Plus Two Maths Chapter 3 Matrices Chapter Wise Question and Answers PDF Download

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Plus Two Maths Chapter 3 Matrices Chapter Wise Question and Answers

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Board	Kerala Board
Study Materials	Chapter wise Question and Answers
For Year	2021
Class	12
Subject	Mathematics
Chapters	Maths Chapter 3 Matrices
Format	PDF
Provider	Spandanam Blog

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Question 1.
Find the value of a, b and c from the following equations;
\(\left[\begin{array}{cc}{a-b} & {2 a+c} \\{2 a-b} & {3 c+d}
\end{array}\right]=\left[\begin{array}{cc}{-1} & {5} \\{0} & {13}\end{array}\right]\).
Answer:
Given;
\(\left[\begin{array}{cc}{a-b} & {2 a+c} \\{2 a-b} & {3 c+d}
\end{array}\right]=\left[\begin{array}{cc}{-1} & {5} \\{0} & {13}\end{array}\right]\)
⇒ a – b = -1, 2a + c = 5, 2a – b = 0, 3c + d = 13
⇒ a – b = -1
2a – b = 0
– a = -1
⇒ a = 1
We have, a – b = -1 ⇒ 1 – b = -1 ⇒ b = 2
⇒ 2a + c = 5 ⇒ 2 + c = 5 ⇒ c = 3
⇒ 3c + d = 13 ⇒ 9 + d = 13 ⇒ d = 4.

Question 2.
Simplify cosx\(\left[\begin{array}{cc}{\cos x} & {\sin x} \\{-\sin x} & {\cos x}\end{array}\right]\) + sinx\(\left[\begin{array}{cc}{\sin x} & {-\cos x} \\{\cos x} & {\sin x}\end{array}\right]\).
Answer:

Question 3.
Solve the equation for x, y z and t; if
\(2\left[\begin{array}{ll}{x} & {z} \\{y} & {t}\end{array}\right]+3\left[\begin{array}{cc}{1} & {-1} \\{0} & {2}\end{array}\right]=3\left[\begin{array}{ll}{3} & {5} \\{4} & {6}\end{array}\right]\).
Answer:

⇒ 2x + 3 = 9 ⇒ x = 3
⇒ 2z – 3 = 15 ⇒ z = 9
⇒ 2y = 12 ⇒ y = 6
⇒ 2t + 6 = 18 ⇒ t = 6.

Question 4.
Find A² – 5A + 6I If A = \(\left[\begin{array}{ccc}{2} & {0} & {1} \\{2} & {1} & {3} \\{1} & {-1} & {0}\end{array}\right]\)
Answer:

A² – 5A + 6I

Question 5.
If A = \(\left[\begin{array}{cc}{3} & {-2} \\{4} & {-2}\end{array}\right]\) find k so that A² = kA – 2I.
Answer:

Given A² = kA – 2I

1 = 3k – 2
⇒ k = 1.

Question 6.
Express A = \(\left[\begin{array}{ccc}{-1} & {2} & {3} \\{5} & {7} & {9} \\{-2} & {1} & {1}
\end{array}\right]\) as the sum of a symmetric and skew symmetric matrix.
Answer:

P = 1/2 (A + A^T) is symmetric.
Q = 1/2 (A – A^T) is skew symmetric.

Question 7.
Find the inverse of the following using elementary transformations.

Answer:
(i) Let A = I A

(ii) Let A = IA

(iii) Let A = IA

(iv) Let A = IA

Question 8.
Find the inverse of the matrix A = \(\left[\begin{array}{cc}{2} & {3} \\{-1} & {5}\end{array}\right]\) using row transformation.
Answer:
A = \(\left[\begin{array}{cc}{2} & {3} \\{-1} & {5}\end{array}\right]\)
Let A = IA

Question 9.
\(A=\left[\begin{array}{ll}{2} & {3} \\{4} & {5} \\{2} & {1}\end{array}\right] B=\left[\begin{array}{ccc}{1} & {-2} & {3} \\{-4} & {2} & {5}\end{array}\right]\)

1. Find AB
2. If C is the matrix obtained from A by the transformation R₁ → 2R₁, find CB

Answer:

(ii) Since C is the matrix obtained from A by the transformation R₁ → 2R₁
⇒ C = \(\left[\begin{array}{ll}{4} & {6} \\{4} & {5} \\{2} & {1}\end{array}\right]\)
Then CB can be obtained by multiplying first row of AB by 2.
CB = \(\left[\begin{array}{ccc}{-20} & {-4} & {42} \\{-16} & {2} & {37} \\{-2} & {-2} & {11}
\end{array}\right]\).

