"Exploring the Essential Functions and Design Considerations of Spillways in ...
Matrices _ Practice Sheet II (12th Board Booster 2023).pdf
1. 1
12th BOARD BOOSTER
Matrices
1. If a matrix has 28 elements, what are the possible
orders it can have? What if it 13 elements?
2. In the matrix 2
1
2 3
2
0 5
5
a x
A x y
= −
−
, write
(i) the order of the matrix A.
(ii) the number of elements.
(iii) elements a23, a31 and a12.
3. Construct a 3 × 2 matrix whose elements are given by
aij = ei.x
sin jx.
4. Find the values of a and b, if A = B, where
4 3
8 6
a b
A
+
=
−
and
2
2
2 2 2
8 5
a b
B
b b
+ +
=
−
Thinking Process:
By using equality of two matrices, we know that each
element of A is equal to corresponding element of B.
5. If
3 1 1
5 2 3
X
−
=
− −
and
2 1 1
7 2 4
Y
−
=
, then find
(i) X + Y.
(ii) 2X – 3Y.
(iii) a matrix Z such that X + Y + Z is a zero matrix.
6. Find the non-zero values of x satisfying the matrix
equation
( )
2
8 24
2 2 8 5
2 2
3 4 4 (10) 6
x
x x
x
x x x
+
+ =
.
7. If
4 8
6 0 6
xy w
z x y
=
+ +
, then find the value of x,
y, z and w.
8. The matrix
0 0 4
0 4 0
4 0 0
P
=
is a
(1) Square matrix (2) Diagonal matrix
(3) Unit matrix (4) None of these
9. Total number of possible matrices of order 3 × 3 with
each entry 2 or 0 is
(1) 9 (2) 27
(3) 81 (4) 512
10. If
1 1
1 1
sin ( ) tan
1
sin cot ( )
x
x
A
x
x
− −
− −
=
and
1 1
1 1
cos ( ) tan
1
sin tan ( )
x
x
B
x
x
− −
− −
−
=
−
, then A – B is equal
to
(1) I (2) 0
(3) 2I (4)
1
2
I
11. If A and B are two matrices of the order 3 × m and
3 × n, respectively and m = n, then order of matrix
(5A – 2B) is
(1) m × 3 (2) 3 × 3
(3) m × n (4) 3 × n
12. If
8 0
4 2
3 6
A
= −
and
2 2
4 2
5 1
B
−
=
−
, then find the
matrix X, such that 2A + 3X = 5B.
13. Find X and Y, if
5 2
0 9
X Y
+ =
and
3 6
0 1
X Y
− =
−
.
Practice Sheet
2. 2
14. Two farmers Ramkishan and Gureharan 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.
September Sales (in Rupees)
Basmati Permal Naura
Ramkishan
10,000 20,000 30,000
Gurcharan Singh
50,000 30,000 10,000
A
=
September Sales (in Rupees)
Basmati Permal Naura
Ramkishan
5000 10,000 6000
Gurcharan Singh
20,000 10,000 10,000
A
=
(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 on October.
15. Find the values of x, y, z and t, if:
1 1 3 5
2 3 3
0 2 4 6
x z
y t
−
+ =
16. If
8 0 2 2
4 2 4 2
3 6 5 1
A B
−
= − =
−
then find the matrix
“x”, of order 3 × 2, such that 2A + 3X = 5B.
17. Let
2 4 1 3 2 5
, ,
3 2 2 5 3 4
A B C
−
= = =
−
Find each of the following
(i) A + B
(ii) A – B
(iii) 3A – C
Evaluate the following:
cos sin sin cos
cos sin
sin cos cos sin
−
+
−
18. Let A and B be non-zero matrices given by
2 2
2 2
A
=
and
2 2
2 2
B
−
=
−
Show that AB and
BA are zero matrices
19. Solve the matrix equation XA = B where
1 2
2 3
A
=
and
2 1
1 3
B
=
20. Consider the following matrices:
2 3 1
1 4 2
A
−
=
− −
and
4 3 1
2 1 2
3 2 1
B
−
= − −
Compute , , ,
T r T T
A B A B A B
and ( )T
AB .
21. Express the matrix A as a sum of a symmetric matrix
and a skew-symmetric matrix.
2 3 1
1 2 4
5 3 5
A
= − −
− −
22. Let
2
2
ab b
A
a ab
=
− −
where 0
ab . Then
(1) 2
A A
= (2) 2
A O
=
(3) 2
A I
= (4) 3
A A
=
23. Find real number a and b such that a + bA + A2
= O
where
3 2
1 1
A
=
. Also find A–1
.
