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Exercise 2.4 [Pages 59 - 60] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices

Exercise 2.4 [Pages 59 - 60]

Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices Exercise 2.4 [Pages 59 - 60]

Exercise 2.4 | Q 1.1 | Page 59

QUESTION

Find AT,  if A = $\left[\begin{array}{cc}1& 3\\ -4& 5\end{array}\right]$

SOLUTION

A = $\left[\begin{array}{cc}1& 3\\ -4& 5\end{array}\right]$

∴ AT = $\left[\begin{array}{cc}1& -4\\ 3& 5\end{array}\right]$.

Exercise 2.4 | Q 1.2 | Page 59

QUESTION

Find AT, if A = $\left[\begin{array}{ccc}2& -6& 1\\ -4& 0& 5\end{array}\right]$

SOLUTION

A = $\left[\begin{array}{ccc}2& -6& 1\\ -4& 0& 5\end{array}\right]$

∴ AT = $\left[\begin{array}{cc}2& -4\\ -6& 0\\ 1& 5\end{array}\right]$.

Exercise 2.4 | Q 2 | Page 59

QUESTION

If [aij]3×3, where aij = 2(i – j), find A and AT. State whether A and AT both are symmetric or skew-symmetric matrices?

SOLUTION

A = ${\left[{\text{a}}_{\text{ij}}\right]}_{3×3}=\left[\begin{array}{ccc}{\text{a}}_{11}& {\text{a}}_{12}& {\text{a}}_{13}\\ {\text{a}}_{21}& {\text{a}}_{22}& {\text{a}}_{23}\\ {\text{a}}_{31}& {\text{a}}_{32}& {\text{a}}_{33}\end{array}\right]$

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

∴ A = $\left[\begin{array}{ccc}0& -2& -4\\ 2& 0& -2\\ 4& 2& 0\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}0& 2& 4\\ -2& 0& -2\\ -4& -2& 0\end{array}\right]$

= $-\left[\begin{array}{ccc}0& -2& -4\\ 2& 0& -2\\ 4& 2& 0\end{array}\right]=-\text{A}$

∴ AT = – A and A = – AT
∴ A and AT both are skew-symmetric matrices.

Exercise 2.4 | Q 3 | Page 59

QUESTION

If A = $\left[\begin{array}{cc}5& -3\\ 4& -3\\ -2& 1\end{array}\right]$, prove that (AT)T = A.

SOLUTION

A = $\left[\begin{array}{cc}5& -3\\ 4& -3\\ -2& 1\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}5& 4& -2\\ -3& -3& 1\end{array}\right]$

∴ (AT)T = $\left[\begin{array}{cc}5& -3\\ 4& -3\\ -2& 1\end{array}\right]$
= A.

Exercise 2.4 | Q 4 | Page 59

QUESTION

If A = $\left[\begin{array}{ccc}1& 2& -5\\ 2& -3& 4\\ -5& 4& 9\end{array}\right]$, prove that AT = A.

SOLUTION

A = $\left[\begin{array}{ccc}1& 2& -5\\ 2& -3& 4\\ -5& 4& 9\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}1& 2& -5\\ 2& -3& 4\\ -5& 4& 9\end{array}\right]$
=A.

Exercise 2.4 | Q 5.1 | Page 59

QUESTION

If A = $\left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right],\text{B}=\left[\begin{array}{cc}2& 1\\ 4& -1\\ -3& 3\end{array}\right],\text{C}=\left[\begin{array}{cc}1& 2\\ -1& 4\\ -2& 3\end{array}\right]$, then show that (A + B)T = AT + BT.