Question 10.
Construct a 3 × 4 matrix whose elements are given by

1. a_y = \(\frac{|-3 i+j|}{2}\) (2)
2. a_ij = 2i – j (2)

Answer:

a₁₃ = 0, a₁₄ = \(\frac{1}{2}\), a₂₁ = \(\frac{5}{2}\), a₂₂ = 2, a₂₃ = \(\frac{3}{2}\), a₂₄ = 1, a₃₁ = 4, a₃₂ = \(\frac{7}{2}\), a₃₃ = 3, a₃₄ = \(\frac{5}{2}\)

a₁₁ = 1, a₁₂ = 0, a₁₃= -1, a₁₄ = -2, a₂₁ = 3, a₂₂ = 2, a₂₃ = 1, a₂₄ = 0, a₃₁ = 5, a₃₂ = 4, a₃₃ = 3, a₃₄ = 2

Question 11.
Express the following matrices as the sum of a Symmetric and a Skew Symmetric matrix.
(i) \(\left[\begin{array}{ccc}{6} & {-2} & {2} \\{-2} & {3} & {-1} \\{2} & {-1} & {3}
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}{3} & {3} & {-1} \\{-2} & {-2} & {1} \\{-4} & {-5} & {2}
\end{array}\right]\)
Answer:

Question 12.
If A = \(\left[\begin{array}{ccc}{2} & {4} & {3} \\{1} & {0} & {6} \\{0} & {-2} & {-3}\end{array}\right]\)

1. Find 3A. (1)
2. Find A^T (1)
3. Evaluate A + A^T , is it symmetric? Justify your answer. (1)

Answer:
1. 3A = \(\left[\begin{array}{ccc}{6} & {12} & {9} \\{3} & {0} & {18} \\{0} & {-6} & {-9}
\end{array}\right]\)

2. A^T = \(\left[\begin{array}{ccc}{2} & {1} & {0} \\{4} & {0} & {-2} \\{3} & {6} & {-3}
\end{array}\right]\)

3. A + A^T

The elements on both sides of the main diagonal are same. Therefore A + A^T is a symmetric matrix.

Plus Two Maths Matrices Four Mark Questions and Answers

Question 1.
Consider the following statement: P(n) : Aⁿ = \(\left[\begin{array}{cc}{1+2 n} & {-4 n} \\{n} & {1-2 n}\end{array}\right]\) for all n ∈ N

1. Write P (1). (1)
2. If P(k) is true, then show that P( k + 1) is also true. (3)

Answer:
1. P(1) : A = \(\left[\begin{array}{cc}{1+2} & {-4} \\{1} & {1-2}\end{array}\right]=\left[\begin{array}{cc}{3} & {-4} \\{1} & {-1}\end{array}\right]\)

2. Assume that P(n) is true n = k

Hence P(k+1) is true n ∈ N.

Question 2.
Find the matrices A and B if 2A + 3B = \(\left[\begin{array}{ccc}{1} & {2} & {-1} \\{0{1} & {2} & {4}\end{array}\right]\) and A + 2B = \(\left[\begin{array}{lll}{2} & {0} & {1} \\{1} & {1} & {2} \\{3} & {1} & {2}\end{array}\right]\).
Answer:

Solving (1) and (2) ⇒ 2 × (2)

Question 3.

1. Construct a 3 × 3 matrix A = [a_ij] where a_ij – 2(i – j) (3)
2. Show that matrix A is skew-symmetric. (1)

Answer:
1.

2.

Therefore A is a skew-symmetric matrix.

Question 4.
Consider the following statement P(n ): Aⁿ = \(\left[\begin{array}{cc}{\cos n \theta} & {\sin n \theta} \\{-\sin n \theta} & {\cos n \theta}\end{array}\right]\) for all n ∈ N

1. Write P(1). (1)
2. If P (k) is true then show that P (k+1) is true (3)

Answer:
1.

