Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Miscellaneous Exercise on Chapter 3
Question 1: Let [Tex]A =\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} [/Tex], show that (aI + bA)n = an I + nan – 1 bA, where I is the identity matrix of order 2 and n ∈ N.
Question 3: If [Tex]A =\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}[/Tex], prove that [Tex]A^n =\begin{bmatrix} 1+2n & -4n\\ n & 1-2n \end{bmatrix} [/Tex] ,where n is any positive integer.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
[Tex]A^1 = A =\begin{bmatrix} 1+2(1) & -4(1)\\ n & 1-2(1) \end{bmatrix}=\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix} [/Tex]
It is true for P(1)
Step 2: Now take n=k
[Tex]A^k =\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}[/Tex]
Question 4. If A and B are symmetric matrices, prove that AB – BA is a skew-symmetric matrix.
Solution:
As, it is mentioned that A and B are symmetric matrices,
A’ = A and B’ = B
(AB – BA)’ = (AB)’ – (BA)’ (using, (A-B)’ = A’ – B’)
= B’A’ – A’B’ (using, (AB)’ = B’A’)
= BA – AB
(AB – BA)’ = – (AB – BA)
Hence, AB – BA is a skew symmetric matrix
Question 5. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.
Solution:
Let’s take A as symmetric matrix
A’ = A
Then,
(B′AB)’ = {B'(AB)}’
= (AB)’ (B’)’ (using, (AB)’ = B’A’)
= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)
= B’A B
As, here (B′AB)’ = B’A B. It is a symmetric matrix.
Let’s take A as skew matrix
A’ = -A
Then,
(B′AB)’ = {B'(AB)}’
= (AB)’ (B’)’ (using, (AB)’ = B’A’)
= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)
= B'(-A) B
= – B’A B
As, here (B′AB)’ = -B’A B. It is a skew matrix.
Hence, we can conclude that B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Question 6. Find the values of x, y, z if the matrix [Tex]A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix} [/Tex] satisfy the equation A′A = I
Solution:
[Tex]A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix} [/Tex]
[Tex]A’ =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}^T=\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} [/Tex]
A’A = [Tex]I =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} [/Tex]
[Tex]\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}[/Tex]
Question 10: A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:
Market
Products
I
10,000
2,000
18,000
II
6,000
20,000
8,000
(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
Solution:
Total revenue in market I and II can be arranged from given data as follows:
Question 12: If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
ABn = AB1 = AB
BnA = B1A = BA
It is true for P(1)
Step 2: Now take n=k
ABk = BkA
Step 3: Let’s check whether, its true for n = k+1
AB(k+1) = ABkB
= BkAB
= Bk+1 A
= P(k+1)
Hence, P(n) is true.
Now, for (AB)n = AnBn
Using mathematical induction,
Step 1: Let’s check for n=1
(AB)1 = AB
B1A1 = BA
It is true for P(1)
Step 2: Now take n=k
(AB)k = AkBk
Step 3: Let’s check whether, its true for n = k+1
(AB)(k+1) = (AB)k(AB)
= AkBk AB
= Ak+1 Bk+1
= (AB)k+1
= P(k+1)
Hence, P(n) is true.
Choose the correct answer in the following questions:
Question 13: If [Tex]A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} [/Tex] is such that A² = I, then
