Sponsor Area
Consider the following information regarding the number of men and women workers in three factories I. II and III
|
Men Workers |
Women Workers |
|
|
I |
30 |
25 |
|
II |
25 |
31 |
|
III |
27 |
26 |
Represent the above information in the form of a 3 x 2 matrix. What does the entry in the third row and second column represent ?
The given information is
|
Factory |
Men Workers |
Women Worker: |
|
I |
, 30 |
25 |
|
II |
' 25 |
31 |
|
III |
27 |
26 |
The information is represented in the form of a 3 X 2 matrix as follows :
The entry in the third row and second column represents the number of women workers in factory III.
In the matrix 
(i) The order of the matrix
(ii) The number of elements,
(iii) Write the elements a13,a21 a33 a24’ a23
( i) Now A has 3 rows and 4 columns and so A is of order 3 x 4.
(ii) Number of elements = 3 x 4 = 12
(iii) a13, =- 19., a21 =35, a33 = 5, J24 = 12, a23 
If a matrix has 8 elements, what arc the possible orders it can have ?
We know' that if a matrix is of order m x n, it has mn elements. Therefore, for finding all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8.
∴ all possible ordered pairs are (1.8), (8, I), (4, 2). (2, 4)
∴ possible orders are I x 8, 8 x 1, 2 x 4, 4 x 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
We know that a matrix of order m x n has mn elements. Therefore, for finding all possible orders of a matrix with 24 elements, w e will find all ordered pairs with products of elements as 24.
∴ all possible ordered pairs are
(1, 24), (24. 1), (2, 12), (12. 2). (3, 8), (8, 3), (4. 6), (6, 4)
∴ possible orders are
1 x 24. 24 x 1, 2 x 12, 12 x 2, 3 x 8, 8 x 3, 4 x 6. 6 x 4
If number of elements =13, then possible orders are 1 x 13, 13 x l.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
We know that a matrix of order m x n has mnelements. Therefore, for finding all possible orders of a matrix with 18 elements, we will find all ordered pairs with products of elements as 18
∴ all possible ordered pairs areall possible ordered pairs areall possible ordered pairs areall possible ordered pairs areall possible ordered pairs areall possible ordered pairs are (I, 18). (18, 1). (2, 9). (9. 2). (3, 6). (0, 3)
∴ possible orders are 1 x 18. 18 x 1. 2 x 9,-9 X 2, 3 x 6. 6 x 3.
If number of elements = 5, then possible orders arc 1 x 5. 5 x I.
Number of elements = 12
∴possible orders of the matrix are
1 x 12, 12 x 1,2 X 6. 6 x 2,3 x 4,4 x 3
If numbers of elements = 7, then possible orders are 1 x 7. 7 x 1
Let A = [a i] be required 3 x3 matrix
where a,ij , = 2 i – 3 j
∴ a11, = 2–3 = –1. a12 = 2–6 = 4. a13,= 2–9 = 7
a21 = 4 — 3 = 1. a22 = 4 — 6 = —2 . a23 = 4 –9 = 5
a31 = 6 – 3 = 3, = 6 – 6 = 0 , a33 = 6 – 9 = –3
Let A = [ aii] be required 3 x 4 matrix where aij= i + j
