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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 3 Matrices – Miscellaneous Exercise on Chapter 3

### Question 1: Let , show that (aI + bA)n = an I + nan – 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)

anI + nan – 1 bA = aI + 1a1 – 1 bA = (aI + bA)

It is true for P(1)

Step 2: Now take n=k

(aI + bA)k = akI + kak – 1 bA …………………(1)

Step 3: Let’s check whether, its true for n = k+1

(aI + bA)k+1 = (aI + bA)k (aI + bA)

= (akI + kak – 1 bA) (aI + bA)

= ak+1I×I + kak bAI + ak bAI + kak-1 b2AA

AA =

= ak+1I×I + kak bAI + ak bAI + 0

= ak+1I + (k+1)ak+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 = ±

Solution:

Solution:

A2 – 5A + 7I =

Hence proved!

Solution:

### (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 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.

### 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 A2 = 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 = I3 + A3 + 3A^2 + 3AI^2 – 7A

= I3 + A3 + 3A2 + 3A – 7A

= I + A3 + 3A2 – 4A

As, A2 = A

A3 = A2A = AA = A

So, I + A3 + 3A2 – 4A = I + A + 3A – 4A = I

Hence, Option (C) is correct.

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