TRACE OF A MATRIX
Thus if A = [aij]n×n
Then tr(A) = ∑i=1n aii = a11 + a22 + ... + ann
2 -1 6
3 1 4 is
(A) 2 (B) 4
(C) -2 (D) 1
Solution: tr (A) = 1 + (-1) + 4 = 4.
Diagonal Matrix
For a square matrix A = [aij]n×n to be a scalar matrix aij = {0, i ≠ j; m, i = j}, where m ≠ 0.
Thus a square matrix A = [aij]n×n is a unit matrix if aij = {1, i = j; 0, i ≠ j}
Triangular Matrix
Given a square matrix A = [aij]n×n, For upper triangular matrix, aij = 0, i > j and for lower triangular matrix, aij = 0, i < j
If A = [aij]m×n and AT = [bij]n×m then bij = aji, ∀ i, j.
Conjugate of a Matrix
(AT)‾ = (Ā)T = Aθ.
If A = [aij]m×n, then Aθ = [bji]n×m where bji = āij
i.e. the (j, i)th element of Aθ = the conjugate of (i, j)th element of A.
3 0 and B = 3 4
1 6 then (AB)T is
(A) 5 9
16 12 (B) 5 9
-16 12
(C) 5 9
16 12 (D) none of these
Solution: (C)
AB = 1×3+2×1 1×4+2×6
3×3+0×1 3×4+0×6 = 5 16
9 12
(AB)T = 5 9
16 12
Do Check: Differential Equation
ALGEBRA OF MATRICES
1 3 2 and B = 1 2
2 1
1 3, AB is
(A) equal to BA (B) not equal to BA
(C) unique matrix (D) none of these
Solution: (B)
A·B = 2 3 1
1 3 2 · 1 2
2 1
1 3 = 2+6+1 4+3+3
1+6+2 2+3+6 = 9 10
9 11
B·A = 1 2
2 1
1 3 · 2 3 1
1 3 2 = 2+2 3+6 1+4
4+1 6+3 2+2
2+3 3+9 1+6 = 4 9 5
5 9 4
5 12 7
Thus A·B ≠ B·A.
Exercise 1: (i) If A = 1 -2 4
2 3 2
3 1 5 and B = 0 -2 4
1 3 2
-1 1 5, then A + B is
(A) 1 -2 4
3 3 2
2 1 5 (B) 1 -2 8
3 3 4
2 1 10 (C) 1 -4 8
3 6 4
2 2 10 (D) none of these
(ii) If A2 = 8A + KI, where A = 1 0
-1 7, then k is
(A) 7 (B) -7 (C) 1 (D) -1
SPECIAL MATRICES
Unitary Matrix
For example: 0 -2+i
2-i 0, 3i -3+2i -1-i
3-2i -2i -2-4i
1+i 2+4i 0 are skew-Hermitian matrices.
a b+ic
b-ic d, 3 3-4i 5+2i
3+4i 5 2-i
5-2i 2+i 2 are Hermitian matrices.
Orthogonal Matrix
Idempotent Matrix
Nilpotent Matrix
5 2 6
-2 -1 -3 is nilpotent matrix of index
(A) 2 (B) 3
(C) 4 (D) 5
Solution: (C) Let A = 1 1 3
5 2 6
-2 -1 -3
⇒ A2 = 0 0 0
3 3 9
-1 -1 -3
⇒ A3 = A2·A = 0 0 0
3 3 9
-1 -1 -3 · 1 1 3
5 2 6
-2 -1 -3 = 0 0 0
0 0 0
0 0 0
∴ A3 = 0 i.e. Ak = 0. Here k = 3. Hence A is nilpotent matrix of index 3.
Adjoint of a Square Matrix
Thus, adjA = [Cij]T ⇒ (adj A)ij = Cji. where Cij denotes the cofactor of aij in A.
∴ adj A = s -r
-q pT = s -q
-r p.
(i). adj (adj A) = |A|n-2 A, where A is a non-singular square matrix.
(ii). |adj (adj A)| = |A|(n-1)², where A is a non-singular square matrix.
2 2 0
2 2 2, then adj (adj A) is
(A) 1 0 0
16 1 1 0
1 1 1 (B) 1 0 0
8 1 1 0
1 1 1
(C) 1 0 0
4 1 1 0
1 1 1 (D) none of these
Solution: (B) adj (adj A) = |A|3-2 A = |A|·A
|A| = 2 0 0
2 2 0
2 2 2 = 8
∴ adj (adj A) = 2 0 0
8 2 2 0
2 2 2 = 16 0 0
16 16 0
16 16 16 = 1 0 0
8 1 1 0
1 1 1.
Inverse of a Matrix
The inverse of A is given by A-1 = (1/|A|) · adj A.
(i). (Reversal Law) If A and B are invertible matrices of the same order, then AB is invertible and (AB)-1 = B-1A-1.