a11 = 1 + 1 = 2, a12 =1+2, a13=1 + 3 = 4, a14 = 1 + 4 = 5
a21 = 1 + 1 = 2, a22 =1+2, a23=1 + 3 = 4, a24 = 1 + 4 = 6
a31 = 1 + 1 = 2, a32 =1+2, a33=1 + 3 = 4, a34 = 1 + 4 = 7
Let A = [ aii] be required 3 x 4 matrix where aij= i - j
a11 = 1 - 1 = 0, a12 =1 - 2= - 1, a13=1 - 3 = - 2, a14 = 1 - 4 = -3
a21 = 2 - 1 = 1, a22 =2 - 2= 0, a23=2 - 3 = -1, a24 = 2 - 4 = - 2
a31 = 3 - 1 = 2, a32 =3 - 2, a33=3 - 3 = 0, a34 = 3 - 4 = - 1 
Let A = [ aii] be required 3 x 4 matrix where aij= i.j
a11 = 1 - 1 = 0, a12 =1 - 2= - 1, a13=1 - 3 = - 2, a14 = 1 - 4 = -3
a21 = 2 - 1 = 1, a22 =2 - 2= 0, a23=2 - 3 = -1, a24 = 2 - 4 = - 2
a31 = 3 - 1 = 2, a32 =3 - 2, a33=3 - 3 = 0, a34 = 3 - 4 = - 1 
Let A = [ aii] be required 3 x 4 matrix where aij= i.j
a11 = 1 . 1 = 1, a12 =1 . 2 = 2, a13 = 1 . 3 = 3, a14 = 1 . 4 = 4
a21 = 2 . 1 = 2, a22 =2 . 2 = 4, a23 = 2 . 3 = 6, a24 = 2 . 4 = 8
a31 = 3 . 1 = 3, a32 =3 . 2 = 6, a33 = 3 . 3 = 9, a34 = 3 . 4 = 12 
Let A = [aij] be required 2 x 3 matrix where aij = i + 2 j
a11 = 1 + 2 = 3, a12 = 1 + 4 = 5, a13 = 1 + 6 = 7
a21 = 2 + 2 = 4, a22 = 2 + 4 = 6, a23 = 2 + 6 = 8
Here aij = 2 i - j
a11 = 2 (1) - 2 = 2 - 1 = 1, a12 = 2 (1) - 2 = 2 - 2 = 0
a13 = 2 (1) - 3 = 2 - 3 = - 1, a14 = 2 (1) - 4 = 2 - 4 = - 2
a21 = 2 (2) - 1 = 4 - 1 = 3, a22 = 2 (2) - 2 = 4 - 2 = 2
a23 = 2 (2) - 3 = 4 - 3 = 1, a24 = 2 (2) - 4 = 4 - 4 = 0 
A = [ai j] is 2 x 2 matrix where , aij = 
A = [aij] is 2 x 2 matrix where aij = 
A = [ aij ] is 2 x 2 matrix where 
Sponsor Area
Let A = [ai j ] be required 3x4 matrix where aij-= 2 i – j
a11 = 2 - 1 = 1, a12 = 2 - 2 = 0, a13 = 2 - 3 = - 1,
a14 = 2 - 4 = - 2
a21 = 4 - 1 = 3, a22 = 4 - 2 = 2, a23 = 4 - 3 = 1,
a24 = 4 - 4 = 0,
a31 = 6 - 1 = 5, a32 = 6 - 2 = 4, a33 = 6 - 3 = 3,
a34 = 6 - 4 = 2
find the values of .x and y.
We are given that 
From the definition of equality of matrices, we have
x + 2y= 4 ...(1)
– y = 3 ...(2)
3 x = 6 ...(3)
From (2). y = –3
From (3), x = 2
Also (1) is satisfied for x = 2, y = – 3 we have x = 2, y = – 3
find the values of x and y.
We are given that 
From the definition of equality of matrices, we have
x+ 2y = 0 -(I)
3.y= – 3 ...(2)
4 x =8 ...(3)
From (2), y = – 1
From (3). x = 2
Also (1) is satisfied for ,x = 2. y = – 1 we have x = 2. y= – 1
, find x. y. z. w
From the definition of equality of matrices, we have,
x–y = – 1 ....(1)
2 x – y = 0 ...(2)
2.x+z = 5 ...(3)
3z + w = 13 . (4)
Subtracting (1) from (2), ,x = 1
∴ from (I), 1 – y = - 1 ;⇒ y = 2
From (3). 2 +z = 5;⇒ z = 3
From (4), 9 +' w=13;⇒ w = 4
∴ x= 1. y = 2, z= 3. w = 4
We are given that
By definition of equality of matrices
4 = y.3 =z,.x = 1
,∴ x = I. y = 4, z = 3
We are given that
By definition of equality of matrices,
x+.y= 6 ....(1)
5 + z =5 ....(2)
x y =8 ...(3)
From (2). z= 0
From (I). y= 6 – x ...(4)
Putting y = 6 –x in (3). we get.