2. Assume that P(n) is true for n = k

P(k+1) = A^k+1

∴ P(k+1) is true. Hence true for all n ∈ N.

Question 5.
A = \(\left[\begin{array}{lll}{1} & {2} & {2} \\{2} & {1} & {2} \\{2} & {2} & {1}\end{array}\right]\), then

1. Find 4A and A² (2)
2. Show that A² -4A = 5I₃ (2)

Answer:
1.

2.

Question 6.
Let A = \(\left[\begin{array}{lll}{2} & {1} & {3} \\{4} & {1} & {0}\end{array}\right]\) and B= \(\left[\begin{array}{cc}{1} & {-1} \\{0} & {2} \\{5} & {0}\end{array}\right]\)

1. Find A^T and B^T (1)
2. Find AB (1)
3. Show that (AB)^T = B^T A^T (2)

Answer:
1.

2.

3.

∴ (AB)^T = B^T A^T.

Question 7.
A = \(\left[\begin{array}{ccc}{1} & {-3} & {1} \\{2} & {0} & {4} \\{1} & {2} & {-2}\end{array}\right]\) Express A as the sum of a symmetric and skew symmetric matrix.
Answer:

\(\frac{1}{2}\) (A + A^T) + \(\frac{1}{2}\) (A – A^T)

Question 8.

1. Consider a 2 × 2 matrix A = [a_ij], where a_ij = \(\frac{(i+j)^{2}}{2}\)
2. Write the transpose of A. (2)
3. Show that A is symmetric. (2)

Answer:
1. A = \(\left[\begin{array}{ll}{2} & {\frac{9}{2}} \\{\frac{9}{2}} & {8}\end{array}\right]\)

2. A^T = \(\left[\begin{array}{ll}{2} & {\frac{9}{2}} \\{\frac{9}{2}} & {8}\end{array}\right]\)

3. A^T = A therefore symmetric matrix.

Question 9.
A = \(\left[\begin{array}{ll}{6} & {5} \\{7} & {6}\end{array}\right]\) is a matrix

1. What is the order of A. (1)
2. Find A² and 12 A. (2)
3. If f(x) = x^T – 12x +1; find f(A). (1)

Answer:
1. Order of A is 2 × 2.

2.

3. f(x) = x² – 12x + 1 ⇒ f(A) = A² – 12A + I

Plus Two Maths Matrices Six Mark Questions and Answers

Question 1.
Let A = \(\left[\begin{array}{ll}{2} & {4} \\{3} & {2}\end{array}\right]\), B = \(\left[\begin{array}{cc}{1} & {3} \\{-2} & {5}\end{array}\right]\), C = \(\left[\begin{array}{rr}{-2} & {5} \\{3} & {4}\end{array}\right]\)
Find each of the following
(i) A + B; A – B
(ii) 3A – C
(iii) AB
(iv) BA
Answer:

Question 2.
Let A = \(\left[\begin{array}{ll}{1} & {2} \\{3} & {4}\end{array}\right]\); B = \(\left[\begin{array}{ll}{2} & {1} \\{4} & {5}\end{array}\right]\); C = \(\left[\begin{array}{ccc}{1} & {-1} \\{0} & {2}\end{array}\right]\)
(i) Find A + B and A – B (2)
(ii) Show that (A + B) + C = A + (B + C) (2)
(iii) Find AB and BA
Answer:

∴ (A + B) + C = A + (B + C)

Question 3.
A = \(\left[\begin{array}{ccc}{-1} & {0} & {2} \\{4} & {0} & {-3}\end{array}\right]\), B = \(\left[\begin{array}{cc}{0} & {2} \\{-1} & {3} \\{0} & {4}\end{array}\right]\)

1. What is the order of matrix AB ? (1)
2. Find A^T, B^T (2)
3. Verify (AB)^T = B^T A^T (3)

Answer:
1. Order of AB is 2 × 2. Since order of A is 2 × 3 and B is 3 × 2.

2.

3.

(AB)^T = B^T A^T.