(ii). If A is an invertible square matrix, then AT is also invertible and (AT)-1 = (A-1)T.
(iii). If A is a non-singular square matrix of order n, then |adjA| = |A|n-1.
(iv). The inverse of the inverse of the matrix is the original matrix itself, i.e. (A-1)-1 = A.
(v). If A is an invertible square matrix, then adj(AT) = (adj A)T.
(vi). If A is a non-singular matrix, then |A-1| = |A|-1.
c 1+bc/a is
(A) 1+bc -b
a
-c a (B) 1+bc -b
a
c a
(C) 1+bc -b
a
-c a (D) none of these
Solution: Given A = a b
c 1+bc/a
|A| = a(1 + bc/a) - bc = 1 + bc - bc = 1 ⇒ |A| ≠ 0 ⇒ A-1 exists
A-1 = (1/|A|) adjA = (1/1) 1+bc -b
-c a
ELEMENTARY TRANSFORMATIONS OR ELEMENTARY OPERATIONS OF A MATRIX
(i). Interchange of any two rows (columns)
If ith row (column) of a matrix is interchanged with the jth row (column), it will be denoted by Ri ↔ Rj (Ci ↔ Cj).
(ii). Multiplying all elements of a row (column) of a matrix by a non-zero scalar.
If the elements of ith row (column) are multiplied by non-zero scalar k, it will be denoted by Ri → Ri (k) [Ci → Ci(k)] or Ri → kRi [Ci → kCi]
(iii). Adding to the elements of a row (column), the corresponding elements of any other row (column) multiplied by any scalar k
If k times the elements of jth row (column) are added to the corresponding elements of the ith row (column), it will be denoted by Ri → Ri +k Rj (Ci → Ci + kCj)
An elementary transformation is called a row transformation or a column transformation according as it is applied to rows or columns.
2 0 1
3 5 0 is
(A) -5/8 -5/4 1/8
3/8 3/4 1/8
-5/4 -3/2 -1/4 (B) -5/8 -5/4 1/8
3/8 3/4 1/8
5/4 3/2 1/4
(C) -5/8 -5/4 1/8
3/8 3/4 1/8
5/4 3/2 1/4 (D) none of these
Solution: (A) We have A = IA
or 3 -1 -2
2 0 1
3 5 0 = 1 0 0
0 1 0
0 0 1A
[Through step-by-step row operations...]
Hence A-1 = -5/8 -5/4 1/8
3/8 3/4 1/8
5/4 3/2 1/4
a11 x1 + a12 x2 + …+ a1n xn = b1
a21 x1 + a22 x2 + …+ a2n xn = b2
. . . .
. . . .
an1 x1 + an2 x2 + …+ ann xn = bn
This system of equation can be written in matrix form as
a11 a12 ...a1n
a21 a22 ...a2n
.
.
.
an1 an2 ...ann x1
x2
.
.
.
xn = b1
b2
.
.
.
bn
or AX = B
The n × n matrix A is called the coefficient matrix of the system of linear equations.
A set of values of the variables x1, x2,...,xn which simultaneously satisfy all the equations is called a solution of the system of equations.
Consistent System
(i). Cramer's Rule is discussed in the Chapter 'Determinants'
(ii). Matrix Method:
Consider the equations
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3 …(i)
If A = a1 b1 c1
a2 b2 c2
a3 b3 c3, X = x
y
z and D = d1
d2
d3, then the equation (i) is equivalent to the matrix equation
A X = D …(ii)
(A) x = 1, y = 2, z = -1 (B) x = 1, y = 2, z = 1
(C) x = -1, y = 2, z = 1 (D) x = 1, y = -2, z = -1
Solution: (A) We can write the given equations as
AX = B …(1)
Where, A = 3 1 2
2 -3 -1
1 2 1, X = x
y
z, B = 3
-3
4
Since, |A| = 3 1 2
2 -3 -1
1 2 1 = 3(-3+2)–1(2+1)+2(4+3) = -3 –3+14= 8 ≠ 0
From (1), we have X = A-1B …(2)
[After calculating cofactors and adjoint matrix...]
A-1 = (1/8) -1 3 5
-3 1 7
7 -5 -11
A-1B = (1/8) -1 3 5
-3 1 7
7 -5 -11 · 3
-3
4 = (1/8) -3-9+20
-9-3+28
21+15-44 = (1/8) 8
16
-8 = 1
2
-1
Hence, from (2) x
y
z = 1
2
-1 ⇒ x = 1, y = 2, z = -1.