SOLUTION

A = $\left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]+\left[\begin{array}{cc}2& 1\\ 4& -1\\ -3& 3\end{array}\right]$

= $\left[\begin{array}{cc}2+2& -3+1\\ 5+4& -4-1\\ -6-3& 1+3\end{array}\right]$

= $\left[\begin{array}{cc}4& -2\\ 9& -5\\ -9& 4\end{array}\right]$

∴ (A + B)T = $\left[\begin{array}{ccc}4& 9& -9\\ -2& -5& 4\end{array}\right]$        ...(i)

Now, AT =

∴ AT + BT = $\left[\begin{array}{ccc}2& 5& -6\\ -3& -4& 1\end{array}\right]+\left[\begin{array}{ccc}2& 4& -3\\ 1& -1& 3\end{array}\right]$

= $\left[\begin{array}{ccc}2+2& 5+4& -6-3\\ -3+1& -4-1& 1+3\end{array}\right]$

= $\left[\begin{array}{ccc}4& 9& -9\\ -2& -5& 4\end{array}\right]$         ...(ii)
From (i) and (ii, we get
(A + B)T = AT + BT.

Exercise 2.4 | Q 5.2 | Page 59

QUESTION

If A = $\left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right],\text{B}=\left[\begin{array}{cc}2& 1\\ 4& -1\\ -3& 3\end{array}\right],\text{C}=\left[\begin{array}{cc}1& 2\\ -1& 4\\ -2& 3\end{array}\right]$, then show that (A – C)T = AT – CT.

SOLUTION

A – C = $\left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ -1& 4\\ -2& 3\end{array}\right]$

= $\left[\begin{array}{cc}2-1& -3-2\\ 5+1& -4-4\\ -6+2& 1-3\end{array}\right]$

= $\left[\begin{array}{cc}1& -5\\ 6& -8\\ -4& -2\end{array}\right]$

∴ (A – C)T = $\left[\begin{array}{ccc}1& 6& -4\\ -5& -8& -2\end{array}\right]$              ...(i)

Now, AT = $\left[\begin{array}{ccc}2& 5& -6\\ -3& -4& 1\end{array}\right]$ and

CT = $\left[\begin{array}{ccc}1& -1& -2\\ 2& 4& 3\end{array}\right]$

∴ AT –  CT = $\left[\begin{array}{ccc}2& 5& -6\\ -3& -4& 1\end{array}\right]-\left[\begin{array}{ccc}1& -1& -2\\ 2& 4& 3\end{array}\right]$

= $\left[\begin{array}{ccc}2-1& 5+1& -6+2\\ -3-2& -4-4& 1-3\end{array}\right]$

= $\left[\begin{array}{ccc}1& 6& -4\\ -5& -8& -2\end{array}\right]$          ...(ii)

From (i) and (ii), we get
(A – C)T = AT – CT.

Exercise 2.4 | Q 6 | Page 59

QUESTION

If A = $\left[\begin{array}{cc}5& 4\\ -2& 3\end{array}\right]$ and B = $\left[\begin{array}{cc}-1& 3\\ 4& -1\end{array}\right]$, then find CT, such that 3A – 2B + C = I, where I is e unit matrix of order 2.

SOLUTION

3A – 2B + C = I
∴ C = I + 2B – 3A

∴ C = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+2\left[\begin{array}{cc}-1& 3\\ 4& -1\end{array}\right]-3\left[\begin{array}{cc}5& 4\\ -2& 3\end{array}\right]$

= $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}-2& 6\\ 8& -2\end{array}\right]-\left[\begin{array}{cc}15& 12\\ -6& 9\end{array}\right]$

= $\left[\begin{array}{cc}1-2-15& 0+6-12\\ 0+8+6& 1-2-9\end{array}\right]$

∴ C = $\left[\begin{array}{cc}-16& -6\\ 14& -10\end{array}\right]$

∴ C = $\left[\begin{array}{cc}-16& 14\\ -6& -10\end{array}\right]$.

Exercise 2.4 | Q 7.1 | Page 59

QUESTION

If A = $\left[\begin{array}{ccc}7& 3& 0\\ 0& 4& -2\end{array}\right],\text{B}=\left[\begin{array}{ccc}0& -2& 3\\ 2& 1& -4\end{array}\right]$, then find AT + 4BT.

