KENDRIYA VIDYALAYA SANGATHAN
AHMEDABAD REGION
QUESTION BANK FOR SLOW LEARNER AND BRIGHT LEARNER
(MATRICES)
FOR SLOW LEARNERS
1. Construct a 2  2 matrix whose elements are given by aij  2i  j .
a12   2 * 1  1 2 * 1  2  1 0
a
SOL: Required matrix is  11
=
=

a21 a22   2 * 2  1 2 * 2  2  3 2
2.If A is a square matrix of order 3 such that adj A 144 , write the value of A .
SOL: adjA =144 => A
31
=144 => A =+12 or -12
 K 2
3. For what value of k,the matrix 
 is singular
 3 4
k 2
3
=o =>4K-6=0 => K= .
2
3 4
SOL:
x  y
4. If 
 5
2  6 2
then find the values of x and y .

xy  5 8 
SOL: From the given matrices we have the equations x+y=6 and xy=8
Put y=6-x in xy = 8 then we get
x (6-x)=8  x 2  6 x  8  0  ( x  4)( x  2)  0  x  4orx  2 , If x=4 then y=2
and if x=2 then y=4.
5. If
, then find the value of a.
 2
 a 4  8   28
 
SOL: By product of the given matrices we have -2a+32=28  -2a=28-32  a= 2
6. Express the following matrix as sum of symmetric and skew symmetric matrix:
7. Using matrix method, solve the following system of linear equations:
3x-2y+3z=8,2x+y-z=1,4x-3y+2z=4
SOL: The given equation can be written in matrix form as
3  2 3   x   8 
2 1  1  y  =  1 

   
4  3 2   z   4 
or AX=B (say)
……….. (1)
now A =3(2-3)+2(4+4)+3(-6-4)
=-17  0
 A-1 exists.
Now,A11=+(2-3)=-1, A12=-(4+4)=-8, A13=+(-6-4)=-10
A21=-(-4+9)=-5, A22=+(6-12)=-6, A23=-(-9+8)=1
A31=+(2-3)=-1, A32=-(-3-6)=9, A33=+(3+4)=7A
  1  5  1
1
 Adj A =   8  6 9   A-1=
(Adj.A)


A
 10 1
7 
  1  5  1
1 
 8  6 9 
=

 17
 10 1
7 
By (1), X=A-1B
  1  5  1  8 
1 
 8  6 9   1 
=
 17 
 10 1
7   4 
 85 4 
1 
 64  6  36
=

17
  80  1  28 
 x
  17 
1 


  y =
 34 =

17
 z 
  51
 x=1, y=2 and z=3
1 
2
 
 3 
(ans.)
8. Solve the system of linear equations:
2x+3y+10z=4, 4x-6y+5z=1, 6x+9y-20z=2
.
Sol: Given system can be written as
2 3
10 𝑥
4
[4 −6
5 ] [𝑦] = [1]or AX=B, |𝐴| = 1200 ≠ 0 → A is invertible
6 9 −20 𝑧
2
𝐴−1 =
𝐴𝑑𝑗 𝐴
|𝐴|
−1
X=𝐴
−2 0 1
= [ 9 2 −3]
6 1 −2
𝑥
75
1
B → [𝑦]=1200 [110
𝑧
72
150
−100
0
1/2
75 4
30 ] [1] = [1/3]
1/5
−24 2
→ x= 1/2, y= 1/3, z = 1/5
9. The cost of 4 chocolates ,3 samosas and 2 apples is Rs 60.The cost of 2 chocolates ,4 samosas
and 6 apples is Rs.90.The cost of 6 chocolates ,2 samosas and 3 apples is Rs. 70.Find the cost of
each item by matrix method. What do you think is the healthiest diet? Suggest an item that
could replace samosa to make the diet healthier.
SOL:. Let the cost of a chocolate, a samosa and an apple be Rs. x, y and z respectively, then
4x + 3y+2z = 60, 2x + 4y+ 6z =90, 6x+ 2y+ 3z =70, or
4x + 3y+2z = 60,x+ 2y+3z=45, 6x+ 2y +3z =70
Given system can be written as
4 3
[1 2
6 2
2 𝑥
60
3] [𝑦] = [45]or AX=B, |𝐴| = 25 ≠ 0 → A is invertible
3 𝑧
70
𝐴−1 =
−1
X=𝐴
Q10
SOL:
𝐴𝑑𝑗 𝐴
|𝐴|
0
−5
5
1
= 25 [ 15
0 −10]
−10 10
5
𝑥
60
0
−5
5
125
1
1
B → [𝑦]=25 [ 15
0 −10] [45] =25 [200] → x= 5, y= 8, z = 8
𝑧
70
−10 10
5
200
FOR BRIGHT LEARNER:
QUS1:
Sol:
=F(x+y)
QUS2: If A=
find A-1 and hence solve the system of linear equations:
x + 2y + z = 4, –x + y + z =0, x – 3y + z=2.
.
SOL: Find |A|=10≠ 0,
2
1 4
[−5 0
10
1 −2
1 4 −5
⇒ (𝐴−1 )′ =
[2 0
10
2 5
=>
×= (𝐴−1 )′ 𝐵
𝐴−1 =
2
5]
3
1
−2]
3
Using ×= (𝐴′ )−1 𝐵
X=9/5, y=2/5, z=7/5
 1 1 0 
 2 2 2


