Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise

Question 1.

Let A = \(\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\). Show that (aI + bA)^{n} + na^{n-1}bA, where I is the identity matrix of order n ∈ N

Solution:

Apply Principle of Mathmatical Induction.

Put P(n) : \(\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\)^{n} = \(\left[\begin{array}{cc} a^{n} & n a^{n-1} b \\ 0 & a^{n} \end{array}\right]\).

For n = 1, P(1) : \(\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\) = \(\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\).

∴ P(n) is true for n = 1.

Let P(n) be true for n = k.

∴ P(k) : \(\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\)^{k} = \(\left[\begin{array}{cc} a^{k} & k a^{k-1} b \\ 0 & a^{k} \end{array}\right]\).

Multiplying both sides by \(\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\), we get :

This shows P(n) is true for n = k + 1 also. Then, by principle of mathematical induction, P(n) is true for all natural numbers n.

Question 2.

If A = \(\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]\), prove that A^{n} = \(\left[\begin{array}{lll} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{array}\right]\), n ∈ N.

Solution:

∴ P(n) is true for n = 1.

Let P(n) be true for n = k.

∴ A^{k} = \(\left[\begin{array}{lll} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{array}\right]\).

Multiplying both sides by A, we get

L.H.S = A^{k}A = A^{k+1}

∴ P(n) is true for n = k + 1.

∴ By principle of mathematical inductiqn, P(n) is true for all n, n ∈ N.

Question 3.

If A = \(\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\), then prove that A^{n} = \(\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]\), where n is any positive integer.

Solution:

Let P(n) : A^{n} = \(\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]\), where A = \(\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\).

∴ P(n) is true for n = 1.

Let P(n) be true for n = k.

∴ A^{k} = \(\left[\begin{array}{cc} 1+2 k & -4 k \\ k & 1-2 k \end{array}\right]\)

Multiplying both sides by A, we get:

This shows that P(n) is true for n = k + 1 also.

Hence, P(n) is true for all n e N.

Question 4.

If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Solution:

A and B are symmetric matrices. So, A’ = A and B’ = B. Also, (AB)’ = B’A’.

Now, (AB – BA)’ = (AB)’ – (BA)’

= B’A’ – A’B’ = BA – AB .

= – (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:

(i) Let A be a symmetric matrix.

⇒ A’ = A

∴ (B’AB)’ = [B'(AB)]’ = (AB) ‘(B)’ = (B’A’)B

= B’AB (∵ A’ = A)

⇒ B’AB is a symmetric matrix.

(ii) Let A be a skew-symmetric matrix.

⇒ A’ = – A.

Now, [B'(AB)]’ = (AB)'(B’)’ = (B’A’)B

= B'(- A)B = – B’AB ( ∵ A’ = A)

⇒ B’AB is a skew symmetric matrix.

Question 6.

Find the values of x, y and z, if the matrix A = \(\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]\) satisfies the equation A’A = I_{3}.

Solution:

Question 7.

For what values of x, \(\left[\begin{array}{lll} 1 & 2 & 1 \end{array}\right]\)\(\left[\begin{array}{lll} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}\right]\)\(\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]\)

Solution:

Question 8.

If A = \(\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]\), show that A² – 5A + 7I = O. Use this result to find A^{4}.

Solution:

Hence, A² – 5A + 7IA = O.

Multiplying both the sides by A, we get:

A³ – 5A² + 7IA = O … (1)

or A³ – 5A² + 7A = O

∴ A³ = 5A² – 7A

Question 9.

Find x, if [x – 5 – 1]\(\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\)\(\left[\begin{array}{l} x \\ 4 \\ 1 \end{array}\right]\) = O.

Solution:

Question 10.

A manufacturer produces three products x, y and z, which he sells in two markets.

Annual sales are indicated below:

(a) If a unit sale price 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.

(b) If the unit cost of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively, find the gross profit.

Solution:

(a) Matrix for the products x, y and z is

Matrix corresponding to sale price of each product

∴ The revenue collected by markets is given by

∴ Revenues in the two markets are ₹ 46000 and ₹ 53000 respectively.

Total revenue = ₹ (46000 + 53000) = ₹ 99000.

(b) The cost price of commodities x, y and z are respectively ₹ 2.00, ₹ 1.00 and ₹ 0.50. Cost price of each market is given below:

Total cost price of the commodities markets I and II are ₹ 31000 and ₹ 36000 respectively.