SOLUTION

A =

∴ AT =

AT + 4BT = $\left[\begin{array}{cc}7& 0\\ 3& 4\\ 0& -2\end{array}\right]+4\left[\begin{array}{cc}0& 2\\ -2& 1\\ 3& -4\end{array}\right]$

= $\left[\begin{array}{cc}7& 0\\ 3& 4\\ 0& -2\end{array}\right]+\left[\begin{array}{cc}0& 8\\ -8& 4\\ 12& -16\end{array}\right]$

= $\left[\begin{array}{cc}7+0& 0+8\\ 3-8& 4+4\\ 0+12& -2-16\end{array}\right]$

= $\left[\begin{array}{cc}7& 8\\ -5& 8\\ 12& -18\end{array}\right]$.

Exercise 2.4 | Q 7.2 | Page 59

QUESTION

If A = $\left[\begin{array}{ccc}7& 3& 0\\ 0& 4& -2\end{array}\right],\text{B}=\left[\begin{array}{ccc}0& -2& 3\\ 2& 1& -4\end{array}\right]$, then find 5AT – 5BT.

SOLUTION

A = $\left[\begin{array}{ccc}7& 3& 0\\ 0& 4& -2\end{array}\right]\text{and}\text{B}=\left[\begin{array}{ccc}0& -2& 3\\ 2& 1& -4\end{array}\right]$

∴ AT = $\left[\begin{array}{cc}7& 0\\ 3& 4\\ 0& -2\end{array}\right]\text{and}{\text{B}}^{\text{T}}=\left[\begin{array}{cc}0& 2\\ -2& 1\\ 3& -4\end{array}\right]$

5AT – 5B = 5(AT – BT)

= $5\left(\left[\begin{array}{cc}7& 0\\ 3& 4\\ 0& -2\end{array}\right]-\left[\begin{array}{cc}0& 2\\ -2& 1\\ 3& -4\end{array}\right]\right)$

= $5\left[\begin{array}{cc}7-0& 0-2\\ 3+2& 4-1\\ 0-3& -2+4\end{array}\right]$

= $5\left[\begin{array}{cc}7& -2\\ 5& 3\\ -3& 2\end{array}\right]$

= $\left[\begin{array}{cc}35& -10\\ 25& 15\\ -15& 10\end{array}\right]$.

Exercise 2.4 | Q 8 | Page 59

QUESTION

If A = , verify that (A + 2B + 3C)T = AT + 2BT+ CT.

SOLUTION

A + 2B + 3C

= $\left[\begin{array}{ccc}1& 0& 1\\ 3& 1& 2\end{array}\right]+2\left[\begin{array}{ccc}2& 1& -4\\ 3& 5& -2\end{array}\right]+3\left[\begin{array}{ccc}0& 2& 3\\ -1& -1& 0\end{array}\right]$

= $\left[\begin{array}{ccc}1& 0& 1\\ 3& 1& 2\end{array}\right]+\left[\begin{array}{ccc}4& 2& -8\\ 6& 1& -4\end{array}\right]+\left[\begin{array}{ccc}0& 6& 9\\ -3& -3& 0\end{array}\right]$

= $\left[\begin{array}{ccc}1+4+0& 0+2+6& 1-8+9\\ 3+6-3& 1+10-3& 2-4+0\end{array}\right]$

∴ A + 2B + 3C = $\left[\begin{array}{ccc}5& 8& 2\\ 6& 8& -2\end{array}\right]$

∴ [A + 2B + 3C]T = $\left[\begin{array}{cc}5& 6\\ 8& 8\\ 2& -2\end{array}\right]$   ...(i)

Now, AT = $\left[\begin{array}{cc}1& 3\\ 0& 1\\ 1& 2\end{array}\right],{\text{B}}^{\text{T}}=\left[\begin{array}{cc}2& 3\\ 1& 5\\ -4& -2\end{array}\right]$

and CT = $\left[\begin{array}{cc}0& -1\\ 2& -1\\ 3& 0\end{array}\right]$

∴ AT + 2BT + 3CT

= $\left[\begin{array}{cc}1& 3\\ 0& 1\\ 1& 2\end{array}\right]+2\left[\begin{array}{cc}2& 3\\ 1& 5\\ -4& -2\end{array}\right]+3\left[\begin{array}{cc}0& -1\\ 2& -1\\ 3& 0\end{array}\right]$