QUS3. ). Given that A   2 3 4  and B   4 2 4 find AB use it to solve equations
 0 1 2 
 2 1 5 
x y 3
2 x  3 y  4 z  17
y  2z  7
SOL: A.B =
2  4  6 0 0 
1  1 2   2
 2 3 4   4 2  4 = 

-1 1


 0 6 0 =6I =>A = 6
0 1 2  2  1 5  0 0 6
2  4
2
  4 2  4


 2  1 5 
If we express the equation in matrix form, we get
2  4  3 
1  1 0   x   3 
2
2 3 4  y  = 17   AX = C X = A-1.C  X = 1  4 2  4 17 

   
 
6 
0 1 4  z   7 
 2  1 5   7 
 6  34  28 
 12 
1 
1  

X =  12  34  28  X =   6  x =2., y = -1., z =4
6
6
 6  17  35 
 24 
QUS4. Residents Welfare association organized a get together cum seminar on protection of
environment.On invitation to different localities 10 men and 8 women from locality A, 7 men
and 6 women from locality B and 12 men and 10 women from C, participated in the seminar. In
this it was decided that each of men participants will educate 3 more people and each of
women participants will educate 2 more people about protection of environment.Using
matrices represent the above data and find the total no. of people educated . Write about
protection of environment.
Sol: Arranging the information as matrix:
A
B
C
MEN
10
7
12
WOMEN
8
6
10
10
7
12
29
A+B+C=[ ] + [ ] + [ ] = [ ]
8
6
10
24
29
Total number of people educated further=[3 2] [ ] = [131]
24
Hence 131 people were educated further.
Environment protection is priority as it controls pollution and balances the nature.
QUS5: Two school A and B decided to award prizes to their students for three values
honesty(x),punctuality (y) and obedience(z).School A decided to award a total of Rs 11000 for
the three values of 5,4 and 3 students respectively while school B decided to award Rs 10700 for
the three values of 4, 3 and 5 students respectively. If all the three prizes together amount to Rs.
2700, then:
i.
ii.
iii.
Represent the above situation by a matrix equation and form linear
equations using matrix multiplication.
Is it possible to solve the system of equation so obtained using matrix?
If yes, find the award money for each value
Which value you prefer to be rewarded most and why?
SOL:
5 4
[4 3
1 1
5x+4y+3z=11000
4x+3y+5z=10700
x +y + z=2700
(1marks)
11000
3 𝑥
5] [𝑦] = [10700]
2700
1 𝑧
for getting |A|= -3
-1 −1
getting A = 3
(1 marks)
−2 −1 11
[1
2 −13]
1 −1 −1
getting x=1000,y=900 and z=800
correct value and reason
QUS 6:
SOL:
(2 marks)
(1 marks)
( 1 mark)
QUS:7
SOL:
QUS 8
SOL: