Table of Contents
Chapter 5 Algebra of Matrices Ex 5.1
1. If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?
Solution:
If a matrix is of order m × n elements, it has m n elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.
m n = 8
Then, ordered pairs m and n will be
m × n be (8 × 1),(1 × 8),(4 × 2),(2 × 4)
Now, if it has 5 elements
Possible orders are (5 × 1), (1 × 5).
Solution:
(i)
Now, Comparing with equation (1) and (2)
a22 = 4 and b21 = – 3
a22 + b21 = 4 + (– 3) = 1
(ii)
Now, Comparing with equation (1) and (2)
a11 = 2, a22 = 4, b11 = 2, b22 = 4
a11 b11 + a22 b22 = 2 × 2 + 4 × 4 = 4 + 16 = 20
3. Let A be a matrix of order 3 × 4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2.
Solution:
Given A be a matrix of order 3 × 4.
So, A = [ai j] 3×4
R1 = first row of A = [a11, a12, a13, a14]
So, order of matrix R1 = 1 × 4
C2 = second column of
Therefore order of C2 = 3 × 1
4. Construct a 2 ×3 matrix A = [aj j] whose elements aj j are given by:
(i) ai j = i × j
(ii) ai j = 2i – j
(iii) ai j = i + j
(iv) ai j = (i + j)2/2
Solution:
(i) Given ai j = i × j
Let A = [ai j]2 × 3
So, the elements in a 2 × 3 matrix are[a11, a12, a13, a21, a22, a23]
a11 = 1 × 1 = 1
a12 = 1 × 2 = 2
a13 = 1 × 3 = 3
a21 = 2 × 1 = 2
a22 = 2 × 2 = 4
a23 = 2 × 3 = 6
Substituting these values in matrix A we get,
(ii) Given ai j = 2i – j
Let A = [ai j]2×3
So, the elements in a 2 × 3 matrix are
a11, a12, a13, a21, a22, a23
a11 = 2 × 1 – 1 = 2 – 1 = 1
a12 = 2 × 1 – 2 = 2 – 2 = 0
a13 = 2 × 1 – 3 = 2 – 3 = – 1
a21 = 2 × 2 – 1 = 4 – 1 = 3
a22 = 2 × 2 – 2 = 4 – 2 = 2
a23 = 2 × 2 – 3 = 4 – 3 = 1
Substituting these values in matrix A we get,
(iii) Given ai j = i + j
Let A = [a i j] 2×3
So, the elements in a 2 × 3 matrix are
a11, a12, a13, a21, a22, a23
a11 = 1 + 1 = 2
a12 = 1 + 2 = 3
a13 = 1 + 3 = 4
a21 = 2 + 1 = 3
a22 = 2 + 2 = 4
a23 = 2 + 3 = 5
Substituting these values in matrix A we get,
(iv) Given ai j = (i + j)2/2
Let A = [ai j]2×3
So, the elements in a 2 × 3 matrix are
a11, a12, a13, a21, a22, a23
Let A = [ai j]2×3
So, the elements in a 2 × 3 matrix are
a11, a12, a13, a21, a22, a23
a11 =
a12 =
a13 =
a21 =
a22 =
a23 =
Substituting these values in matrix A we get,
5. Construct a 2 × 2 matrix A = [ai j] whose elements ai j are given by:
(i) (i + j)2 /2
(ii) ai j = (i – j)2 /2
(iii) ai j = (i – 2j)2 /2
(iv) ai j = (2i + j)2 /2
(v) ai j = |2i – 3j|/2
(vi) ai j = |-3i + j|/2
(vii) ai j = e2ix sin x j
Solution:
(i) Given (i + j)2 /2
Let A = [ai j]2×2
So, the elements in a 2 × 2 matrix are
a11, a12, a21, a22
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
(ii) Given ai j = (i – j)2 /2
Let A = [ai j]2×2
So, the elements in a 2 × 2 matrix are
a11, a12, a21, a22
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
(iii) Given ai j = (i – 2j)2 /2
Let A = [ai j]2×2
So, the elements in a 2 × 2 matrix are
a11, a12, a21, a22
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
(iv) Given ai j = (2i + j)2 /2
Let A = [ai j]2×2
So, the elements in a 2 × 2 matrix are
a11, a12, a21, a22
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
(v) Given ai j = |2i – 3j|/2
Let A = [ai j]2×2
So, the elements in a 2×2 matrix are
a11, a12, a21, a22
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
(vi) Given ai j = |-3i + j|/2