x(6 – x) = 8 or 6 x – x2 –8 = 0
∴x2–6x + 8 = 0 ⇒ (.x –2) (x–4) = 0;⇒ .x=2.4
;∴ from (4). y = 6 – 2. 6 – 4 = 4, 2
∴we have
x = 2, ,y = 4, z = 0 ; .x = 4, y = 2, z = 0
We are given that
By definition of equality of matrices.
x.+ y + z = 9 ,..(1)
.x + z = 5 ...(2)
y + z = 7 ...(3)
Subtracting (2) from (1), y = 4
Subtracting (3) from (1). x = 2
∴from (2), 2 + z = 5 ;⇒ z =3
∴x = 2, .y = 4, z = 3
We are given that
By definition of equality of mgtijices,
a – h = - 1 ...(1)
2a – h = 0 ...(2)
2a + c‘ - 5 , ...(3)
3 c + d= 13 ...(4)
Subtracting (1) from (2), a = 1
∴ from (1). 1– b = – 1 ;⇒ b= 2
From (3), 2 + c = 5 ;⇒ c = 3
From (4). 9 + d = 13 ;⇒ d = 4
∴a = 1, b= 2, c = 3.d = 4
If 
find x, y, z w.
Wea are given that 
From the definition of equality of matrices, we have,
x = 3 ...(1)
3 .x – y = 2 .... (2)
2 .x + z =4 ...(3)
3y – W'=7 ...(4)
From (1) and (2), 9- y = 2 ;⇒ y= 7
From (1) and (3), b + z=4 ;⇒ z = –2
From (4), 21 –w = 7 ⇒ w= = 14
∴ x = 3. y = 7,z = -2, w= 14
We are given that
From the definition of equality of matrices, we have,
2 x – 3 v = 1 ...(I)
x + 4 y = 6 ...(2)
a – b= – 2 ...(3)
3a + 4 b = 29 ...(4)
(1) and (2) can be writta/i«s
2 x – 3 y – I = 0
x + 4 y – 6 = 0
(3) and (4) can be written as
a–b+2-0
3 a + 4 b – 29 = 0
If 
find the values of a and b.
Here
⇒a + b = 6 ... (1)
and a b = 8 (2)
From (1). b = 6 – a ...(3)
∴from (2), a (6 –a) = 8 ;⇒ – a2 + (6 a – 8 = 0
⇒ a – 6 a+ 8 = 0;⇒ (a – 2) (a – 4) = 0
⇒ a = 2, 4
∴ from (3). b = 6 – 2. 6–4 = 4. 2
∴we have a-=2. b = 4 : a = 4. b = 2
We are given that 
From the definition of equality of matrices, we have,
x + 10 = 3x + 4, y2+2y=3, -4= y2 –5y
Now x + 10 = 3 x + 4 ⇒2x= 6 ⇒ ,x = 3
y2 + 2 y – =3 ;⇒ y2+2y – 3 =0 ⇒ (y + 3) ( y –1) = 0 ⇒ y = 1. –3
– 4 = y 2 – 5 y ⇒ y2 – 5 y + 4 = 0 ⇒ ( y – 1) ( y – 4) = 0 ;⇒ y – 1. 4
Since both the equations y2 + 2 y =3 and - 4 ='y2 -5 y are to be satisfied, so we have y'= 1
∴ x = 3. y = I.
Here
Now, A = B
∴ by the definition of equality of matrices.
2 x + 1 = .x + 3, 3 y = + 2 , y2 –5 y = – 6
Now 2 x + I = x + 3 ⇒ x= 2
3 x = y2 +2 ⇒ y2 -3 y + 2 = 0 ⇒ ( y –1) (y – 2) = 0 ⇒ y = I, 2
y2 –5 y =-6 ;⇒ -5 y + 6 = 0 ;⇒ (y–2)(y–3) = 0 ;⇒ y = 2.3
Since both the equations 3 y = y2 +2 and y2 –5 y = – 6 arc to be satisfied
simultaneously, so we have y = 2.
∴A – B when x = 2, y – 2.
If 
find .x. y. z .w.
