(i) a ij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2, n = 3
To construct 2 × 3 matrices.
(ii) a ij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3, n = 4
To construct a 3 × 4 matrices.
Equating the corresponding entries
⇒ p 2 – 1 = 1
⇒ p 2 = 1 + 1 = 2
p = ± \(\sqrt{2}\)
-31 – q 3 = -4
-q 3 = -4 + 31 = 27
q 3 = -27 = (-3) 3
⇒ q = -3
r + 1 = \(\frac{3}{2}\)
⇒ r = \(\frac{3}{2}\) – 1 = \(\frac{3-2}{2}\) = \(\frac{1}{2}\)
s – 1 = π
⇒ s = – π + 1 (i.e.,) s = 1 – π
So, p = ± \(\sqrt{2}\), q = -3, r = 1/2 and s = 1 – π
\(\left[\begin{array}{cc}{2 x+y} & {4 x} \\ {5 x-7} & {4 x}\end{array}\right]=\left[\begin{array}{cc}{7} & {7 y-13} \\ {y} & {x+6}\end{array}\right]\)
⇒ 2x + y = 7 ………….. (1)
4x = 7y – 13 ………….. (2)
5x – 7 = y …………… (3)
4x = x + 6 ……………. (4)
from (4) 4x – x = 6
3x = 6 ⇒ x = \(\frac{6}{3}\) = 2
Substituting x = 2 in (1), we get
2(2) + y = 7 ⇒ 4 + y = 7 ⇒ y = 7 – 4 = 3
So x = 2 and y = 3
∴ x + y = 2 + 3 = 5
(ii) Find all possible real values α satisfying the condition A α + A α T = I
Equating the corresponding entries – 2 + k = 0 ⇒ k = 2
∴ The required value of k is k = 2
(i) A and B such that AB ≠ BA
(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.
(iii) A and B such that AB = 0 and BA ≠ 0
Given A 2 = A
So 7A – (I + A) 3 = 7A – (I + 3A + 3A 2 + A 3 ]
= 7A – I – 3A – 3 A 2 – A 3
Given A 2 = A
7A – I – 3A – 3A – A 3 = -I + A – A 3
= -I + A – (A 2 × A)
= -I + A – (A × A) = -I + A – A 2
= -I + A – A = -I
So the value of 7A – (I + A) 3 = -I.
(i) (A + B) T = A T + B T = B T + A T
(ii) (A – B) T = A T – B T
(iii) (B T ) T = B
A is a matrix of order 3 × 4
So AT will be a matrix of order 4 × 3
AT B will be defined when B is a matrix of order 3 × n
BA T will be defined when B is of order m × 4
from (1) and (2) we see that B should be a matrix of order 3 × 4
A = \(\frac{1}{2}\)(A + A T ) + \(\frac{1}{2}\)(A – A T
Thus A is expressed as a sum of a symmetric and skew-symmetric matrix.
(ii) \(\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right] \)
Equating like entries
2a – d = -1, 2b – e = – 8, 2c – f = -10
a = 1, b = 2, c = -5
2a – d = -1 ⇒ 2 × 1 – d = – 1
⇒ 2 + 1 = d ⇒ d = 3
2b – e = – 8 ⇒ 2 × 2 – e = – 8
⇒ 4 + 8 = e ⇒ e = 12
2c – f = -10 ⇒ 2 × – 5 – f = -10
⇒ – 10 – f = -10 ⇒ f = 0
Equating the corresponding entries
x + 4 + 2y = 0 ………… (1)
2x + 2 – 2y = o ………… (2)
x + 4 + 2y = 0 ………… (3)
2x + 2 – 2y = 0 ………… (4)
x 2 + 4 + y 2 = 9 ………… (5)
Substituting the value of y in equation (1) we have
x + 4 + 2x – 1 = 0
x + 4 – 2 = 0 ⇒ x = – 2
Substituting x = – 2 and y = – 1 in equation(5) we have
(5) ⇒ (-2) 2 + 4 + (- 1) 2 = 9
4 + 4 + 1 = 9
9 = 9
∴ The required values of x and y are
x = – 2 and y = – 1
The matrix A is skew-symmetric if A = – A T
Equating the corresponding entries
x 3 – 3 = 0
x 3 = 3 ⇒ x = 3 1/3
(ii) If \(\left[ \begin{matrix} 0 & p & 3 \\ 2 & { q }^{ 2 } & -\quad 1 \\ r & 1 & 0 \end{matrix} \right] \) is skew – symmetric find the values of p, q and r.