Let A = [ai j]2×2
So, the elements in a 2 × 2 matrix are
a11, a12, a21, a22
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
(vii) Given ai j = e2ix sin x j
Let A = [ai j]2×2
So, the elements in a 2 × 2 matrix are
a11, a12, a21, a22,
a11 =
a12 =
a21 =
a22 =
Substituting these values in matrix A we get,
6. Construct a 3×4 matrix A = [ai j] whose elements ai j are given by:
(i) ai j = i + j
(ii) ai j = i – j
(iii) ai j = 2i
(iv) ai j = j
(v) ai j = ½ |-3i + j|
Solution:
(i) Given ai j = i + j
Let A = [ai j]2×3
So, the elements in a 3 × 4 matrix are
a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34
A =
a11 = 1 + 1 = 2
a12 = 1 + 2 = 3
a13 = 1 + 3 = 4
a14 = 1 + 4 = 5
a21 = 2 + 1 = 3
a22 = 2 + 2 = 4
a23 = 2 + 3 = 5
a24 = 2 + 4 = 6
a31 = 3 + 1 = 4
a32 = 3 + 2 = 5
a33 = 3 + 3 = 6
a34 = 3 + 4 = 7
Substituting these values in matrix A we get,
A =
(ii) Given ai j = i – j
Let A = [ai j]2×3
So, the elements in a 3×4 matrix are
a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34
A =
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 = 1
a33 = 3 – 3 = 0
a34 = 3 – 4 = – 1
Substituting these values in matrix A we get,
A =
(iii) Given ai j = 2i
Let A = [ai j]2×3
So, the elements in a 3×4 matrix are
a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34
A =
a11 = 2×1 = 2
a12 = 2×1 = 2
a13 = 2×1 = 2
a14 = 2×1 = 2
a21 = 2×2 = 4
a22 = 2×2 = 4
a23 = 2×2 = 4
a24 = 2×2 = 4
a31 = 2×3 = 6
a32 = 2×3 = 6
a33 = 2×3 = 6
a34 = 2×3 = 6
Substituting these values in matrix A we get,
A =
(iv) Given ai j = j
Let A = [ai j]2×3
So, the elements in a 3×4 matrix are
a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34
A =
a11 = 1
a12 = 2
a13 = 3
a14 = 4
a21 = 1
a22 = 2
a23 = 3
a24 = 4
a31 = 1
a32 = 2
a33 = 3
a34 = 4
Substituting these values in matrix A we get,
A =
(vi) Given ai j = ½ |-3i + j|
Let A = [ai j]2×3
So, the elements in a 3×4 matrix are
a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34
A =
a11 =
a12 =
a13 =
a14 =
a21 =
a22 =
a23 =
a24 =
a31 =
a32 =
a33 =
a34 =
Substituting these values in matrix A we get,
A =
Multiplying by negative sign we get,
7. Construct a 4 × 3 matrix A = [ai j] whose elements ai j are given by:
(i) ai j = 2i + i/j
(ii) ai j = (i – j)/ (i + j)
(iii) ai j = i
Solution:
(i) Given ai j = 2i + i/j
Let A = [ai j]4×3
So, the elements in a 4 × 3 matrix are
a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, a43
A =
a11 =
a12 =
a13 =
a21 =
a22 =
a23 =
a31 =
a32 =
a33 =
a41 =
a42 =
a43 =
Substituting these values in matrix A we get,
A =
(ii) Given ai j = (i – j)/ (i + j)
Let A = [ai j]4×3
So, the elements in a 4 × 3 matrix are
a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, a43
A =
a11 =
a12 =
a13 =
a21 =
a22 =
a23 =
a31 =
a32 =
a33 =
a41 =
a42 =
a43 =
Substituting these values in matrix A we get,
A =
(iii) Given ai j = i
Let A = [ai j]4×3
So, the elements in a 4 × 3 matrix are
a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, a43
A =
a11 = 1
a12 = 1
a13 = 1
a21 = 2
a22 = 2
a23 = 2
a31 = 3
a32 = 3
a33 = 3
a41 = 4
a42 = 4
a43 = 4
Substituting these values in matrix A we get,
A =
8. Find x, y, a and b if
Solution:
Given
Given that two matrices are equal.
We know that if two matrices are equal then the elements of each matrices are also equal.