We are given that
From the definition of equality of matrices, we have
x –y = –1 ..(I)
2.x–y=0 ...(2)
z = 4 ...(3)
w = 5 ..(4)
Subtracting (1) from (2). we get,
x= 1
∴from (1). I -y = –l ⇒ y = 2
∴.x – 1. y = 2. z =-4. w- = 5
Not possible to find
B.
Not possible to find
We are given that 
By definition of equality of matrices.
3.x + 7 = 0. 5-= y – 2. y + I = 8, 2 – 3 .r = 4
∴ (B) is correct answer.
C.
m = nFor A = [a I J ]m x n to be square matrix.
number of row s = number of columns
∴m = n
∴ (C) is correct.
D.
512A matrix of order 3 x 3 has 9 elements. Now each element can be 0 or 1.
∴ 9 places can be filled up in 29 ways
required number of matrices = 29
=2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
=512 ,
∴ (D) is correct answer.
Sponsor Area
If 
compute (A + B) and (B – C). Also verify that A+ (B – C) = (A + B) – C.
then find a matrix X such that 3A - 2B + 4X=O, where 
We have
Now 3A - 2B + 4X = O 
4X = -3A + 2B
.
....(1)
....(2)
By definition of equality of matrices, we have
2.x = 4, 2 y –6 = 12
∴-x = 2, 2y = 18
∴.x = 2, y = 9.
The given matrix equation is :
By definition of equality of matrices,
2 + y= 5, 2 .x + 2 = 8
∴ y = 5-2, 2x = 8 –'2
∴y = 3, .x = 3
∴.x =- 3, y = 3.
The given matrix equation is
By definition of equality of matrices, we get,
2x = 6, 2y= 12, 2z = 18,2 t = 12
∴x = 3, y = 6, z = 9, t = 6
find the values of x and y.
By definition of equality of matrices, we get,
2x – y= 10 ..,(1)
3x +y = 5 .,.(2)
Adding (1) and (2), we get,
5 x = 15 or x = 3
∴ from (1), 6 – y=-10 ⇒ y = – 4
∴. x = 3, y = – 4.
, find the values of .x, y, z and w.
By definition of equality of matrices, we get,
3 x = .x + 4 ...(1)
3y = 6+x+ y -(2)
3: = – I +z + w ’ ...(3)
3 w’ = 2 w + 3 ....(4)
From (1), 3 .x – .x = 4 or 2 .x = 4 ⇒ x = 2
From (4), 3 w – 2 w = 3 or w = 3
Putting x = 2 in (2), we get,
3 y = 6 + 2 + y or 2 y = 8 ⇒ y = 4
Putting w = 3 in (3), we get,
3z= – l + z + 3 or 2 : = 2 ⇒ z = l
∴ we have
x = 2, y = 4, z = 1, w = 3.
(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 cach variety sold in October.
X, Z are matrices of order 2 x 2 X p respectively.
If n = p, then the order of the matrix 7X – 5Z is :
p x 2
2 x n
B.
2 x n
X, Z are matrices of order 2 X n, 2 X p respectively. Also n = p. So X, Z are matrices of the same type 2 X n
∴ 7X - 5Z is defined and of type 2 X
∴ (B) is correct answer.
Find AB, if A = 
Here 
The matrix A is of type 2 x 2 and B is of type 2 x 3.
Number of columns of A = number of rows of B = 2. So AB is defined
Note 1. Here AB is defined. But BA is not defined as number of columns of B is not equal to number of rows of A.
Note 2. If A and B are square matrices of the same order, then AB and BA are both defined.
then find AB, BA.
Show that AB ≠ BA.
A is of type 2 x 3 and B is of type 3 x 2.
Since number of columns of A = number of rows of B =3
∴ AB is defined.
Again number of columns of B = number of rows of A = 2
∴ BA is defined.
Note 1. From above example, it is clear that though AB and BA are defined, yet AB ;≠ BA. Here A and B are of different types.
Note 2. Even if A and B are of same order, there are chances that AB≠ BA.
∴ we can say that matrix multiplication is not commutative.
Note 3. Even if AB ≠ BA in many cases, there are cases when AB = BA.