A is skew-symmetric if A = – A T
Equating the corresponding entries.
p = – 2, r = – 3
q 2 = -q 2 ⇒ q 2 + q 2 = 0
⇒ 2q 2 = 0 ⇒ q = 0
∴ The required values are
p = – 2, q = 0, r = – 3
a ij = i – j
a 11 = 1 – 1 = 1
a 12 = 1 – 2 = – 1
a 13 = 1 – 3 = – 2
a 21 = 2 – 1= 1
a 22 = 2 – 2 = 1
a 23 = 2 – 3 = – 1
a 31 = 3 – 1= 2
a 32 = 3 – 2 = 1
a 33 = 3 – 3 = 0
Given A and B two symmetric matrices.
∴ A = A T and B = B T
First, let us assume AB = BA.
Let us prove AB is a symmetric matrix.
∴ AB is a symmetric matrix.
conversely let us assume that AB is a symmetric matrix.
we prove AB = BA
AB is symmetric then
Given A and B are symmetric matrices
⇒ – A T = A and B T = B
(i) To prove AB + BA is a symmetric matrix.
Proof: Now (AB + BA) T = (AB) T + (BA) T = B T A T + A T B T
= BA + AB = AB + BA
i.e. (AB + BA) T = AB + BA
⇒ (AB + BA) is a symmetric matrix.
(ii) To prove AB – BA is a skew symmetric matrix.
Proof: (AB – BA) T = (AB) T – (BA) T = B T A T – A T B T = BA – AB
i.e. (AB – BA) T = – (AB – BA)
⇒ AB – BA is a skew symmetric matrix.
Let us consider 50 gm of cashew nuts as one unit, 50 gms of raisins as one unit 50 gm of almonds as one unit.
∴ The Gift pack matrix becomes
Also given 50 gms of Cashew nuts cost = Rs. 50
50 gms of Raisins cost = Rs. 10
50 gms of Almonds cost = Rs. 60
∴ Cost matrix is B = [50 10 60]
∴ Cost of cash gift pack = BA
∴ Cost of I gifts Pack = Rs. 180
Cost of II gift Pack = Rs.340
Cost of III gift Pack = Rs. 480
= s (a 2 + b 2 + c 2 ) × 0
since two columns are equal.
= 0
= 2abc [0 – b(0 – ac) + c(ab – 0)]
= 2 abc [ abc + abc ]
= 2 abc × 2abc
Δ = 4 a 2 b 2 c 2
= a [b(1 + c) + c (1)] – 0 – c [0 – b]
= a[b + bc + c] + bc
= ab + abc + ac + bc
= abc + ab + bc + ac
= abc
Let Δ = \(\left| \begin{matrix} x\quad +\quad 2a & y\quad +\quad 2b & z\quad +\quad 2c \\ x & y & z \\ a & b & c \end{matrix} \right| \)
A square matrix A = [ a ij ]3 × 3 is a skew – symmetric matrix if a ij = – a ij for all i,j and the elements on the main diagonal of a skew – symmetric matrix are zero.
= 0 – a 12 (0 + a 13 a 23 ) + a 13 (a 12 a 23 – 0)
= – a 12 a 13 a 23 + a 13 a 12 a 23
= 0
Hence the determinant of a skew – symmetric matrix is 0.
Given a, b, c are p th, q th and r th terms of an A.P.
t p = a = A + (p – 1)D,
t q = b = A + (q – 1)D,
t r = c = A + (r – 1) D
where A – first term, D – Common difference of the AP.
Given a, b, c are the p th, q th and r th terms of a G.P.
∴ a = AR p-1, b = AR q-1, c = AR r-1
where A is the first term, R – common ratio.
(i) \(\left| \begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} x\quad +\quad y & y\quad +\quad z & z\quad +\quad x \\ z & x & y \\ 1 & 1 & 1 \end{matrix} \right| [
|A| = 2 (Given) |A T | = 2
Now |AA T | = |A| |A T | = 2 × 2 = 4.
Given |A| = -1: |B| = 3
Given A and B are square matrices of order 3.
