Students can Download Maths Chapter 3 Matrices Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise

Question 1.
Let \(A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\), Show that (al +bA)n + nan – 1 ba where 1 is the identity matrix of order 2 and n ∈ N.
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 1

Question 2.
If\(A=\left[\begin{array}{lll}{1} & {1} & {1} \\{1} & {1} & {1} \\{1} & {1} & {1}\end{array}\right]\)
,prove that
An = \(\left[\begin{array}{lll}{3^{n-1}} & {3^{n-1}} & {3^{n-1}} \\{3^{n-1}} & {3^{n-1}} & {3^{-1}} \\{3^{n-1}} & {3^{n-1}} & {3^{n-1}}\end{array}\right], n \in N\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 2
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 3

KSEEB Solutions

Question 3.
\(\mathbf{A}=\left[\begin{array}{ll}{\mathbf{3}} & {-4} \\{\mathbf{1}} & {\mathbf{1}}\end{array}\right]\)
then prove that
An = \(\left[\begin{array}{cc}{1+2 n} & {-4 n} \\{n} & {1-2 n}\end{array}\right]\) where n is any positive integer.
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 4
∴ P (k +1) is true
P(n) is true for all the values of +ve n.

Question 4.
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Answer:
Since A &B are symmetric matrices A = a’ & B = B’
To prove (AB – BA)’ = – (AB – BA)
L.H.S = (AB – BA)’
(AB)’ – (BA)’
= B’A’-A’B’
= BA – AB
= – (AB – BA) = R.H.S
Thus proved.

KSEEB Solutions

Question 5.
Show that the matrix g’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 5

Question 6.
find the values of x, y, z if the matrix
A = \(\left[\begin{array}{rrr}{0} & {2 y} & {x} \\{x} & {y} & {-z} \\{x} & {-y} & {z}\end{array}\right]\) satisfy the equation A’ A = 1
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 6

2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 7

Question 7.
For what values of x:
\(\left[\begin{array}{lll}{1} & {2} & {1}\end{array}\right]\left[\begin{array}{rrr}{1} & {2} & {0} \\{2} & {0} & {1} \\{1} & {0} & {2}\end{array}\right]\left[\begin{array}{l}{0} \\{2} \\{x}\end{array}\right]=0 ?\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 8

KSEEB Solutions

Question 8.
\(\mathbf{A}=\left[\begin{array}{rr}{3} & {1} \\{-1} & {2}\end{array}\right]\) show that A2 – 5A + 7I = 0
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 9
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 10

Question 9.
Find x, if
\(\left[\begin{array}{ccc}{\mathbf{x}} & {-5} & {-1}\end{array}\right]\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{2}} \\{\mathbf{0}} & {\mathbf{2}} &{\mathbf{1}} \\{\mathbf{2}} & {\mathbf{0}} &{\mathbf{3}}\end{array}\right]\left[\begin{array}{l}{\mathbf{x}} \\{\mathbf{4}} \\ {\mathbf{1}}\end{array}\right]=\mathbf{0}\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 11

KSEEB Solutions

Question 10.
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 12
(a) If unit sale prices of, 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 costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Answer:
Matrix representing the sales is
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 13
Profit in market I is
₹ 46,000 – ₹ 31,000
= ₹ 15,000

Profit in market 11 is
₹ 53,000 – ₹36,000
= ₹ 17,000

Gross profit
= ₹ 15,000 + ₹17,000
= ₹ 32,000.

Question 11.
Find the matrix X so that
\(\mathbf{X}\left[\begin{array}{lll}{1} & {2} & {3} \\{4} & {5} & {6}\end{array}\right]_{2 \times 3}=\left[\begin{array}{rrr}{-7} & {-8} & {-9} \\{2} & {4} & {6}\end{array}\right]_{2 \times 3}\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 14

KSEEB Solutions

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 = An Bn for all n ∈ N.
Answer:
(i) Let P(n) be the statement
P(n) – ⇒ Ab n = BnA
P(1) ⇒ AB = BA
∴ P( 1) is true
Let P(k) be true
P(k) ⇒ ABk = BkA
P(k +1) ⇒ ABk+1
= A(BkB) = (ABk)B
∵ matrix multiplication is associative.
⇒ (BkA) B ⇒ Bk (AB) = Bk (BA)
= (BkB) A = Bk+1 A
Hence P(k + 1) is true.
∴ P(n) is true for all the values of n ∈ N

(ii) Let P(n) be the statement
P(n) ⇒ (AB)n = AnBn
P(1) = (AB)1 = AB
∴ P(1) is true Let P(k) be true
P(k) ⇒ (AB)k
= AkBk P(k+1)
⇒ (AB)k+1 = (AB)kAB = (AkBk) AB
= Ak (BkA)B = Ak (ABk) B
(∵ ABn = BnA whenever Ab = BA)
= Ak+1 Bk+1
∴ P (k+1) is true
Hence P(n) is true for all the natures of n ∈ N

Choose correct answer in the following questions:

Question 13.
If \(\mathbf{A}=\left[\begin{array}{ll}{\alpha} & {\beta} \\{\gamma} & {-\alpha}\end{array}\right]\)
is such that A2= I ,then
(A) \(1+\alpha^{2}+\beta \gamma=0\)
(B) \(1-\alpha^{2}+\beta \gamma=0\)
(C) \(1-\alpha^{2}-\beta \gamma=0\)
(D) \(1+\alpha^{2}-\beta \gamma=0\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 15
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 16

KSEEB Solutions

Question 14.
If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal martix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Answer:
Let A be a square matrix such that A is both symmetric & skew-symmetric
⇒ A’ = A – (i) & A’ = – A -(ii)
⇒ A + A = 0  from (i) & (ii)
2A = 0
⇒ A = o
∴ (b) is correct

Question 15.
If A is square matrix such that A2 = A, then (I + A)3 – 7A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Answer:
A2 = A
(I + A)3 – 7A = (I + A) (I+A)2 – 7A
= (I + A) {(I + A) (I+A)} -7A
= (I + A) (I2 + 2A+A2) – 7A
[v IA = AI = A]
= (I + A) (I+2A+A) – 7A (A2= A)
= (I + A) (I + 3A) – 7A
= I2 + AI + 3IA + 3A2 – 7A
= I + A + 3A + 3A – 7A
[∵ A2 = A & IA = AI = A]
=1
∴ (C) is correct.

2nd PUC Maths Matrices Miscellaneous Exercise Additional Questions and Answers

One Mark Questions:

Question 1.
If \(\mathbf{A}=\left(\begin{array}{ll}{\mathbf{0}} & {\mathbf{1}} \\{\mathbf{1}} & {\mathbf{0}}\end{array}\right), \mathbf{F} \text { ind } \mathbf{A}^{4}\)
(UP CET 2010)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 17

Question 2.
If
\(\mathbf{A}=\left(\begin{array}{rr}{0} & {2} \\{3} & {-4}\end{array}\right), \mathbf{K A}=\left(\begin{array}{ll}{0} & {3 \mathbf{a}} \\{2 \mathbf{b}} & {24}\end{array}\right)\),find k, a and b.
Answer
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 18

KSEEB Solutions

Question 3.
Find X and Y
\(2 \mathbf{X}-\mathbf{Y}=\left[\begin{array}{rrr}{6} & {-6} & {0} \\{-4} & {2} & {1} \end{array}\right]\)
\(\mathbf{X}+2 \mathbf{Y}=\left[\begin{array}{rrr}{3} & {2} & {5} \\{-2} & {1} & {-7}\end{array}\right]\)
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 19
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 20

Question 4.
\(\left[\begin{array}{rrr}{1} & {1} & {1} \\{1} & {-2} & {-2} \\{1} & {3} &{2}\end{array}\right]\left[\begin{array}{l}{x} \\{y} \\{z}\end{array}\right]=\left[\begin{array}{l}{0} \\{3} \\{4}\end{array}\right], \text { then }\left[\begin{array}{l}{x} \\{y} \\{z}\end{array}\right]\)(I I T 2006)
Answer:

2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 21

Question 5.
\(\mathbf{A}=\left[\begin{array}{lll}{6} & {8} & {7} \\{4} & {2} & {3} \\{9} & {7} & {1}\end{array}\right]\)
is the sum of symmentric and skew symmentric matrix, Find B
Answer:
2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise 22

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