Therefore by equating them we get,
3x + 4y = 2 …… (1)
x – 2y = 4 …… (2)
a + b = 5 …… (3)
2a – b = – 5 …… (4)
Multiplying equation (2) by 2 and adding to equation (1), we get
3x + 4y + 2x – 4y = 2 + 8
⇒ 5x = 10
⇒ x = 2
Now, substituting the value of x in equation (1)
3 × 2 + 4y = 2
⇒ 6 + 4y = 2
⇒ 4y = 2 – 6
⇒ 4y = – 4
⇒ y = – 1
Now by adding equation (3) and (4)
a + b + 2a – b = 5 + (– 5)
⇒ 3a = 5 – 5 = 0
⇒ a = 0
Now, again by substituting the value of a in equation (3), we get
0 + b = 5
⇒ b = 5
∴ a = 0, b = 5, x = 2 and y = – 1
9. Find x, y, a and b if
Solution:
We know that if two matrices are equal then the elements of each matrices are also equal.
Given that two matrices are equal.
Therefore by equating them we get,
2a + b = 4 …… (1)
And a – 2b = – 3 …… (2)
And 5c – d = 11 …… (3)
4c + 3d = 24 …… (4)
Multiplying equation (1) by 2 and adding to equation (2)
4a + 2b + a – 2b = 8 – 3
⇒ 5a = 5
⇒ a = 1
Now, substituting the value of a in equation (1)
2 × 1 + b = 4
⇒ 2 + b = 4
⇒ b = 4 – 2
⇒ b = 2
Multiplying equation (3) by 3 and adding to equation (4)
15c – 3d + 4c + 3d = 33 + 24
⇒ 19c = 57
⇒ c = 3
Now, substituting the value of c in equation (4)
4 × 3 + 3d = 24
⇒ 12 + 3d = 24
⇒ 3d = 24 – 12
⇒ 3d = 12
⇒ d = 4
∴ a = 1, b = 2, c = 3 and d = 4
10. Find the values of a, b, c and d from the following equations:
Solution:
Given
We know that if two matrices are equal then the elements of each matrices are also equal.
Given that two matrices are equal.
Therefore by equating them we get,
2a + b = 4 …… (1)
And a – 2b = – 3 …… (2)
And 5c – d = 11 …… (3)
4c + 3d = 24 …… (4)
Multiplying equation (1) by 2 and adding to equation (2)
4a + 2b + a – 2b = 8 – 3
⇒ 5a = 5
⇒ a = 1
Now, substituting the value of a in equation (1)
2 × 1 + b = 4
⇒ 2 + b = 4
⇒ b = 4 – 2
⇒ b = 2
Multiplying equation (3) by 3 and adding to equation (4)
15c – 3d + 4c + 3d = 33 + 24
⇒ 19c = 57
⇒ c = 3
Now, substituting the value of c in equation (4)
4 × 3 + 3d = 24
⇒ 12 + 3d = 24
⇒ 3d = 24 – 12
⇒ 3d = 12
⇒ d = 4
∴ a = 1, b = 2, c = 3 and d = 4
Chapter 5 Algebra of Matrices Ex 5.2
1. Compute the following sums:
Solution:
(i) Given
Corresponding elements of two matrices should be added
Therefore, we get
Therefore,
(ii) Given
Therefore,
Find each of the following:
(i) 2A – 3B
(ii) B – 4C
(iii) 3A – C
(iv) 3A – 2B + 3C
Solution:
(i) Given
First we have to compute 2A
Now by computing 3B we get,
Now by we have to compute 2A – 3B we get
Therefore
(ii) Given
First we have to compute 4C,
Now,
Therefore we get,
(iii) Given
First we have to compute 3A,
Now,
Therefore,
(iv) Given
First we have to compute 3A
Now we have to compute 2B
By computing 3C we get,
Therefore,
(i) A + B and B + C
(ii) 2B + 3A and 3C – 4B
Solution:
(i) Consider A + B,
A + B is not possible because matrix A is an order of 2 x 2 and Matrix B is an order of 2 x 3, so the Sum of the matrix is only possible when their order is same.
Now consider B + C
(ii) Consider 2B + 3A
2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrix.