Exercise 2: (i) If A = 2 0 -1
5 1 0
0 1 3, then A-1 – A2 + 6A – 11I is
(A) 11I (B) 10I (C) I (D) 0
(ii) The solution of the system of equation x + y – 2z = 1; x – y + z = 0; 2x + 3y – 4z = 2
(A) x = 1/3, y = -1, z = -1/3 (B) x = 1/3, y = 0, z = -1/3
(C) x = -1/3, y = 1, z = -1/3 (D) x = 1/3, y = 0, z = 1/3
(iii) If A = 0 -4 1
2 λ -3
1 2 -1, then A-1 exists (i.e. A is invertible) if
(A) λ ≠ 4 (B) λ ≠ 8
(C) λ = 4 (D) none of these
Exercise 2: (i) D (ii) B (iii) B
2. Properties of Transposes:
(i) (AT)T = A
(ii) (A + B)T = AT + BT, A and B being conformable matrices
(iii) (αA)T = αAT, α being scalar
(iv) (AB)T = BTAT, A and B being conformable for multiplication
3. Any square matrix A of order n is said to be orthogonal if AAT = ATA = In.
4. A square matrix A is called idempotent provided it satisfies the relation A2 = A.
5. A matrix such that A2 = I is called involutary matrix.
6. A square matrix A is called a nilpotent matrix if there exists a positive integer m such that Am = 0. If m is the least positive integer such that Am = 0, then m is called the index of the nilpotent matrix A.
7. The inverse of A is given by A-1 = (1/|A|) · adj A
2 4-x 1
2 4 1-x is singular
(A) x = 1, 2 (B) x = 0, 2
(C) x = 0, 1 (D) x = 0, 3.
Sol. (D). Since, the given matrix is singular
3-x 2 2
2 4-x 1
2 4 1-x = 0
R2 + R3 ⇒ 3-x 2 2
0 x -x
2 4 1-x = 0
3-x 2 2
x 0 1 1
2 4 1-x = 0
⇒ x {(3 – x) (1 + x – 4) – 0 + 2 (2 – 2)} = 0 ⇒ x (3 – x) (x – 3) = 0 ⇒ x = 0, 3.
-3 1 2
1 2 1, then A-1 is equal to
(A) -3/4 1/4 7/4
-1/4 1/4 5/4
5/4 1/4 13/4 (B) -3/4 1/4 5/4
-1/4 1/4 1/4
7/4 5/4 13/4
(C) -3 1 7
-1 1 5
5 1 13 (D) none of these
Sol. (A). Since, |A| = 2 5 3
-3 1 2
1 2 1 = 2(1 – 4) –5 (–3 –2 ) + 3(–6 –1 ) = –6 –5 + 15 =15 – 11 = 4 ≠ 0
∴ A is a non-singular matrix. Now,
A11 = -3, A12 = 1, A13 = -5
A21 = 1, A22 = -1, A23 = 1
A31 = 7, A32 = -5, A33 = 13
Let, B = -3 1 -5
1 -1 1
7 -5 13
adj (A) = BT = -3 1 7
1 -1 -5
-5 1 13
Hence A-1 = (adj A)/|A| = (1/4) -3 1 7
1 -1 -5
-5 1 13 = -3/4 1/4 7/4
1/4 -1/4 -5/4
-5/4 1/4 13/4
3 x²-2 4 1
-1 -2 x-3 1
2 0 4 x²-6 is 0, then x is equal to
(A) {–2, 3} (B) (–2, 3)
(C) {–3, 2} (D) (–3, 2)
Sol. (C). Trace of matrix is defined as ∑i=1n aii = 2x² + 2x –12 = 0 ⇒ x = –3, 2
(A) (AB)T = ATBT (B) adj(AB) = adj(A) adj(B)
(C) (AB)T = BTAT (D) AB = O ⇒ A = O or B = O
Sol. (C). Multiplication of square matrices is a square matrix of same order and by property of transpose.