24. If the product of the matrices
1 1 1 2 1 3 1 1 378
...
0 1 0 1 0 1 0 1 0 1
n
=
Then n is equal to
(1) 27 (2) 26
(3) 376 (4) 378
3. 3
25. If
2
4 , 1 3 6
5
A B
−
= = −
, verify that (AB)' = B'A'.
26. For any square matrix A with real number entries,
A + A' is a symmetric matrix and A – A' is a skew
symmetric matrix.
27. Any square matrix can be expressed as the sum of a
symmetric and a skew symmetric matrix.
28. If
cos sin
sin cos
A
−
=
, and ' 1
A A
+ = , then the
value of is
(1)
6
(3)
3
(3) (4)
3
2
29. If A is a square matrix such that A2
= I, then find the
simplified value of (A – I)3
+ (A + I)3
– 7A.
30. If A is a square matrix such that A2
= A, then write the
value of 7A – (I + A)3
, where I is an identity matrix.
5. 5
Hints and Solutions
1. We know that, if matrix is of order m × n, it has mn
elements, where m and n are natural number.
We have, m × n = 28
(m, n) = {(1, 28), (2, 14), (4, 7), (7, 4), (14, 2),
(28, 1)}
So, the possible orders are 1 × 28, 2 × 14, 4 × 7,
7 × 4, 14 × 2, 28 × 1.
Also, if it has 13 elements, then m × n = 13
(m, n) = {(1, 13), (13, 1)}
Hence, the possible order are 1 × 13, 13 × 1.
2. We have, 2
1
2 3
2
0 5
5
a x
A x y
= −
−
(i) the order of matrix A = 3 × 3
(ii) the number of elements = 3 × 3 = 9
[since, the number of elements in an m × n
matrix will be equal to m × n = mn]
(iii) 2
23 31 12
, 0, 1
a x y a a
= − = =
[since, we know that aij, is a representation of
element lying in the i th
row and jth
column]
3. Since, A = [aij]m × n 1 ≤ i ≤ m and 1 ≤ j ≤ n. i. j ∈ N
.
3 2
[ sin ] ;1 3;1 2
i x
A e jx i j
=
1.
11
1.
12
2. 2
21
2. 2
22
3. 3
31
3. 3
32
.sin1. sin
.sin 2. sin 2
.sin1. sin
.sin 2. sin 2
sin1. sin
.sin 2. sin 2
x x
x x
x x
x x
x x
x x
a e x e x
a e x e x
a e x e x
e e x e x
e e x e x
a e x e x
= =
= =
= =
= =
= =
= =
2 2
3 3
3 2
sin sin 2
sin sin 2
sin sin 2
=
x x
x x
x x
e x e x
A e x e x
e x e x
4. We have,
2
2
2 2 2 2
4 3 2 2 2
and
8 6 8 5
+ + +
= =
−
−
a b a b
A B
b b
Also, A = B
By equality of matrices we know that each element
of A is equal to the corresponding element of B, that
is aij = bij for all i and j.
11 11 4 2 2 2
= + = + =
a b a a a
2 2
12 12 3 2 3 2
= = + = −
a b b b b b
and 2
22 22 6 5
= − = −
a b b b
6 3 2 5
− = − −
b b 2
b 3b 2
= −
2 4 2
= =
b b
2 and 2
= =
a b
5. We have,
2 3 2 3
3 1 1 2 1 1
and
5 2 3 7 2 4
− −
= =
− −
X Y
(i)
3 2 1 1 1 1
X Y
5 7 2 2 3 4
+ + − −
+ =
+ − + − +
5 2 2
12 0 1
−
=
(ii)
3 1 1 6 2 2
2 2
5 2 3 10 4 6
− −
= =
− − − −
X
and
2 1 1 6 3 3
3 3
7 2 4 21 6 12
− −
= =
Y
(iii)
6 6 2 3 2 3
2 3
10 21 4 6 6 12
− − − +
− =
− − − − −
X Y
0 1 1
11 10 18
−
=
− − −
Also,
0 0 0
0 0 0
+ + =
X Y Z
We see that Z is the additive inverse of (X + Y) or
negative of (X + Y).