Find AB, if A = 
Note : In the case of real numbers a, b if a b = 0 then either a = 0 or b = 0.
But if the product of two matrices is a zero matrix, then it is not necessary that one of the matrices is a zero matrix.
Sponsor Area
If 
then find AB and BA. Is AB= BA ? What conclusion do you draw?
By definition of equality of matrices, we get,
3 x – 4z = 1 ...(1)
x + 2 z = 0 ...(2)
3 y – 4 u= 0 ...(3)
– y + 2u = l ...(4)
Multiplying (1) by 1 and (2) by 2, we get,
3 x – 4 z = 1 ...(5)
– 2x + 4z = 0 ...(6)
Adding (5) and (6), we get,
x = 1Putting x = 1 in (5), we get,
Multiplying (3) by 1 and (4) by 2, we get,
3y – 4 u = 0 ...(7)
–2y + 4 u = 2 ..(8)
Adding (7) and (8), we get,
y = 2
Putting y = 2 in (3), we get,
such that 
by definition of equality of matrices,
5 x – 7 z = – 16 ...(1)
–2x + 3z = 7 ...(2)
5y–7u = -6 .... ..(3)
–2y’ + 3u = 2 ...(4)
Equations (1) and (2) can be written as
5 x – 7 z + 16 = 0
–2x – 3z + 7 = 0 
Let x be of m X n type.
Since number of columns of x = number of rows of 
By the definition of equality of matrices,
a + 4b = –7 ...(1)
2a + 5 b =–8 ..(2)
3a + 6 b = –9 –(3)
c + 4 d = 2 ......(4)
2c + 5 d = 4 (5)
3 c + 6 d = 6 ....(6)
Multiplying (1) by 2 and (2) by 1, we get,
2a + 8 b = –14 ...(7)
2a + 5 b = –8 .(8)
Subtracting (8) from (7), we get,
3b = –6 or b = – 2
∴ from (1), a - 8 = – 7 or a = 1
Now a = 1, b = -2 also satisfy (3)
∴ we have a= 1, b = – 2
Multiplying (4) by 2 and (5) by 1, we get
2 c + 8 d = 4 (9)
2;c + 5 d = 4 ...(10)
Subtracting (10) from (9), we get
3d = 0 or d = 0
∴from (4), c + 0 = 2 or c = 2
Now c = 2, d = 0 satisfy (6)
∴ we have c = 2, d = 0
Sponsor Area
Matrix X has a + b rows and a + 2 columns
Matrix Y has b + 1 rows and a + 3 columns.
Since XY is defined
∴ number of columns of X = number of rows of Y
∴ a + 2 = b + 1
∴ a – b + 1 = 0 ...(1)
Again YX is defined
∴number of columns of Y = number of rows of X
∴a + 3 = a + b ⇒ b = 3
Putting b = 3 in (1), we get,
a –3 + l = 0 ⇒ a = 2
∴ we have a = 2, b = 3
∴ X is of type 5 x 4 and Y is of type 4 x 5.
∴ XY is of type 5 x 5 and YX is of type 4 x 4.
So XY and YX are not of the same type and so XY ≠ YX,
. Below is given the network of bus routes connecting the different bus stands of two cities X and Y : 
(a) Find the network may ices connecting the cities :
(i) X to Y (ii) Y to X
(b)What is the relation between the far network matrices obtained in (a) above ?
(a) (i) We describe the number of bus routes from X to Y as below :
This is written in the matrix form as
(ii) We describe the number of bus routes from Y to X as below
This is written in the matrix form as
(b) Matrix for the far net work
|
Market |
Products |
||
|
I |
10,000 |
2,000 |
18,000 |
|
II |
6,000 |
20,000 |
8,000 |
Assume Y, Z, W and P are the matrices of order 3 x k x 2 x p, n x 3 and p x k, respectively.
The restrictions on n, k and p so that PY + WY will be defined are :
(A) k = 3, p = n (B) A:is arbitrary,p = 2
(C) p is arbitrary, k = 3 (D)k = 2, p = 3
Y, Z, W, P are of type 3 x k, 2 x p, n x 3, p x k respectively.
