Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7
Solution & Step-by-Step Answer:
i. A = ∴ AT =
ii. A = ≤ft[{array}{ccc}
-4 & 0 & 5
2 & -6 & 1 \\
{array}
∴ AT= ≤ft[{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
{array}
[Note: Answer given in the textbook is AT= ≤ft[{array}{cc}
2 & -4 \\
6 & 0 \\
1 & 5
{array}. However, as per our calculation it is AT= ≤ft[{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
{array}. ]
Solution & Step-by-Step Answer:
A = [aij]3×3 = Given aij = 2 (i — j) ∴ a11 = 2(1-1) = 0, a12 = 2(1-2) = -2, a13 = 2(1-3) = -4, a21 = 2(2-1) = 2, a22 = 2(2-2) = 0, a23=2(2-3) = -2, a31 = 2(3-1) = 4, a32 = 2(3-2) = 2, a33=2(3-3) = 0 ∴ AT = -A and A = -AT ∴ A and AT both are skew-symmetric matrices.
Questionn 3.
If A = ≤ft[{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
{array}, prove that (2A)T= 2AT.
Solution:
From (i) and (ii), we get
(2A)T= 2AT
Solution & Step-by-Step Answer:
From (i) and (ii), we get (3A)T = 3AT
Solution & Step-by-Step Answer:
Solution & Step-by-Step Answer:
From (i) and (ii), we get (A + B)T = AT + BT [Note: The question has been modified.]
From (i) and (ii), we get
(A – C)T= AT– CT</sup
Solution & Step-by-Step Answer:
3A – 2B + C = I ∴ C = I + 2B – 3A
Solution & Step-by-Step Answer:
ii. ii. 5AT– 5BT= 5(AT– BT)
Solution & Step-by-Step Answer:
A + 2B + 3C ∴ AT + 2BT + 3CT = From (i) and (ii), we get (A + 2B + 3C)T = AT + 2BT + 3CT
Solution & Step-by-Step Answer:
From (i) and (ii), we get (A + BT)T = AT + B
Solution & Step-by-Step Answer:
∴ (A + AT)T = A + AT, i.e., A + AT = (A + AT)T ∴ A + AT is a symmetric matrix. ∴ (A – AT)T = – (A – AT), i.e., A – AT = -(A – AT)T ∴ A – AT is skew symmetric matrix.
∴ (A + AT)T= A + AT, i.e., A + AT= (A + AT)T
∴ A + ATis a symmetric matrix.
∴ (A – AT)T= – (A – AT),
i.e., A – AT= -(A – AT)T
∴ A – ATis skew symmetric matrix.
Solution & Step-by-Step Answer:
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as A = (A + AT) + (A – AT) P is symmetric matrix …[∵ aij = aji] and Q is a skew symmetric matrix [∵ -aij = -aji] A = P + Q A =
∴ P is symmetric matrix …[∵ aij= aji]
and Q is a skew symmetric matrix [∵ -aij= -aji]
∴ A = P + Q
∴ A = ≤ft[{array}{cc}
4 & \\
& -5
{array}]+≤ft[{array}{ll}
0 & \\
& 0
{array}
Solution & Step-by-Step Answer:
From (i) and (ii), we get (AB)T = BTAT From (i) and (ii) we get (BA)T = ATBT
Solution & Step-by-Step Answer:
∴ ATA = I, where I is the unit matrix of order 2.