5 2 2
12 0 1
Z
− −
− −
=
[ ( )]
Z X Y
= − +
6. 6
6. Given that,
( )
2
8 24
2 2 8 5
2 2
3 4 4 10 6
x
x x
x
x x x
+
+ =
2 2
2
2 2 16 10 2 16 48
8 8 20 12
3
+
+ =
x x x x
x x
x x
2 2
2
2 16 2 10 2 16 48
20 12
3 8 8
+ + +
=
+ +
x x x x
x
x x x
2 10 48
+ =
x x
12 48
=
x
48
4
12
x = =
7. We have,
4 8
6 0 6
=
+ +
xy w
z x y
By equality of matrix x + y = 6 and xy = 8
6 and(6 ) 8
= − − =
x y y y
2
6 8 0
− + =
y y
2
4 2 8 0
− − + =
y y y
( 2)( 4) 0
− − =
y y
2 or 4
= =
y y
6 2 4
= − =
x
or 6 4 2
= − =
x [ x = 6 – y]
also, z + 6 = 0
6 and 4
= − =
z w
2, 4 or 4, 2, 6 and 4
= = = = = − =
x y x y z w
8. (1)
We known that, in a square matrix number of rows
are equal to the number of columns, so the matrix
0 0 4
0 4 0
4 0 0
=
P is a square matrix.
9. (4)
Total number of possible matrices of order 3 × 3 with
each entry 2 or 0 is 29
i.e., 512.
10. (4)
We have,
1 1
1 1
1 1
sin tan
1 1
sin cot
x
x
A
x
x
− −
− −
=
and
1 1
1 1
1 1
cos tan
1 1
sin tan
x
x
B
x
x
− −
− −
−
=
( )
1 1 1 1
1 1 1 1
1 1
sin cos tan tan
1 1
sin sin cot tan
x x
x x
A B
x x
x x
− − − −
− − − −
+ −
− =
− +
1
0
2
1
0
2
=
π
π
π
π
1 1
1 1
sin cos
2
and tan cot
2
x x
x x
− −
− −
+ =
+ =
1 0
1
0 1
2
=
1
2
= I
11. (4) A3×m and B3×n are two matrices. If m = n, then A
and B have same orders as 3 × n each, so the order of
(5A – 2B) should be same as 3 × n.
12. We have 2A + 3X = 5B
Or 2 3 2 5 2
A X A B A
+ − = −
Or 2 2 3 5 2
A A X B A
− + = −
(Matrix addition is commutative)
Or 3 5 2
O X B A
+ = −
(–2A is the additive inverse of 2A)
Or 3 5 2
X B A
= − (O is the additive identity)
Or
1
(5 2 )
3
X B A
= −
Or
2 2 8 0
1
4 2 2 4 2
3
5 1 3 6
X
−
= − −
−
10 10 16 0
1
20 10 8 4
3
25 5 6 12
− −
= + −
− − −
10 16 10 0 6 10
1 1
20 8 10 4 12 14
3 3
25 6 5 12 31 7
− − + − −
= − + =
− − − − −
10
2
3
14
4
3
31 7
3 3
−
−
=
− −
7. 7
13. We have
5 2 3 6
( ) ( )
0 9 0 1
X Y X Y
= + + − = +
−
Or
8 8 8 8
( ) ( ) 2
0 8 0 8
X X Y Y X
+ + − = =
Or
8 8 4 4
1
0 8 0 4
2
X
= =
Also
5 2 3 6
( ) ( )
0 9 0 1
X Y X Y
+ − − = −
−
Or
5 3 2 6 2 4
( ) ( ) 2
0 9 1 0 10
X X Y Y Y
− − −
− + + = =
+
Or
2 4 1 2
1
0 10 0 5
2
Y
− −
= =
14. Combined sales in September and October for each
farmer in each variety is given by
Basmati Permal Naura
Ramkishan
15,000 30,000 36,000
Gurcharan Singh
70,000 40,000 20,000
A B
+ =
(ii) Change in sales from September to October is given
by
Basmati Permal Naura
Ramkishan
5000 10,000 24,000
Gurcharan Singh
30,000 20,000 0
A B
− =
(iii) 2% of
2
0.02
100
B B B
= =
Basmati Permal Naura
Ramkishan
0.02 5000 10,000 6000
Gurcharan Singh
20,000 10,000 10,000
= =
Basmati Permal Naura
Ramkishan
100 200 120
Gurcharan Singh
400 200 200
=
15. We have:
1 1 3 5
2 3 3
0 2 4 6
x z
y t
−
+ =
2 2 3 3 9 15
2 2 0 6 12 18
x z
y t
−
+ =
2 3 2 3 9 15
2 2 6 12 18
x z
y t
+ −
=
+
2 3 9
x
+ = ………… (1)
2 3 15
z − = ………….. (2)
2 12
y = ………….. (3)
2 6 18
t + = ………..(4)
From (1) 2 9 3
x
= −
2 6
x
=
3
x
= .