∴ |kAB| = k 3 |AB|
Here k = 3 ∴ |3AB| = 3 3 |AB|
= 27 |AB|
= 27 (-1) (3)
= -81
Expanding along the first row
Δ = 0 + 4 [4 × 0 – (- 1 ) ( 13)] + [4 × -13 – 0 × – 1]
= 4 [0 + 13] + 1 [- 52 + 0]
= 52 – 52 = 0
\(\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & { 5x }^{ 2 } \end{matrix} \right| \) = 0 ………… (1)
Put x = -1 then (1) ⇒
∴ x = – 1 satisfies equation (1)
Hence x = – 1 is a root of equation (1)
Put x = 2 then ……….. (1)
[Property 4: If two rows (columns) of a determinant are identical then its determinant value is zero.]
∴ x = 2 satisfies equation (1)
Hence x = 2 is a root of equation (1)
Hence the required roots are x = -1, 2
= 4 [-10 (0 – 9 × 19) – 5 (0 + 17 × 19) + 1 (32 × 9 + 17 × 26)]
= 4 [1710 – 5 × 323 + 288 + 442]
= 4 [1710 – 1615 + 730]
= 4 [2440 – 1615]
= 4 × 825
det (AB) = 3300 …….. (1)
= 4(0 – 21) – 3 (- 5 – 14) – 2 (3 – 0)
= -84 – 3 × – 19 – 6
= -84 + 57 – 6
= -90 + 57
det A = -33 ………… (2)
= 1 (20 – 0) – 3 (- 10 – 0) + 3 (-14 – 36)
= 20 + 30 + 3 × – 50
= 50 – 150
det A = – 100 ……….. (3)
From equations (2) and (3)
(det A) (det B) = – 33 × – 100
(detA) (det B) = 3300 ………… (4)
From equations (1) and (4), we have
det (AB) = (det A) (det B)
|A| = a 21 A 21 + a 22 A 22 + a 23 A 23
= 2 × 7 + 0 × 7 + 1 × – 7
= 14 – 7
|A| = 7
By putting x = a, we have three rows of |A| are identical. Therefore (x – a) 2 is a factor of |A|
Put x = – 2a in |A|
∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a) 2 (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x. x. x is 3.
∴ The other factor is the contant factor k.
a 3 [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k. 4a 3
a 3 [o + 2 + 2 ] = 4 ka 3
4a 3 = 4 ka 3
k = 1
Free handy Remainder Theorem Calculator tool displays the remainder of a difficult polynomial expression in no time.
since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|
since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|
since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.
Put x = 0
x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)
∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0, – (a + b + c)
Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a, b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c). b. c 2 is 4.
Therefore, the other factor is k (a + b + c).
5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1
Put x = 0
∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.
Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.
∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0, 0, – 12
|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.
The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z 2 is 3. Therefore, the other factor is the constant factor k.
Put x = 0, y = 1, z = -1 we get
Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1
The given points are (0, 0), (1, 2) and (4, 3)
Area of the triangle with vertices
(x 1, y 1 ), (x 2, y 2 ) and (x 3, y 3 ) is
∴ The area of the triangle with vertices
(0, 0), (1, 2) and (4, 3) is
Area cannot be negative. Taking positive value, we have
Required area Δ = \(\frac{5}{2}\) sq.units.
Given Area of the triangle with vertices (k, 2), (2, 4) and (3, 2) is 4 square units.
The area of the triangle with vertices
(x 1, y 1 ), (x 2, y 2 ) and (x 3, y 3 ) is
Given Δ = 4, (x 1, y 1 ) = (k, 2), (x 2, y 2 ) = (2, 4) and (x 3, y 3 ) = (3, 2)
± 4 = k(4 – 2) – 2 (2 – 3) + 1(4 – 12)
± 4 = k × 2 – 2 × – 1 – 8
± 4 = 2k + 2 – 8
± 4 = 2k – 6
2k – 6 = 4 or 2k – 6 = -4
2k = 4 + 6 or 2k = – 4 + 6
2k = 10 or 2k = 2
k = 5 or k = 1
Required values of k are 1, 5.
(i) \(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right] \)
|A| = 1 (45 – 48) – 2(36 – 42) + 3(32 – 35)
|Al = – 3 – 2 × – 6 + 3 × – 3
|A| = – 3 + 12 – 9
|A| = – 12 + 12 = 0
∴ A is a singular matrix.
(ii) \(\left[ \begin{matrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{matrix} \right] \)
|B| = 2(0 – 20) + 3 (- 42 – 4) + 5(30 – 0)
|B| = -40 + 3 × – 46 + 150
|B| = -40 – 138 + 150
|B| = -178 + 150 ≠ 0
∴ B is non singular.