Since PY is defined
∴number of columns of P = number of rows of Y
∴ k = 3
Now PY is of type p x k and WY is of type n x k
Since PY + WY is defined
∴ p = n
∴ (A) is correct answer.
If A is a 3 x 3 matrix |3A| = k|A|, then write the value of k.
|3A| = k |A|
|3A| = 27|A|
k = 27
A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However, typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?
Let charges for typing one English page be Rs. x.
Let charges for typing one Hindi page be Rs.y.
Thus from the given statements, we have,
10x + 3y = 145
3x + 10y = 180
Thus the above system can be written as,
Humanity is reflected in this problem.
Show that the four points A(4,5,1), B(0,-1,-1), C(3,9,4) and D(-4,4,4)
are coplanar.
Given four points A(4,5,1), B(0,-1,-1), C(3,9,4) and D(-4,4,4).
therefore,
If A is a square matrix such that 
then write the value of 
where I is an identity matrix.
Given that 
We need to find the value of 
Where I is the identity matrix.
Thus, 
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award x each, y each and z each for the three respective values to 3, 2 and 1 students respectively with a total award money of 1,600. School B wants to spend 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount for one prize on each value is 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
From the given data, we write the following equations:
From above system, we get:
3x+2y+z = 1600
4x+y+3z = 2300
x+y+z = 900
Thus we get:
This is of the form
Therefore, 
Therefore, 
Awards can be given for discipline.
Find the value of a if 
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
Let the award money given for honesty, regularity and hard work be Rs. x, Rs. y and Rs. z respectively.
Since total cash award is Rs. 6,000.
Three times the award money for hard work and honesty amounts to Rs.11,000.
Award money for honesty and hard work is double that given for regularity. 
The above system of equations can be written in matrix form AX = B as:
Here, 
Thus, A is non-singular. Hence, it is invertible.
Hence, x = 500, y = 2000, and z = 3500.
Thus, award money given for honesty, regularity and hardwork is Rs. 500, Rs.2000 and Rs. 3500 respectively.
The school can include awards for obedience.
Show that all the diagonal elements of a skew symmetric matrix are zero.
Let A [aij] be a skew symmetric matrix.
so,
aij =-aji for all i,j
⇒aii -aii for all values of i
⇒2aii =0
⇒aii =0 for all values of i
⇒a11 = a22 = a33 =..... ann =0
Let 
find a matrix D such that CD – AB = O.
By equality of matrices we get,
2p +5r-3 = 0 ....(1)
3p +8r-43 = 0 ..(2)
2q +5s = 0 ......(3)
3q +8s-22 = 0 ..(4)
By solving (1) and (2) we get p = -191 and r = 77
Similarly, on solving (3) and (4) we get q = - 110 and s = 44
If the matrix is skew symmertic, find the values of 'a' and 'b'.
Given, A is skew-symmetric,
∴
If Use it solve the system of equations.
2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y- 2z =-3
Now, A11 0, A12 = 2., A12 =1
A21 = -1, A22 = -9, A23 = -5
A31 = 2, A32 = 23, A33 = 13
Now, the given system of equations can be written in the form of AX = B, where
Find the value of x and y if:
On comparing the corresponding elements of the matrices on both sides, We get:
2 + y = 5 y = 3
2 ( x + 1 ) = 8 x = 3
Let Express A as sum of two matrices such that one is symmetric and the other is skew symmetric.
Now, A can be written as:
If A is an invertible matrix of order 3 and |A| = 5, then find |adj. A|.
where n is order of square matrix
Given A is an invertible matrix of order 3
If matrix A = (1, 2, 3), write AA’, where A’ is the transpose of matrix A.
Given: A = [ 1 2 3 ]
Using matrices, solve the following system of equations:
2x – 3y + 5 = 11
3x + 2y – 4z = -5
x + y – 2z= -3
2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y - 2z = -3
System of equations can be written as AX = B
So, x = 1, y = 2, z = 3.
If A = , then for what value of α is A an identity matrix?
Matrix A is a matrix of order 2.
Identity matrix of second order is
For A to be an identity matrix,
If , then write the value of k.