∴ Total cost price = ₹ (31000 + 36000) = ₹ 67000

∴ Gross profit = Revenue – Cost price

= ₹ (99000 – 67000) = ₹ 32000

Question 11.

Find the matrix X so that X\(\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right]\) = \(\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \end{array}\right]\).

Solution:

Let X be of order m x n. Since \(\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right]\) is of the order 2 x 3, so n = 2

Since \(\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \end{array}\right]\) is of the order 2 x 3. So, m = 2

∴ X is of the order 2 x 2.

Let X = \(\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\)

Equating the corresponding elements,

a + 4b = – 7 … (1)

2a + 5b = – 8 … (2)

3a + 6b = – 9 … (3)

Multiplying (1) by (2), we get

2a + 8b = – 14 … (4)

and 2a + 5b = – 8 … (2)

Subtracting (2) from (4),

3b = -6.

∴ b = – 2

Putting value of b in (1),

a – 8 = – 7 ∴ a = 8 – 7 = 1.

a = 1, b = – 2 satisfies equation (3) also.

Equating the elements of second row,

c + 4d = 2 … (5)

2c + 5d = 4 … (6)

3c + 6d = 6 … (7)

Multiplying equation (5) by 2,

2c + 8d = 4 … (8)

Subtracting (6) from (8),

3d = 4 – 4 = 0. ∴ d = 0

∴ From c + 4d = 2, we get ∴ c = 2

c = 2, d = 0 satisfies equation (7) also.

Thus, a = 1, b = – 2, c = 2, d = 0

Hence, X = \(\left[\begin{array}{ll} 1 & -2 \\ 2 & 0 \end{array}\right]\).

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:

Let P(n) : AB^{n} = B^{n}A.

But n = 1, AB = BA (Given)

∴ P(n) is true for n = 1.

Let P(n) be true for n = k.

∴ AB^{k} = B^{k}A.

Multiplying both sides by B, we get

L.H.S. = AB^{k}.B = A(B^{k}B) = AB^{k+1} (Associative property)

and R.H.S. = (B^{k}A)B = B^{k}(AB)

= B^{k}(BA) (∵AB = BA)

= (B^{k}B)A

= B^{k+1}A.

So, P(n) is true for n = k + 1 also.

∴ By principle of mathematical induction.

P(n) is true for all n e N.

(ii) Let P(n) : (AB)^{n} = A^{n}B^{n}

For n = 1, L.H.S. = AB and R.H.S. = AB.

∴ P(n) is true for n = 1.

Let P(n) be true for n = k.

So, (AB)^{k} = A^{k}B^{k}

Multiply both sides by AB. We get

L.H.S. = (AB)^{k}(AB) = (AB)^{k+1}

and R.H.S. = A^{k}B^{k}(AB) = A^{k}B^{k}(BA) (∵ AB = BA)

= A^{k}(B^{k}.B)A = A^{k}(B^{k+1}A) [∵ AB^{n} = B^{n}A]

= A^{k}(AB^{k+1}) = A^{k+1}B^{k+1}

⇒ P(n) is true for n = k + 1 also.

By principle of mathematical induction,

∴ P(n) is true for all n ∈ N.

Choose the correct answer in the following questions:

Question 13.

If A = \(\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right]\) is such that A² = I, then

(A) 1 + α² + ßγ = 0

(B) 1 – α² + ßγ = 0

(C) 1 – α² – ßγ = 0

(D) 1 + α² – ßγ = 0

Solution:

Question 14.

If a marix is both symmetric and skew symmetric matrix, then

(A) A is a diagonal matrix

(B) A is a zero matrix

(C) A is a square matrix

(D) None of these

Solution:

A is a symmetric matrix, if a_{ij}. = a_{ji}

A is a skew symmetric matrix, if a_{ij} = – a_{ji}

If a_{ij} = a_{ji} = – a_{ji}.

⇒ A is a zero matrix.

∴ Part (B) is the correct answer.

Question 15.

If A is a square matrix such that A² = A, then (1 + A)³ – 7A is equal to

(A) A

(B) 1 – A

(C) 1

(D) 3A

Solution:

A³ = A².A [ ∵ A² = A]

= A.A = A² = A

(1 + A)³ – 7A = (1 + 3A + 3A² + A³) – 7A

= (1 + 3A + 3A + A) – 7A

= (1 + 7A) – 7A .

∴ Hence, Part (C) is the correct answer.