Question 4.
Let A = \(\left[\begin{array}{rrr}{1} & {2} & {-3} \\{2} & {1} & {-1}\end{array}\right]\), B = \(\left[\begin{array}{ll}{2} & {3} \\{5} & {4} \\{1} & {6}\end{array}\right]\)
(i) FindAB. (1)
(ii) Find A^T, B^T & (AB)^T (3)
(iii) Verify that (AB)^T = B^T A^T (2)
Answer:

Question 5.
If A = \(\left[\begin{array}{c}{-2} \\{4} \\{5}\end{array}\right]\), B = \(\left[\begin{array}{lll}{1} & {3} & {6}\end{array}\right]\)
(i) Find A^T, B^T (1)
(ii) Find (AB)^T (2)
(iii) Verify (AB)^T = B^T A^T (3)
Answer:

Question 6.
Let A = \(\left[\begin{array}{cc}{3} & {1} \\{-1} & {2}\end{array}\right]\)
(i) Find A² (1)
(ii) Show that A² – 5A + 7I = 0 (1)
(iii) Using this result find A^-1 (2)
(iv) Slove the following equation using matrix: 3x + y = 1, – x + 2y = 2.
Answer:

(iii) A² – 5A + 7I = 0 ⇒ A² – 5A = -7I,
multiplying by A^-1 on both sides,
⇒ A – 5I = -7 A^-1

(iv) The equation can be represented in matrix form as follows, AX = B ⇒ X = A^-1B

Question 7.
A = \(\left[\begin{array}{ccc}{1} & {2} & {3} \\{3} & {-2} & {1} \\{4} & {2} & {1}
\end{array}\right]\)
(i) Show that A³ – 23A – 40I = 0 (3)
(ii) Hence find A^-1 (3)
Answer:

A³ – 23A – 40I = 0

(ii) A^-1A³ – 23 A^-1A – 40A^-1I = 0
⇒ A² – 23I – 40A^-1 = 0

Question 8.
A is a third order square matrix and \(a_{i j}=\left\{\begin{aligned}-i+2 j & \text { if } i=j \\i \times j & \text { if } i \neq j\end{aligned} \text { and } B=\left[\begin{array}{lll}{2} & {1} & {1} \\{1} & {1} & {5} \\{1} & {5} & {2}\end{array}\right]\right.\)

1. Construct the matrix A. (1)
2. Interpret the matrix A. (1)
3. Find AB – BA. (3)
4. Interpret the matrix AB – BA. (1)

Answer:
1. a₁₁ = 1, a₁₂ = 2, a₁₃ = 3, a₂₁ = 2, a₂₂ = 2, a₂₃ = 6, a₃₁ = 3, a₃₂ = 6, a₃₃ = 3
A = \(\left[\begin{array}{lll}{1} & {2} & {3} \\{2} & {2} & {6} \\{3} & {6} & {3}\end{array}\right]\)

2. Now,

Therefore A is symmetric matrix.

3.

4.

= -(AB – BA)
∴ skew symmetric matrix.

Question 9.
Find x and y if

Answer:

Question 10.
Given that A + B = \(\left[\begin{array}{ll}{2} & {5} \\{7} & {8}\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}{6} & {8} \\{4} & {3}\end{array}\right]\)

1. Find 2A. (1)
2. Find A² – B². (3)
3. Is it equal to (A + B) (A – B)? Give reason (2)

Answer:
1. 2A = A + B + A – B

2.

3. (A + B)(A – B)

(A + B)(A – B) = A² + AB – BA – B²
≠ A² – B²
∵ AB ≠ BA.

Question 11.
(i) Consider A = \(\left[\begin{array}{lll}{1} & {x} & {1}\end{array}\right]\), B = \(\left[\begin{array}{ccc}{1} & {3} & {2} \\{2} & {5} & {1} \\{15} & {3} & {2}
\end{array}\right]\), C = \(\left[\begin{array}{l}{1} \\{2} \\{x}\end{array}\right]\) (2)

A – Matrix	B – Order
A	3 × 1
B	1 × 1
BC	2 × 2
ABC	3 × 3
1 × 3

(ii) Find x if ABC = 0 (4)
Answer:
(i)

A – Matrix	B – Order
A	1 × 3
B	3 × 3
BC	3 × 1
ABC	1 × 1

(ii) Given, ABC = 0

⇒ x² + 16x + 28 = 0
⇒ (x + 14)(x + 2) = 0
⇒ x = -14, -2.

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