Sol. (A). AB = -2 3 1
-1 2 1
-6 9 4 1 3 1
2 2 1
3 0 1 = 2+6+3 -4+2+6 -6+18+12
-6+6+0 -3+4+0 -18+18+0
-2+3+1 -1+2+1 -6+9+4 = 1 0 0
0 1 0
0 0 1
; BA = 1 3 1
2 2 1
3 0 1 -2 3 1
-1 2 1
-6 9 4 = -2+6+3 -4+2+6 -6+0+6
-3+6+9 -6+4+9 -9+0+9
-6+6+0 -2+2+4 -3+0+4 = 1 0 0
0 1 0
0 0 1
∴ AB = BA
-tanx 1, then the value of |ATA-1| is
(A) cos4x (B) sec2x
(C) – cos4x (D) 1
Sol. (D). AT = 1 -tanx
tanx 1 ; A-1 = (1/(1+tan²x)) 1 -tanx
tanx 1 ; ATA-1 = cos²x -sin²x
sin²x cos²x
| ATA-1| = 1
3 4 and A² – KA – 5I = 0, then K is equal to
(A) 5 (B) 3
(C) 7 (D) none of these
Sol. (A). Given A² = KA – 5I = 0 ; 10 15
15 25 = K1 3
3 4 - 5 0
0 5 = 5 15
15 20
; K1 3
3 4 = 5 15
15 20 + 5 0
0 5 = 10 15
15 25 = 51 3
3 4
∴ K = 5
tanθ 1 1 tanθ
-tanθ 1 = a -b
b a, then
(A) a = cos 2θ, b = sin 2θ (B) a = 1, b = 1
(C) a = sin 2θ, b = cos 2θ (D) None of these
Sol. (C). 1 -tanθ
tanθ 1 1 tanθ
-tanθ 1 = 1+tan²θ -2tanθ
2tanθ 1+tan²θ = 1/(1+tan²θ) -2tanθ/(1+tan²θ)
2tanθ/(1+tan²θ) 1/(1+tan²θ)
from given relation
∴ (1/(1+tan²θ)) 1-tan²θ -2tanθ
2tanθ 1-tan²θ = a -b
b a ⇒ a = (1-tan²θ)/(1+tan²θ) = cos2θ and
b = 2tanθ/(1+tan²θ) = sin2θ
2x+3 x+2 is a symmetric matrix, tan A is equal to
(A) 4 (B) 3
(C) -4 (D) -3
Sol. (C). A is symmetric ⇒ A = AT ; 3 x-1
2x+3 x+2 = 3 2x+3
x-1 x+2
⇒ x – 1 = 2x + 3 ⇒ x = -4
5 -4 2
2 2 8, then A (adj A) equals
(A) 36 -36 18
-36 36 -18
-18 18 9 (B) -36 36 18
-36 36 -18
-18 18 9
(C) 0 0 0
0 0 0
0 0 0 (D) none of these
Sol. (D). We know that A (adj A) = |A| I, so A (adj A) must be a scalar matrix.
1 1, then for all natural numbers n, An is equal to
(A) 1 0
n 1 (B) 1 0
n n
(C) 1 0
-n 1 (D) none of these
Sol. (A). A² = AA = 1 0
2 1 ; A³ = 1 0
3 1 ...... so on Hence, An = 1 0
n 1 ∀ n ∈ ℕ
ASSIGNMENT
1. 7 1 2
9 2 1 3
4
5 + 2 4
2 is equal to
(A) 43
44 (B) 43
45 (C) 45
44 (D) none of these
2. If A = 0 0 0
0 0 0
0 1 0, then A is
(A) an invertible matrix (B) an idempotent matrix
(C) a nilpotent matrix (D) none of these
3. If A = 1 2 -1
-1 1 2
2 -1 1, then det (adj (adjA)) is
(A) (14)4 (B) (14)3
(C) (14)2 (D) 14
4. Let A be a skew-symmetric matrix of order n then
(A) |A| = 0 if n is even (B) |A| = 0 if n is odd
(C) |A| = 0 for all n ∈ ℕ (D) none of these
5. If A is a skew-symmetric matrix, then trace of A is
(A) 0 (B) –1
(C) 1 (D) none of these
6. If A = a b
b a, A² = α β
β α, then
(A) α = 2ab, β = a² + b² (B) α = a² + b², β = a² - b²
(C) α = a² + b², β = 2ab (D) α = a² + b², β = ab
ANSWERS TO ASSIGNMENT
Frequently Asked Questions
A scalar matrix is a diagonal square matrix whose diagonal entries are all the same non-zero number m
. Example: diag(5, 5, 5)
is scalar. A unit (identity) matrix is a special scalar matrix with m = 1
. It’s denoted by In
. Every identity matrix is scalar, but not every scalar matrix is the identity (unless m = 1).
A square matrix A is invertible (non-singular) iff its determinant is non-zero: det(A) ≠ 0. If det(A) = 0, it is singular and has no inverse. Example: For A = [[a, b],[c, d]], A is invertible when ad − bc ≠ 0.
The trace of a square matrix is the sum of its main diagonal entries. If A
is n×n, then tr(A) = a11 + a22 + … + ann
. Transpose does not change the diagonal, so tr(A) = tr(AT)
.
A square matrix A
is nilpotent if there exists a positive integer m
such that Am = 0
(the zero matrix). The index of A
is the smallest such m
. Example: A = [[0,1],[0,0]] has A² = 0, so it’s nilpotent of index 2.
Write the system as AX = B
, where A
is the coefficient matrix.
- If
det(A) ≠ 0
, thenA
is invertible and the unique solution isX = A⁻¹B
. - If
det(A) = 0
, use ranks to decide consistency:- Consistent when
rank(A) = rank([A|B])
. If this common rank < number of unknowns, there are infinitely many solutions. - Inconsistent when
rank(A) ≠ rank([A|B])
(no solution).
- Consistent when