= $\left[\begin{array}{cc}1& 3\\ 0& 1\\ 1& 2\end{array}\right]+\left[\begin{array}{cc}4& 6\\ 2& 10\\ -8& -4\end{array}\right]+\left[\begin{array}{cc}0& -3\\ 6& -3\\ 9& 0\end{array}\right]$

= $\left[\begin{array}{cc}1+4+0& 3+6+3\\ 0+2+6& 1+10-3\\ 1-8+9& 2-4+0\end{array}\right]$

∴ AT + 2BT + 3CT = $\left[\begin{array}{cc}5& 6\\ 8& 8\\ 2& -2\end{array}\right]$     ...(iii)

From (i) and (ii), we get
[A + 2B + 3C]T = AT + 2BT + 3CT.

Exercise 2.4 | Q 9 | Page 59

QUESTION

If A = $\left[\begin{array}{ccc}-1& 2& 1\\ -3& 2& -3\end{array}\right]$ and B = $\left[\begin{array}{cc}2& 1\\ -3& 2\\ -1& 3\end{array}\right]$, prove that (A + BT)T = AT + B.

SOLUTION

A = $\left[\begin{array}{ccc}-1& 2& 1\\ -3& 2& -3\end{array}\right]\text{and B}=\left[\begin{array}{cc}2& 1\\ -3& 2\\ -1& 3\end{array}\right]$

∴ AT = $\left[\begin{array}{cc}-1& -3\\ 2& 2\\ 1& -3\end{array}\right]{\text{and B}}^{\text{T}}=\left[\begin{array}{ccc}2& -3& -1\\ 1& 2& 3\end{array}\right]$

∴ A + BT = $\left[\begin{array}{ccc}-1& 2& 1\\ -3& 2& -3\end{array}\right]+\left[\begin{array}{ccc}2& -3& 1\\ 1& 2& 3\end{array}\right]$

= $\left[\begin{array}{ccc}-1+2& 2-3& 1-1\\ -3+1& 2+2& -3+3\end{array}\right]$

= $\left[\begin{array}{ccc}1& -1& 0\\ -2& 4& 0\end{array}\right]$

∴ (A + BT)T = $\left[\begin{array}{cc}1& -2\\ -1& 4\\ 0& 0\end{array}\right]$       ...(i)

Now, AT + B = $\left[\begin{array}{cc}-1& -3\\ 2& 2\\ 1& -3\end{array}\right]+\left[\begin{array}{cc}2& 1\\ -3& 2\\ -1& 3\end{array}\right]$

= $\left[\begin{array}{cc}-1+2& -3+1\\ 2-3& 2+2\\ 1-1& -3+3\end{array}\right]$

= $\left[\begin{array}{cc}1& -2\\ -1& 4\\ 0& 0\end{array}\right]$                 ...(ii)
From (i) and (ii), we get
(A + BT)T = AT + B.

Exercise 2.4 | Q 10.1 | Page 59

QUESTION

Prove that A + AT is a symmetric and A – AT is a skew symmetric matrix, where A = $\left[\begin{array}{ccc}1& 2& 4\\ 3& 2& 1\\ -2& -3& 2\end{array}\right]$

SOLUTION

A = $\left[\begin{array}{ccc}1& 2& 4\\ 3& 2& 1\\ -2& -3& 2\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}1& 3& -2\\ 2& 2& -3\\ 4& 1& 2\end{array}\right]$

∴ A + AT = $\left[\begin{array}{ccc}1& 2& 4\\ 3& 2& 1\\ -2& -3& 2\end{array}\right]+\left[\begin{array}{ccc}1& 3& -2\\ 2& 2& -3\\ 4& 1& 2\end{array}\right]$

= $\left[\begin{array}{ccc}1+1& 2+3& 4-2\\ 3+2& 2+2& 1-3\\ -2+4& -3+1& 2+2\end{array}\right]$

∴ A + AT = $\left[\begin{array}{ccc}2& 5& 2\\ 5& 4& -2\\ 2& -2& 4\end{array}\right]$

∴ (A + AT)T = $\left[\begin{array}{ccc}2& 5& 2\\ 5& 4& -2\\ 2& -2& 4\end{array}\right]$

∴ (A + AT)T = A + AT i.e., A + AT = (A + AT)T
∴ A + AT is a symmetric matrix.