Now using matrix multiplication in L.H.S., we get
Now on equating the corresponding elements we get the value of k = 17
Using elementary row operations, find the inverse of the following matrix:
We can write A = IA
For a 2 x 2 matrix, A = [ aij ] whose elements are given by aij = , write the value of a12 .
For a12, the value of i = 1 and j = 2.
For what value of x, the matrix is singular?
Thus, when x = 3, the given matrix A is singular.
If P = 
is the adjoint of a 3 x3 matrix A and |A| = 4, then α is equal to
4
11
5
0
B.
11
Given, 
|P| = 1(12-12)-α (4-6) +3(4-6)
= 2α -6.
∵ P =adj(A)
∴ |P| = |adj A | = |A|3-1 = |A|2 = 16
[∵ |adj A| = |A|n-1 order is 3 x3
∴ 2α -6 = 16
2α = 22
α = 11
Let A and B be two symmetric matrices of order 3.
Statement-1: A(BA) and (AB)A are symmetric matrices.
Statement-2: AB is symmetric matrix if matrix multiplication of A with B is commutative.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
B.
Statement-1 is true, Statement-2 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
A' = A , B' = A
P = A(BA)
P' = (A(BA))'
= (BA)' A'
= (A'B') A'
= (AB) A
= A(BA)
∴A(BA) is symmetric
similarly (AB) A is symmetric
Statement(2) is correct but not correct explanation of statement (1).
Let f : R → R be defined by 
If f has a local minimum at x = - 1 then a possible value of k is
1
0
-1/2
-1
C.
-1/2
k – 2x > 1 k + 2 = 1
k > 1 + 2x k = -1
k > 1 + 2(-1)
k > -1
If S is the set of distinct values of 'b' for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is
a singleton
an empty set
an infinite set
a finite set containing two or more elements
A.
a singleton
but at a = 1 and b =1
First two equations are x +y+ z =1
and third equations is x + y +z = 0
⇒ There is no solution
therefore, b = {1}
⇒ it is a singleton set
Let ω be a complex number such that 2ω +1 = z where z = √-3. if
then k is equal to
1
-z
z
-1
B.
-z
Here ω is complex cube of unity
Let A be a 2 × 2 matrix
Statement 1 : adj (adj A) = A
Statement 2 : |adj A| = |A|
Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1
Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.
Statement–1 is true, statement–2 is false.
Statement–1 is false, Statement–2 is true
A.
Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1
adj(adjA) = |A|n – 2A,
where |A| = determinant of A but n = 2
⇒ A also |adj A| = |A|n – 1
⇒ |A| Statement–1 is true and Statement–2 is also true and Statement–2 is correct explanation of Statement–1.
If A and B are square matrices of size n × n such that A2 − B2 = (A − B) (A + B), then which of the following will be always true?
A = B
AB = BA
either of A or B is a zero matrix
either of A or B is an identity matrix
B.
AB = BA
A2 − B2 = (A − B) (A + B)
A2 − B2 = A2 + AB − BA − B2
⇒ AB = BA
If A2 – A + I = 0, then the inverse of A is
A + I
A
A – I
I – A
D.
I – A
Given A2 – A + I = 0
A–1A2 – A–1A + A–1 – I = A–1⋅0 (Multiplying A–1 on both sides)
⇒ A - I + A-1 = 0 or A–1 = I – A.
Sponsor Area
Sponsor Area
, Find A + B.
, find 2A - 3B.
, find A + B and A - B.
, then find 2A - B.
then computer 3A-5B
.find the matrix C such that A + B + C is a zero matrix.
X= - 2A - B
, then find the matrix X. such that 2A + 3X = 5B.
show that (A-2I)(A-3I)=O.
verify that 
,
, find K such that 
, 
Show that 
of the matrix 
whose elements 
are given by 
then find the matrix A. 
. If u1 and u2 are column matrices such that Au1 = 
and Au2 = 
, then u1 +u2 is equal to
,then the value of 'n' is 
then f(x) is a polynomial of degree
The only correct statement about the matrix A is
.If B is the inverse of matrix A, then α is