Now consider 3C – 4B,
Solution:
Given
Now we have to compute 2A – 3B + 4C
5. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find
(i) A – 2B
(ii) B + C – 2A
(iii) 2A + 3B – 5C
Solution:
(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)
(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)
We have to find B + C – 2A
Here,
Now we have to compute B + C – 2A
(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)
Now we have to find 2A + 3B – 5C
Here,
Now consider 2A + 3B – 5C
6. Given the matrices
Verify that (A + B) + C = A + (B + C)
Solution:
Given
Now we have to verify (A + B) + C = A + (B + C)
First consider LHS, (A + B) + C,
Now consider RHS, that is A + (B + C)
Therefore LHS = RHS
Hence (A + B) + C = A + (B + C)
7. Find the matrices X and Y,
Solution:
Consider,
Now by simplifying we get,
Therefore,
Again consider,
Now by simplifying we get,
Therefore,
Solution:
Given
Now by transposing, we get
Therefore,
Solution:
Given
Now by multiplying equation (1) and (2) we get,
Now by adding equation (2) and (3) we get,
Now by substituting X in equation (2) we get,
Solution:
Consider
Now, again consider
Therefore,
And
Chapter 5 Algebra of Matrices Ex 5.3
1. Compute the indicated products:
Solution:
(i) Consider
On simplification we get,
(ii) Consider
On simplification we get,
(iii) Consider
On simplification we get,
2. Show that AB ≠ BA in each of the following cases:
Solution:
(i) Consider,
Again consider,
From equation (1) and (2), it is clear that
AB ≠ BA
(ii) Consider,
Now again consider,
From equation (1) and (2), it is clear that
AB ≠ BA
(iii) Consider,
Now again consider,
From equation (1) and (2), it is clear that
AB ≠ BA
3. Compute the products AB and BA whichever exists in each of the following cases:
Solution:
(i) Consider,
BA does not exist
Because the number of columns in B is greater than the rows in A
(ii) Consider,
Again consider,
(iii) Consider,
AB = 0+(−1)+6+6
AB = 11
Again consider,
(iv) Consider,
4. Show that AB ≠ BA in each of the following cases:
Solution:
(i) Consider,
Again consider,
From equation (1) and (2), it is clear that
AB ≠ BA
(ii) Consider,
Again consider,
From equation (1) and (2) it is clear that,
AB ≠ BA
5. Evaluate the following:
Solution:
(i) Given
First we have to add first two matrix,
On simplifying, we get
(ii) Given,
First we have to multiply first two given matrix,
= 82
(iii) Given
First we have subtract the matrix which is inside the bracket,
Solution:
Given
We know that,
Again we know that,
Now, consider,
We have,
Now, from equation (1), (2), (3) and (4), it is clear that A2 = B2= C2= I2
Solution:
Given
Consider,
Now we have to find,
Solution:
Given
Consider,
Hence the proof.
Solution:
Given,
Consider,
Again consider,
Hence the proof.
Solution:
Given,
Consider,
Hence the proof.
Solution:
Given,
Consider,
We know that,
Again we have,
Solution:
Given,
Consider,
Again consider,
From equation (1) and (2) AB = BA = 03×3
Solution:
Given
Consider,
Again consider,
From equation (1) and (2) AB = BA = 03×3
Solution:
Given
Now consider,
Therefore AB = A
Again consider, BA we get,
Hence BA = B
Hence the proof.
Solution:
Given,
Consider,
Now again consider, B2
Now by subtracting equation (2) from equation (1) we get,
16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC)
Solution:
(i) Given
Consider,
Now consider RHS,
From equation (1) and (2), it is clear that (AB) C = A (BC)
(ii) Given,
Consider the LHS,
Now consider RHS,
From equation (1) and (2), it is clear that (AB) C = A (BC)
17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.
Solution:
(i) Given
Consider LHS,
Now consider RHS,
From equation (1) and (2), it is clear that A (B + C) = AB + AC
(ii) Given,
Consider the LHS
Now consider RHS,
Solution:
Given,
Consider the LHS,
Now consider RHS
From the above equations LHS = RHS
Therefore, A (B – C) = AB – AC.
19. Compute the elements a43 and a22 of the matrix:
Solution:
Given
From the above matrix, a43 = 8and a22 = 0
Solution:
Given
Consider,
Again consider,
Now, consider the RHS
Therefore, A3 = p I + q A + rA2
Hence the proof.
21. If ω is a complex cube root of unity, show that
Solution:
Given
It is also given that ω is a complex cube root of unity,
Consider the LHS,
We know that 1 + ω + ω2 = 0 and ω3 = 1
Now by simplifying we get,
Again by substituting 1 + ω + ω2 = 0 and ω3 = 1 in above matrix we get,
Therefore LHS = RHS
Hence the proof.