A – AT = $\left[\begin{array}{ccc}1& 2& 4\\ 3& 2& 1\\ -2& -3& 2\end{array}\right]-\left[\begin{array}{ccc}1& 3& -2\\ 2& 2& -3\\ 4& 1& 2\end{array}\right]$

= $\left[\begin{array}{ccc}1-1& 2-3& 4+2\\ 3-2& 2-2& 1+3\\ -2-4& -3-1& 2-2\end{array}\right]$

∴ A – AT = $\left[\begin{array}{ccc}0& -1& 6\\ 1& 0& 4\\ -6& -4& 0\end{array}\right]$

∴ (A – AT)T $\left[\begin{array}{ccc}0& 1& -6\\ -1& 0& -4\\ 6& 4& 0\end{array}\right]$

= $-\left[\begin{array}{ccc}0& -1& 6\\ 1& 0& 4\\ -6& -4& 0\end{array}\right]$

∴ (A – AT)T = –  (A – AT)
i.e., A – AT = –  (A – AT)T
∴ A – AT  is a skew symmetric matrix.

Exercise 2.4 | Q 10.2 | Page 59

QUESTION

Prove that A + AT is a symmetric and A – AT is a skew symmetric matrix, where A = $\left[\begin{array}{ccc}5& 2& -4\\ 3& -7& 2\\ 4& -5& -3\end{array}\right]$

SOLUTION

A = $\left[\begin{array}{ccc}5& 2& -4\\ 3& -7& 2\\ 4& -5& -3\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}5& 3& 4\\ 2& -7& -5\\ -4& 2& -3\end{array}\right]$

∴ A + AT = $\left[\begin{array}{ccc}5& 2& -4\\ 3& -7& 2\\ 4& -5& -3\end{array}\right]+\left[\begin{array}{ccc}5& 3& 4\\ 2& -7& -5\\ -4& 2& -3\end{array}\right]$

= $\left[\begin{array}{ccc}5+5& 2+3& -4+4\\ 3+2& -7-7& 2-5\\ 4-4& -5+2& -3-3\end{array}\right]$

∴ A + AT = $\left[\begin{array}{ccc}10& 5& 0\\ 5& -14& -3\\ 0& -3& -6\end{array}\right]$

∴  (A + AT)T = A + AT i.e., A + AT = (A + AT)T
∴ A + AT = is a symmetric matrix.

A – AT = $\left[\begin{array}{ccc}5& 2& -4\\ 3& -7& 2\\ 4& -5& -3\end{array}\right]-\left[\begin{array}{ccc}5& 3& 4\\ 2& -7& -5\\ -4& 2& -3\end{array}\right]$

= $\left[\begin{array}{ccc}5-5& 2-3& -4-4\\ 3-2& -7+7& 2+5\\ 4+4& -5-2& -3+3\end{array}\right]$

= $\left[\begin{array}{ccc}0& -1& -8\\ 1& 0& 7\\ 8& -7& 0\end{array}\right]$

∴ (A – AT)T = $\left[\begin{array}{ccc}0& 1& 8\\ -1& 0& -7\\ -8& 7& 0\end{array}\right]$

= $\left[\begin{array}{ccc}0& -1& -8\\ 1& 0& 7\\ 8& -7& 0\end{array}\right]$

∴ (A – AT)T = – (A – AT)
i.e., A – AT = – (A – AT)T
∴ A – AT  is a skew symmetric matrix.

Exercise 2.4 | Q 11.1 | Page 59

QUESTION

Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix $\left[\begin{array}{cc}4& -2\\ 3& -5\end{array}\right]$.