(iii) \(\left[ \begin{matrix} 0 & a\quad -\quad b & k \\ b-\quad a & 0 & 5 \\ -k & -5 & 0 \end{matrix} \right] \)
|C| = 0 – (a – b) (0 + 5k) + k(-5 (b – a) – 0)
|C| = -5k (a – b) – 5k (b – a)
|C| = -5k (a – b) + 5k(a – b)
|C| = o
∴ C is a singular matrix.
(i) A = \(\left[ \begin{matrix} 7 & 3 \\ -2 & a \end{matrix} \right] \)
|A| = \(\left[ \begin{matrix} 7 & 3 \\ -2 & a \end{matrix} \right] \)
|A| = 7a + 6
Given that A is singular
∴ |A| = 0
7a + 6 = 0 ⇒ a = \(\frac{-6}{7}\)
(ii) B = \(\left[ \begin{matrix} b\quad -\quad 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{matrix} \right] \)
|B| = \(\left[ \begin{matrix} b\quad -\quad 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{matrix} \right] \)
= (b – 1 )(4 + 4) – 2(12 – 2) + 3(- 6 – 1)
= 8 (b – 1) – 20 – 21
= 8b – 8 – 41
|B| = 8b -49
Given that B is singular
∴ |B| = 0
8b – 49 = 0 ⇒ b = \(\frac{49}{8}\)
Given cos 2θ = 0
(1)
Explaination:
(1) a scalar matrix
Explaination:
Let A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
(1) a scalar matrix – not true
(2) a diagonal matrix – true
(3) an upper triangular matrix – true
(4) a lower triangular matrix – true
[(1) A square matrix A = [a ij ] m × n is called a diagonal matrix if a ij = 0 whenever i ≠ j
(2) A diagonal matrix whose entries along the principle diagonal are equal is called a scalar matrix.
(3) A square matrix is said to be an upper triangular matrix if all the elements below the main diagonal are zero.
(4) A square matrix is said to be a lower triangular matrix if all the elements above the main diagonal are zero.]
(2) A and B are square matrices of same order
Explaination:
Given A and B are two matrices such that A + B and AB are both defined.
A + B defined means A and B are same order.
AB defined means, Number of columns of A equal to Number of rows of B.
A + B and AB are simultaneously defined.
Therefore A and B are of same order.
(2) ± 1
Explaination:
(2) a = 1, b = 4
Explaination:
L.2
Equating the like entrices
(a – 1) 2 = a 2 + b – 1 ………. (3)
o = a – 1 ………. (4)
2a+ 2 + ab + b = ab – b ………. (5)
4 = b ………. (6)
a = 1, b = 4
(4) (- 2, – 1)
Explaination:
Equating the like entries
a + 2b + 4 = 0 …………. (1)
2a – 2b + 2 = 0
a – b + 1 = 0 …………. (2)
a 2 + b 2 + 4 = 9 …………. (3)
3b = – 3 ⇒ b = -1
Substituting in equation (1) we get
a + 2 × – 1 + 4 = 0
a – 2 + 4 = 0
a = – 2
Substituting the values a = – 2, b = – 1 in equation (3)
we have
(- 2) 2 + (-1) 2 + 4 = 9
4 + 1 + 4 = 9
9 = 9
∴ The required value of the ordered pair (a, b) is
(a, b) = (- 2, – 1)
(4) AA T
Explaination:
Given A is a square matrix.
A square matrix A is symmetric if A T = A
(1) A + A T
(A + A T ) T = A T + (A T ) T
= A T + A = A + A T
∴ A + A T is symmetric.
(2) AA T
(AA T ) T = (A T ) T A T
= AA T
∴ AA T is symmetric.
(3) A T A
(A T A) T = A T (A T ) T
= A T A
∴ A T A is symmetric.
(4) A – A T
(A – A T ) T = A T – (A T ) T
= A T A
∴ A – A T is not symmetric.
(2) A + B is symmetric
Explaination:
Given A and B are symmetric matrices of order n.
∴ A T = A and B T = B
A matrix A is skew symmetric if A T = – A
(1)(A + B) T = A T + B T = A + B
A + B is not skew symmetric.
(2)(A + B) T = A T +B T = A + B
∴ A + B is symmetric.
(3) A + B is a diagonal matrix is incorrect.
(4) A + B is a zero matrix is incorrect.