Solution:
Given,
Consider A2
Therefore A2 = A
Solution:
Given
Consider A2,
Hence A2 = I3
Solution:
(i) Given
= 2x+1+2+x+3 = 0
= 3x+6 = 0
= 3x = -6
x = -6/3
x = -2
(ii) Given,
On comparing the above matrix we get,
x = 13
Solution:
Given
⇒ (2x+4)x+4(x+2)–1(2x+4) = 0
⇒ 2x2 + 4x + 4x + 8 – 2x – 4 = 0
⇒ 2x2 + 6x + 4 = 0
⇒ 2x2 + 2x + 4x + 4 = 0
⇒ 2x (x + 1) + 4 (x + 1) = 0
⇒ (x + 1) (2x + 4) = 0
⇒ x = -1 or x = -2
Hence, x = -1 or x = -2
Solution:
Given
By multiplying we get,
Solution:
Given
Now we have to prove A2 – A + 2 I = 0
Solution:
Given
Solution:
Given
Hence the proof.
Solution:
Given
Hence the proof.
Solution:
Given
Solution:
Given
Solution:
Given
Solution:
Given
Solution:
Given
Solution:
Given
I is identity matrix, so
Also given,
Now, we have to find A2, we get
Now, we will find the matrix for 8A, we get
So,
Substitute corresponding values from eqn (i) and (ii), we get
And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal
Hence,
Therefore, the value of k is 7
Solution:
Given
To show that f (A) = 0
Substitute x = A in f(x), we get
I is identity matrix, so
Now, we will find the matrix for A2, we get
Now, we will find the matrix for 2A, we get
Substitute corresponding values from eqn (ii) and (iii) in eqn (i), we get
So,
Hence Proved
Solution:
Given
So
Now, we will find the matrix for A2, we get
Now, we will find the matrix for λ A, we get
But given, A2 = λ A + μ I
Substitute corresponding values from equation (i) and (ii), we get
And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal
Hence, λ + 0 = 4 ⇒ λ = 4
And also, 2λ + μ = 7
Substituting the obtained value of λ in the above equation, we get
2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1
Therefore, the value of λ and μ are 4 and – 1 respectively
39. Find the value of x for which the matrix product
Solution:
We know,
is identity matrix of size 3.
So according to the given criteria
Now we will multiply the two matrices on LHS using the formula cij = ai1b1j + ai2b2j + … + ain bnj, we get
And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal
So we get
So the value of x is
Chapter 5 Algebra of Matrices Ex 5.4
(i) (2A)T = 2 AT
(ii) (A + B)T = AT + BT
(iii) (A – B)T = AT – BT
(iv) (AB)T = BT AT
Solution:
(i) Given
Consider,
Put the value of A
L.H.S = R.H.S
(ii) Given
Consider,
L.H.S = R.H.S
Hence proved.
(iii) Given
Consider,
L.H.S = R.H.S
(iv) Given
So,
Solution:
Given
L.H.S = R.H.S
So,
(i) A + B)T = AT + BT
(ii) (AB)T = BT AT
(iii) (2A)T = 2 AT
Solution:
(i) Given
Consider,
L.H.S = R.H.S
So,
(ii) Given
Consider,
L.H.S = R.H.S
So,
(iii) Given
Consider,
L.H.S = R.H.S
So,
Solution:
Given
Consider,
L.H.S = R.H.S
So,
Solution:
Given
Now we have to find (AB)T
So,
Chapter 5 Algebra of Matrices Ex 5.5
Solution:
Given
Consider,
… (i)
… (ii)
From (i) and (ii) we can see that
A skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,
X = – XT
So, A – AT is a skew-symmetric.
Solution:
Given
Consider,
… (i)
… (ii)
From (i) and (ii) we can see that
A skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,
X = – XT
So, A – AT is a skew-symmetric matrix.
Solution:
Given,
is a symmetric matrix.
We know that A = [aij]m × n is a symmetric matrix if aij = aji
So,
Hence, x = 4, y = 2, t = -3 and z can have any value.
4. Let. Find matrices X and Y such that X + Y = A, where X is a symmetric and y is a skew-symmetric matrix.
Solution:
Given,
Then
Now,
Now,
X is a symmetric matrix.
Now,
-Y T = Y
Y is a skew symmetric matrix.
Hence, X + Y = A
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