2), (-1,4) ] = [ ( 1+0 +1 , 3+2-4), ( 2+0 -3, 6+0+12), ( 3+0-2, 9-2+8) ] = [ (2,1), (-1,18), (1,15) ]

(AB) (C) = = [ (2,1), (-1,18), (1,15) ] [ (1,2,3,-4), ( 2,0,-2,1),] = [ (2+2, 4+0 , 6-2 , -8 +1 ), ( -1+36 , -2+0 , -3-36 , 4 +18 ), ( 1+30, 2+0 , 3-30 , -4+15) ]

= [ ( 4,4,4,-7), ( 35, -2,-39 ,22), (31,2,-27 , 11) ]

Now BC = [ (1,3), (0,2), (-1,4) ] [ (1,2,3,-4), ( 2,0,-2,1) ] = [ (1+6 , 2+0 , 3-6 , -4+3), ( 0+4, 0 +0 , 0-4, 0+2), ( -1+8 , -2+0 , -3-8, 4+4) ]

= [ ( 7,2,-3,-1), (4,0,-4,2) , ( 7,-2,-11,8) ]

Therefore A(BC) = [ (1,1,-1), (2,0,3), ( 3,-1,2) ] [ (7,2,-3,-1), ( 4,0,-4,2), ( 7,-2,-11, 8) ]

= [ ( 7+4-7 , 2+0 +2, -3-4+11, -1+2-8 ), ( 14 + 0 + 21, 4 + 0 -6 , -6+0 -33 , -2 + 0 + 24), ( 21 -4 +14 , 6+0 -4 , -9 +4-22 , -3-2+16) ]

= [ ( 4,4,4,-7), ( 35,-2,-39,22), (31,2,-27 ,11) ] Clearly, (AB) C = A (BC)
Q 3124180051

If A = [ (0,6,7 ), (-6,0 ,8 ), (7, -8 , 0 ) ] , B = [ (0,1,1 ), (1,0,2), (1,2, 0 ) ] ,C = [ (2), (-2), (3) ]

Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC
Class 12 Chapter 3 Example 17
Solution:

Now , A+ B = [ (0,7,8 ), (-5,0,10 ), ( 8,-6,0) ]

so,  (A+B ) C = [ ( 0,7,8) , (-5,0,10 ), ( 7, -8 , 0 ) ] [ (2), (-2), (3) ] = [ ( 0 -12 +21),(-12+0+24) ,(14+16 +0) ] = [ ( 9), (12), (30) ]

and BC = [ (0,1,1), ( 1,0,2), ( 1,2, 0) ] [ (2,), (-2), (3) ] = [ (0-2+3), ( 2+0+6), ( 2-4 + 0) ] = [ (1), (8), (-2) ]

So AC + BC = [ (9), (12), ( 30) ] + [ (1), (8), (-2) ] = [ (10), (20), (28) ]

Clearly, (A + B) C = AC + BC
Q 3154380254

If A= [ (1,2,3), (3,-2,1), (4,2,1) ] , then show that A^3 – 23A – 40 I = O
Class 12 Chapter 3 Example 18
Solution:

We have A^2 = A * A= [ (1,2,3), (3,-2,1), (4,2,1) ] [ (1,2,3), (3,-2,1), (4,2,1) ] = [ (19,4,8), (1,12,8), ( 14,6,15 ) ]

So, A^3 = A A^2 = [ (1,2,3), (3,-2,1), (4,2,1) ] [ (19,4,8 ), ( 1,12,8 ), ( 14,6, 15) ] = [ (63,46,69 ), ( 69,-6, 23), ( 92,46,63) ]

Now,

A^3 -23 A -40 I = [ (63,46 ,69), ( 69 , -6,23), (92,46,63) ] -23 [ (1,2,3), ( 3,-2,1), (4,2,1) ] -40 [ (1,0,0 ), (0,1,0 ), ( 0 ,0 ,1) ]

= [ (63,46 ,69 ), (69,-6 ,23 ), ( 92,46,63) ] + [ (-23 , -46, -69), ( -69 ,46, -23), ( -92,-46, -23) ] + [ (-40,0,0 ), ( 0, -40 , 0 ), ( 0,0, -40 ) ]

= [ ( 63-23-40 ,46 -46 +0 , 69-69 +0 ), (69 -69 +0 , -6 +46-40 , 23-23 +0 ), ( 92-92+0 , 46 -46 +0 , 63 -23 -40) ]

= [ (0,0,0), (0,0,0), (0,0,0) ] =0
Q 3114480350

In a legislative assembly election, a political group hired a public relations
firm to promote its candidate in three ways: telephone, house calls, and letters. The
cost per contact (in paise) is given in matrix A as

The number of contacts of each type made in two cities X and Y is given by
Class 12 Chapter 3 Example 19
Solution:

We have

So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i.e., Rs 3400 and Rs 7200, respectively.