NCERT Solutions Class 12 Maths (matrices) Miscellaneous Exercise
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Question 1: Let,show that (aI + bA)^{n} = a^{n} I + na^{n – 1} bA, where I is the identity matrix of order 2 and n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s
check for n=1
(aI + bA)^{n} =
(aI + bA)^{1} = (aI + bA)
a^{n}I
+ na^{n – 1} bA = aI +
1a^{1 – 1} bA = (aI +
bA)
It is true for P(1)
Step 2: Now
take n=k
(aI + bA)^{k} =
a^{k}I + ka^{k –
1} bA …………………(1)
Step 3: Let’s
check whether, its true for n = k+1
(aI + bA)^{k+1} =
(aI + bA)^{k} (aI + bA)
= (a^{k}I
+ ka^{k – 1} bA) (aI +
bA)
= a^{k+1}I×I
+ ka^{k} bAI + a^{k} bAI
+ ka^{k-1} b^{2}AA
AA =
= a^{k+1}I×I
+ ka^{k} bAI + a^{k} bAI
+ 0
= a^{k+1}I
+ (k+1)a^{k+1-1} bA
= P(k+1)
Hence,
P(n) is true.
Question 2: If , prove that
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
It is true for P(1)
Step 2: Now take n=k
Step 3: Let’s check whether, its true for n = k+1
= P(k+1)
Hence, P(n) is true.
Question 3: If , prove that
,where n is any positive
integer.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
It is true for P(1)
Step 2: Now take n=k
Step 3: Let’s check whether, its true for n = k+1
= P(k+1)
Hence, P(n) is true.
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
satisfy the equation A′A = I
Solution:
A’A =
By evaluating the values, we have
2x2 = 1
x = ±
6y2 = 1
y = ±
3z2 = 1
z = ±
Question 7: For what values of x:
Solution:
Question 8: If ,show that A^{2} – 5A + 7I = 0.
Solution:
A2 – 5A + 7I =
Hence proved!
Question 9: Find x, if
Solution:
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:
After multiplication, we get
Hence, the total revenue in Market I and market II are ₹ 46,000 and ₹ 53,000 respectively
(b)
If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50
paise respectively. Find the gross profit.
Solution:
Total cost prices of all the products in market I and market II can be arranged from given data as follows:
After multiplication, we get
As, Profit earned = Total revenue – Cost price
Profit earned
Profit earned =
Hence, profit earned in Market I and market II are ₹ 15,000 and ₹ 17,000 respectively. Which is equal to ₹ 32,000
Question 11. Find the matrix X so that
Solution:
Here, the RHS is a 2×3 matrix and LHS is 2×3. So, X will be 2×2 matrix.
Let’s take X as,
Now solving the matrix, we have
Equating each of them, we get
p+4q = -7 ………..(1)
2p+5q = -8 ………….(2)
3p + 6q = -9
r + 4s = 2 …………(3)
2r + 5s = 4 ……………(5)
3r + 6s = 6
Solving (1) and (2), we get
p = 1 and q = -2
Solving (3) and (4), we get
r = 2 and s = 0
Hence, matrix X is
Question 12: If A and B are
square matrices of the same order such that AB = BA, then prove by induction
that AB^{n} = B^{n}A. Further, prove that (AB)^{n} = A^{n}B^{n} for all n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s
check for n=1
AB^{n} =
AB^{1} = AB
B^{n}A
= B^{1}A = BA
It is true for P(1)
Step 2: Now
take n=k
AB^{k} =
B^{k}A
Step 3:
Let’s check whether, its true for n = k+1
AB^{(k+1)} =
AB^{k}B
= B^{k}AB
= B^{k+1} A
= P(k+1)
Hence, P(n) is true.
Now, for (AB)^{n} =
A^{n}B^{n}
Using mathematical induction,
Step 1:
Let’s check for n=1
(AB)^{1} =
AB
B^{1}A^{1} =
BA
It is true for P(1)
Step 2:
Now take n=k
(AB)^{k} =
A^{k}B^{k}
Step 3:
Let’s check whether, its true for n = k+1
(AB)^{(k+1)} =
(AB)^{k}(AB)
= A^{k}B^{k} AB
= A^{k+1} B^{k+1}
= (AB)^{k+1}
= P(k+1)
Hence,
P(n) is true.
Choose
the correct answer in the following questions:
Question 13: If
is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
Solution:
As, A2 = I
α² + βγ = 1
1 – α² – βγ = 0
Hence, Option (C) is correct.
Question
14. If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Solution:
If the matrix A is both symmetric and skew
symmetric, then
A = A’
and A = -A
Only zero matrix satisfies both the conditions.
Hence,
Option (B) is correct.
Question 15. If A is square
matrix such that A^{2} = A, then (I + A)³ – 7 A
is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Solution:
(I + A)³ – 7 A = I^{3} +
A^{3} + 3A^2 +
3AI^2 – 7A
= I3 + A3 + 3A^{2} +
3A – 7A
= I + A^{3} +
3A^{2} – 4A
As, A^{2} =
A
A3 = A^{2}A
= AA = A
So, I + A3 + 3A2 – 4A = I + A + 3A – 4A = I
Hence, Option (C) is correct.
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