SOLUTION

A square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as

A = $\frac{1}{2}\left(\text{A}+{\text{A}}^{\text{T}}\right)+\frac{1}{2}\left(\text{A}-{\text{A}}^{\text{T}}\right)$

Let A = $\left[\begin{array}{cc}4& -2\\ 3& -5\end{array}\right]$

∴ AT = $\left[\begin{array}{cc}4& 3\\ -2& -5\end{array}\right]$

∴ A + AT = $\left[\begin{array}{cc}4& -2\\ 3& -5\end{array}\right]+\left[\begin{array}{cc}4& 3\\ -2& -5\end{array}\right]$

= $\left[\begin{array}{cc}4+4& -2+3\\ 3-2& -5-5\end{array}\right]$

= $\left[\begin{array}{cc}8& 1\\ 1& -10\end{array}\right]$

Also, A – AT = $\left[\begin{array}{cc}4& -2\\ 3& -5\end{array}\right]-\left[\begin{array}{cc}4& 3\\ -2& -5\end{array}\right]$

= $\left[\begin{array}{cc}4-4& -2-3\\ 3+2& -5+5\end{array}\right]$

= $\left[\begin{array}{cc}0& -5\\ 5& 0\end{array}\right]$

Let P = $\frac{1}{2}\left(\text{A}+{\text{A}}^{\text{T}}\right)$

= $\frac{1}{2}\left[\begin{array}{cc}8& 1\\ 1& -10\end{array}\right]$

= $\left[\begin{array}{cc}4& \frac{1}{2}\\ \frac{1}{2}& -5\end{array}\right]$
and
Q = $\frac{1}{2}\left(\text{A}-{\text{A}}^{\text{T}}\right)$

= $\frac{1}{2}\left[\begin{array}{cc}0& -5\\ 5& 0\end{array}\right]$

= $\left[\begin{array}{cc}0& -\frac{5}{2}\\ \frac{5}{2}& 0\end{array}\right]$

∴ P is a symmetric matrix         ...[∵ aij = aij]

and Q is a skew-symmetric matrix.   ...[∵ aij = – aij]
∴ A = P + Q

∴ A = $\left[\begin{array}{cc}4& \frac{1}{2}\\ \frac{1}{2}& -5\end{array}\right]+\left[\begin{array}{cc}0& -\frac{5}{2}\\ \frac{5}{2}& 0\end{array}\right]$.

Exercise 2.4 | Q 11.2 | Page 59

QUESTION

Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix $\left[\begin{array}{ccc}3& 3& -1\\ -2& -2& 1\\ -4& -5& 2\end{array}\right]$.

SOLUTION

A square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as

A = $\frac{1}{2}\left(\text{A}+{\text{A}}^{\text{T}}\right)+\frac{1}{2}\left(\text{A}-{\text{A}}^{\text{T}}\right)$

Let A = $\left[\begin{array}{ccc}3& 3& -1\\ -2& -2& 1\\ -4& -5& 2\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

∴ A + AT = $\left[\begin{array}{ccc}3& 3& -1\\ -2& -2& 1\\ -4& -5& 2\end{array}\right]+\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

= $\left[\begin{array}{ccc}3+3& 3-2& -1-4\\ -2+3& -2-2& 1-5\\ -4-1& -5+1& 2+2\end{array}\right]$

= $\left[\begin{array}{ccc}6& 1& -5\\ 1& -4& -4\\ -5& -4& 4\end{array}\right]$

Also, A – AT = $\left[\begin{array}{ccc}3& 3& -1\\ -2& -2& 1\\ -4& -5& 2\end{array}\right]-\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

= $\left[\begin{array}{ccc}3-3& 3+2& -1+4\\ -2-3& -2+2& 1+5\\ -4+1& -5-1& 2-2\end{array}\right]$

= $\left[\begin{array}{ccc}0& 5& 3\\ -5& 0& 6\\ -3& -6& 0\end{array}\right]$

Let P = $\frac{1}{2}\left(\text{A}+{\text{A}}^{\text{T}}\right)$

= $\frac{1}{2}\left[\begin{array}{ccc}6& 1& -5\\ 1& -4& -4\\ -5& -4& 4\end{array}\right]$

and Q = $\frac{1}{2}\left(\text{A}-{\text{A}}^{\text{T}}\right)$

= $\frac{1}{2}\left[\begin{array}{ccc}0& 5& 3\\ -5& 0& 6\\ -3& -6& 0\end{array}\right]$

∴ P is a symmetric matrix          ...[∵ aij = aij]

and Q is a skew symmetric matrix.  ...[∵ aij = –  aij]
∴ A = P + Q

∴ A = $\frac{1}{2}\left[\begin{array}{ccc}6& 1& -5\\ 1& -4& -4\\ -5& -4& 4\end{array}\right]+\frac{1}{2}\left[\begin{array}{ccc}0& 5& 3\\ -5& 0& 6\\ -3& -6& 0\end{array}\right]$.