From (3) 2y = 12
6.
y
=
From (2), 2 3 15
z
− =
2 18
z
=
9
z
=
From (4), 2t + 6 = 18
2 12
t
=
6
t
= .
Hence, x = 3, y = 6, z = 9 and t = 6.
16. We have: 2A + 3X = 5B
2 3 2 5 2
A X A B A
+ − = −
2 2 3 5 2
A A X B A
− + = −
(2 2 ) 3 5 2
A A X B A
− + = −
0 3 5 2
X B A
+ = −
[ 2 is theinverse of 2 ]
A A
−
3 5 2
X B A
= − .
[ O is the additive identity]
Hence, ( )
1
5 2
3
X B A
= −
2 2 8 0
1
5 4 2 2 4 2
3
5 1 3 6
−
= − −
−
10 10 16 0
1
20 10 8 4
3
25 5 6 12
− −
= + −
− − −
10 16 10 0
1
20 8 10 4
3
25 6 5 12
− − +
= − +
− − −
6 10 2 10 / 3
1
12 14 4 14 / 3
3
31 7 31/ 3 7 / 3
− − − −
= =
− − − −
8. 8
17. We have
cos sin sin cos
cos sin
sin cos cos sin
−
+
−
2
2
cos cos sin
cos sin cos
=
−
2
2
sin sin cos
sin cos sin
−
+
2 2
2 2
cos sin 0
0 cos sin
+
=
+
2
1 0
0 1
I
= =
18. We have
2 2 2( 2) 2( 2) 2·2 0 0
2 2 2( 2) 2( 2) 2·2 0 0
AB
+ − − +
= =
+ − − +
And
2 2 ( 2)2 2·2 ( 2)2 0 0
( 2)2 2 2 ( 2)2 2 2 0 0
BA
+ − + −
= =
− + − +
19. Since A and B are 2 × 2 matrices, to satisfy the
equation XA = B, X also must be a 2 × 2 matrix.
Let
a b
X
c d
=
From XA = B, we have
2 2 3 2 1
2 2 3 1 3
a b a b
c d c d
+ +
=
+ +
Equating the corresponding entries on both sides, we
have
2 2; 2 1; 2 3 1; 2 3 3
a b c d a b c d
+ = + = + = + =
Solving these, we get that 4, 3, 3
a b c
= − = = and
1
d = − .
Therefore
4 3
3 1
X
−
=
−
20. Since A and B are of order 2 × 3 and 3 ×3,
respectively, AB is defined and is of order 2 × 3. We
have
1
2 1
3 4
1 2
A
−
= −
−
and 1
4 2 3
3 1 2
1 2 1
B
= − − −
− −
Therefore
1 6
5 0
2 9
( )T
A B
−
−
+ =
Also
2 ( 1) 4 ( 2)
3 2 1 1
1 ( 3) 5 4
T T
A B
+ − − + −
+ = + − +
+ − +
1 6
5 0 ( )
2 9
T
A B
−
= = +
−
5 5 7
2 3 7
AB
=
− −
5 2
( )' 5 3
7 7
AB
−
= −
Also
8 6 3 4 8 6 5 2
6 3 2 3 4 4 5 3
2 6 1 1 8 2 7 7
T T
B A
− + − + − −
= − + − − + = −
+ − − + −
9. 9
21. To do this, we should compute
( ) ( )
1 1
and
2 2
T T
A A A A
+ −
Now transpose of A is given by
2 1 5
3 2 3
1 4 5
T
A
= − −
−
For AT
, first row of A becomes the first column of AT
,
the ith
row of A becomes the ith
column of AT
. Now
2 2 3 ( 1) 1 5 4 2 6
1 3 2 ( 2) 4 ( 3) 2 4 1
5 1 3 4 5 ( 5) 6 1 10
T
A A
+ + − +
+ = − + − + − + − = −
+ − + − + − −
( )
2 1 3
1 1
1 2
2 2
1
3 5
2
T
A A
+ = −
−
(8.12)
Again
2 2 3 ( 1) 1 5 0 4 4
1 3 2 ( 2) 4 ( 3) 4 0 7
5 1 3 4 5 ( 5) 4 7 0
T
A A
− − − − −
− = − − − − − − − = −
− − − − − − −
( )
0 2 2
1 7
2 0
2 2
7
2 0
2
T
A A
−
− = −
−
(8.13)
By Theorem 8.19 and using Eqs. (8.12) and (8.13) we
have
( ) ( )
1 1
2 2
T T
A A A A A
= + + −
2 1 3 0 2 2
1 7
1 2 2 0
2 2
1 7
3 5 2 0
2 2
−
= − + −
−
−
Symmetric skew-symmetric
22. We have
2 2
2
2 2
ab b ab b
A
a ab a ab
=
− − − −
2 2 2 2 3 3
3 3 2 2 2 2