(4) (a 2 – 1) 2
Explaination:
= (a 2 + x 2 ) (y 2 + a 2 ) – (ax + ay) (ax + ay)
= a 2 y 2 + a 4 + x 2 y 2 + a 2 x 2 – ((ax) 2 + (ay) 2 + 2 (ax)(ay))
= a 2 x 2 + a 2 y 2 + a 4 + (xy) 2 – a 2 x 2 – a 2 y 2 – 2a 2 xy
= a 4 + (1) 2 – 2a 2 (1)
= a 4 – 2a 2 + 1
|AA T | = (a 2 – 1) 2
(2) 8
Explaination:
(4) 3
Explaination:
Let the given points be (x 1, y 1 ) = (x, – 2),
(x 2, y 2 ) = (5, 2) and (x 3, y 3 ) = (8, 8)
The condition for the three points (x 1, y 1 ), (x 2, y 2 ) and (x 3, y 3 ) to be collinear is
\(\frac{1}{2}\) [x(2 – 8) + 2(5 – 8) + 1(40 – 16)] = 0
x × – 6 + 2 × – 3 + 1 × 24 = 0
– 6x – 6 + 24 = 0
– 6x + 18 = 0
6x = 18 ⇒ x = \(\frac{18}{6}\) = 3
x = 3
(3) \(\frac{1}{8}\)
Explaination:
(1) 1 + α 2 + βγ = 0
Explaination:
– α 2 – βγ = 1
α 2 + βγ + 1 = 0
(3) 0
Explaination:
0 = 0
x = 0 satisfies the given equation.
Hence the root of the given equation is x = 0.
(3) 0
Explaination:
= 0 – a(0 – bc) – b (ac – 0)
= abc – abc = 0
(4) collinear
Explaination:
Given x 1, x 2, x 3, as well as y 1, y 2, y 3, are in geometric progression with the same common ratio.
∴ x 1 = a, x 2 = ar, x 3 = ar 2,
y 1 = b, y 2 = br, y 3 = br 2
(x 1, y 1 ) = (a, b),
(x 2, y 2 ) = (ar,br)
and (x 3, y 3 ) = (ar 2, br 2 )
Area of the triangle whose vertices are
(x 1, y 1 ),(x 2, y 2 ) and (x 3 y 3 ) is
= \(\frac{1}{2}\)ab[1(r – r 2 ) – 1 (r – r 2 ) + 1 (r 3 – r 3 )]
= \(\frac{1}{2}\)ab[r – r 2 – r + r 2 + 0]
= \(\frac{1}{2}\)ab × 0 = 0
∴ The points (x 1, y 1 ),(x 2, y 2 ) and (x 3 y 3 ) are collinear.
(1)
Explaination:
(3) b 3
Explaination:
(a – 6) (b 2 – ac) = 0
Since a ≠ 6, we have a – 6 ≠ 0
∴ b 2 – ac = 0
b 2 = ac
b 3 = abc
(2) B = – 4A
Explaination:
(3) a zero matrix of order 1
Explaination:
Given A is a skew symmetric matrix of order n.
∴ A T = – A
C is a column matrix of order n × 1
C T is of order 1 × n
∴ C T A is of order (1 × n) × (n × n) = (1 × n)
C T AC is of order (1 × n) × (n × 1) = (1 × 1)
Since A is a skew – symmetric matrix, we have
A T = -A
(C T AC) T = C T A T (C T ) T = C T (-A) C
= -C T AC
∴ C T AC is a skew – symmetric matrix.
A matrix of order 1 is skew – symmetric if A is zero matrix.
Since C T AC is a skew – symmetric and its order is 1.
∴ C T AC is a zero matrix.
(3)
Explaination:
x + 3z = 1 ——– (1)
y + 3t = 1 ——- (2)
z = 0
t = – 1
(1) ⇒ x + 3 × 0 = 1 ⇒ x = 1
(2) ⇒ y + 3 × – 1 = 1 ⇒ y = 4
∴ A = \(\left[ \begin{matrix} 1 & 4 \\ 0 & -1 \end{matrix} \right] \)
(1)
Explaination:
(2) AB is a symmetric matrix
Explaination:
Given A and B are two symmetric matrices of the same order.
A = A T, B = B T
(1)(A+B) T = A T + B T = A + B
A + B is symmetric.
(2) (AB) T = B T A T BA
Thus (AB) T ≠ AB
Hence, AB is not symmetric.
(3) AB = (BA) T
= A T B T = AB
Statement is true.
(4) A T B = AB T Since A T = A
B = B T
Statement is true.