Exercise 2.4 | Q 12.1 | Page 60

QUESTION

If A = $\left[\begin{array}{cc}2& -1\\ 3& -2\\ 4& 1\end{array}\right]\text{and B}=\left[\begin{array}{ccc}0& 3& -4\\ 2& -1& 1\end{array}\right]$, verify that (AB)T = BTAT.

SOLUTION

A = $\left[\begin{array}{cc}2& -1\\ 3& -2\\ 4& 1\end{array}\right]\text{and B}=\left[\begin{array}{ccc}0& 3& -4\\ 2& -1& 1\end{array}\right]$

∴ AT = $\left[\begin{array}{ccc}2& 3& 4\\ -1& -2& 1\end{array}\right]{\text{and B}}^{\text{T}}=\left[\begin{array}{cc}0& 2\\ 3& -1\\ -4& 1\end{array}\right]$

AB = $\left[\begin{array}{cc}2& -1\\ 3& -2\\ 4& 1\end{array}\right]\left[\begin{array}{ccc}0& 3& -4\\ 2& -1& 1\end{array}\right]$

= $\left[\begin{array}{ccc}0-2& 6+1& -8-1\\ 0-4& 9+2& -12-2\\ 0+2& 12-1& -16+1\end{array}\right]$

= $\left[\begin{array}{ccc}-2& 7& -9\\ -4& 11& -14\\ 2& 11& -15\end{array}\right]$

∴ (AB)T = $\left[\begin{array}{ccc}-2& -4& 2\\ 7& 11& 11\\ -9& -14& -15\end{array}\right]$     ...(i)

BTAT = $\left[\begin{array}{cc}0& 2\\ 3& -1\\ -4& 1\end{array}\right]\left[\begin{array}{ccc}2& 3& 4\\ -1& -2& 1\end{array}\right]$

= $\left[\begin{array}{ccc}0-2& 0-4& 0+2\\ 6+1& 9+2& 12-1\\ -8-1& -12-2& -16+1\end{array}\right]$

= $\left[\begin{array}{ccc}-2& -4& 2\\ 7& 11& 11\\ -9& -14& -15\end{array}\right]$          ...(ii)
From (i) and (ii), we get
(AB)T = BTAT.

Exercise 2.4 | Q 12.2 | Page 60

QUESTION

If A = $\left[\begin{array}{cc}2& -1\\ 3& -2\\ 4& 1\end{array}\right]\text{and B}=\left[\begin{array}{ccc}0& 3& -4\\ 2& -1& 1\end{array}\right]$, verify that (BA)T = ATBT.

SOLUTION

BA = $\left[\begin{array}{ccc}0& 3& -4\\ 2& -1& 1\end{array}\right]\left[\begin{array}{cc}2& -1\\ 3& -2\\ 4& 1\end{array}\right]$

= $\left[\begin{array}{cc}0+9-16& 0-6-4\\ 4-3+4& -2+2+1\end{array}\right]$

∴ BA = $\left[\begin{array}{cc}-7& -10\\ 5& 1\end{array}\right]$

∴ (BA)T = $\left[\begin{array}{cc}-7& 5\\ -10& 1\end{array}\right]$          ...(i)

ATBT = $\left[\begin{array}{ccc}2& 3& 4\\ -1& -2& 1\end{array}\right]\left[\begin{array}{cc}0& 2\\ 3& -1\\ -4& 1\end{array}\right]$

= $\left[\begin{array}{cc}0+9-16& 4-3+4\\ 0-6-4& -2+2+1\end{array}\right]$

= $\left[\begin{array}{cc}-7& 5\\ -10& 1\end{array}\right]$                ...(ii)
From (i) and (ii)
(BA)T = ATBT.

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