a b a b ab ab
a b a b a b a b
− −
=
− + − +
0 0
0 0
O
= =
23. First
2
3 2 3 2
1 1 1 1
A A A
= =
3 3 2 1 3 2 2 1 11 8
1 3 1 1 1 2 1 1 4 3
− + +
= =
+ +
Now, suppose that a and b are real numbers such that
2
0 a bA A
= + +
0 3 2 11 8
0 4 3
a b b
a b b
= + +
3 11 2 8
4 3
a b b
b a b
+ + +
=
+ + +
Therefore,
3 11 0;2 8 0
a b b
+ + = + =
4 0; 3 0
b a b
+ = + + =
Solving these we get b = – 4 and a = 1. Therefore
2
1 4 0
A A
− + =
1 1 2 1
4 0
A AA A A
− − −
− + =
This gives the inverse as
1
4 0 3 2 1 2
4
0 4 1 1 1 3
A A
−
−
= − = − =
−
24. We have
1 1 1 2 1 3 1 1 2
0 1 0 1 0 1 0 1
+
= =
Again
1 3 1 3 1 6 1 1 2 3
0 1 0 1 0 1 0 1
+ +
= =
By induction,
1
1 1 378
0 1
0 1
n
k
k
LHS =
=
Therefore
( )
1
378
2
n n +
= or ( )
1 27 28
n n + =
Hence n = 27.
10. 10
25. We have
2
4 , 1 3 6
5
A B
−
= = −
Then
2 2 6 12
4 1 3 6 4 12 24
5 5 15 30
AB
− − −
= − = −
−
Now
1
, ' 3
2 4 5
6
' B
− =
−
1 2 4 5
' 3 2 4 5 6 12 15 ( )
6 12 24
'
30
'
B A AB
−
= − = − =
− − −
Clearly ( )' ' '
AB B A
=
26. Let B = A + A', then
( )
' ' '
B A A
= +
( ) ( )
' ' ' as ( )' ' '
A A A B A B
= + + = +
( )
( )
' as ' '
A A A A
= + =
'(as )
A A A B B A
= + + = +
= B
Therefore '
B A A
= + is a symmetric matrix
Now let '
C A A
= −
( ) ( )
' ' ' ' ' '
C A A A A
= − = − (Why?)
'
A A
= − (Why?)
( )
'
A A C
= − − = −
Therefore '
C A A
= − is a skew symmetric matrix.
27. Let A be a square matrix, then we can write
( ) ( )
1 1
' '
2 2
A A A A A
= + + −
From the Theorem 1, we know that ( )
'
A A
+ is a
symmetric matrix and ( )
'
A A
− is a skew symmetric
matrix. Since for any matrix A, ( )' '
kA kA
= , it follows
that ( )
1
'
2
A A
+ is symmetric matrix and ( )
1
'
2
A A
− is
skew symmetric matrix. Thus, any square matrix can
be expressed as the sum of a symmetric and a skew
symmetric matrix.
28. The correct answer is (2)
cos sin cos sin 1 0
sin cos sin cos 0 1
−
+ =
−
2cos 0 1 0
0 2cos 0 1
=
To find the value of , equate the corresponding
terms, we get
2cos 1
=
cos cos
3
=
3
=
29. Given, 2
7
A = ……… (i)
Now 3 3
( ) ( ) 7
A I A I A
− + + −
( ) ( )
3 2 2 3 2 2 3
3 3 3 3 7
A A I AI I A A I AI I A
= − + − + + + + −
3 2 3 2
3 3 3 3 1 7
A A AI I A A AI A
= − + − + + + + −
2 2 3 3
and
A I A I I I
= = =
3 2
2 6 7 2 6 7 [ ]
A AI A A A A A AI A
= + − = + − =
= 2IA – A [from equation (i)]
2 [ ]
A A A IA A
= − = =
30. Given, 2
A A
=
Now 3 3 3
7 (1 ) 7 | 3| 4(1 )
A A A A A
− + = − + + +
3 3 3
( ) 3 ( )
y x y x y xy x y
+ = + + +
2
7 1 3 (1 )
A A A A A
= − + + +
3
and
I IA A
= =
I
( )
2
7 , 3 3
A I A A AI A
= − + + +
2
A A
=
7 ( 3 3 )
A I A A A
= − + + +
2
and
AI A A AI
= =
7 ( 7 ) 1
A I A
= = + = −
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