Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.1

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 1.
Construct an m × n matrix A = [aij], where aij is given by
(i) aij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2 , n = 3
(ii) aij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3 , n = 4
Answer:
(i) aij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2 , n = 3
To construct 2 × 3 matrices.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 1
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 2

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

(ii) aij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3 , n = 4
To construct a 3 × 4 matrices.
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

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Question 2.
Find the value of p, q, r and s if
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 7
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 8
Equating the corresponding entries
⇒ p2 – 1 = 1
⇒ p2 = 1 + 1 = 2
p = ± \(\sqrt{2}\)
-31 – q3 = -4
-q3 = -4 + 31 = 27
q3 = -27 = (-3)3
⇒ q = -3
r + 1 = \(\frac{3}{2}\)
⇒ r = \(\frac{3}{2}\) – 1 = \(\frac{3-2}{2}\) = \(\frac{1}{2}\)
s – 1 = π
⇒ s = – π + 1 (i.e.,) s = 1 – π
So, p = ± \(\sqrt{2}\), q = -3, r = 1/2 and s = 1 – π

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 3.
Determine the value of x + y if
\(\left[ \begin{matrix} 2x\quad +\quad y & 4x \\ 5x\quad -\quad 7 & 4x \end{matrix} \right] \) = \(\left[ \begin{matrix} 7 & 7y\quad -\quad 13 \\ y & x\quad +\quad 6 \end{matrix} \right]\)
Answer:
\(\left[\begin{array}{cc}{2 x+y} & {4 x} \\ {5 x-7} & {4 x}\end{array}\right]=\left[\begin{array}{cc}{7} & {7 y-13} \\ {y} & {x+6}\end{array}\right]\)
⇒ 2x + y = 7 ………….. (1)
4x = 7y – 13 ………….. (2)
5x – 7 = y …………… (3)
4x = x + 6 ……………. (4)
from (4) 4x – x = 6
3x = 6 ⇒ x = \(\frac{6}{3}\) = 2
Substituting x = 2 in (1), we get
2(2) + y = 7 ⇒ 4 + y = 7 ⇒ y = 7 – 4 = 3
So x = 2 and y = 3
∴ x + y = 2 + 3 = 5

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 4.
Determine the matrices A and B if they satisfy 2A – B + \(\left[ \begin{matrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{matrix} \right] \) = 0 and A – 2B = \(\left[ \begin{matrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{matrix} \right] \)
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 12

Question 5.
If A = \(\left[ \begin{matrix} 1 & a \\ 0 & 1 \end{matrix} \right] \), then compute A4
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 6.
Consider the matrix Aα = \(\left[ \begin{matrix} cos\quad α & -\quad sin\quad α \\ sin\quad α & cos\quad α \end{matrix} \right] \)
(i) Show that AαAβ = A(α+β)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 14

(ii) Find all possible real values α satisfying the condition Aα + AαT = I
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 16

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 7.
If A = \(\left[ \begin{matrix} 4 & 2 \\ -1 & x \end{matrix} \right] \) and such that (A – 2I) (A – 3I) = 0, find the value of x.
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

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Question 8.
If A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{matrix} \right] \), show that A2 is a unit matrix.
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 21

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 9.
If A = \(\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right]\) and A3 – 6A2 + 7A + kI = 0, find the value of k.
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

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Equating the corresponding entries – 2 + k = 0 ⇒ k = 2
∴ The required value of k is k = 2

Question 10.
Give your own examples of matrices satisfying the following conditions in each case:
(i) A and B such that AB ≠ BA
(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.
(iii) A and B such that AB = 0 and BA ≠ 0
Answer:
(i) A and B such that AB ≠ BA
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 26

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 28

(iii) A and B such that AB = 0 and BA ≠ 0
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 29

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 11.
Show that f(x) f(y) = f(x + y) , where f(x) = \(\left[ \begin{matrix} cos\quad x & -\quad sin\quad x & 0 \\ sin\quad x & cos\quad x & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 12.
If A is a square matrix such that A2 = A, find the value of 7A – (I + A )3
Answer:
Given A2 = A
So 7A – (I + A)3 = 7A – (I + 3A + 3A2 + A3]
= 7A – I – 3A – 3 A2 – A3
Given A2 = A
7A – I – 3A – 3A – A3 = -I + A – A3
= -I + A – (A2 × A)
= -I + A – (A × A) = -I + A – A2
= -I + A – A = -I
So the value of 7A – (I + A)3 = -I.

Question 13.
Verify the property A (B + C) = AB + AC, when the matrices A, B, and C are given by
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 31
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 32
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 14.
Find the matrix A which satisfies the matrix relation A\(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] \) = \(\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] \)
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 37

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 15.
If AT = \(\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \)
verify the following
(i) (A + B)T = AT + BT = BT + AT
(ii) (A – B)T = AT – BT
(iii) (BT)T = B
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 38
(i) (A + B)T = AT + BT = BT + AT
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

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(ii) (A – B)T = AT – BT
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

(iii) (BT)T = B
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Question 16.
If A is a 3 × 4 matrix and B is a matrix such that both ATB and BAT are defined, what is the order of the matrix B?
Answer:
A is a matrix of order 3 × 4
So AT will be a matrix of order 4 × 3
AT B will be defined when B is a matrix of order 3 × n
BAT will be defined when B is of order m × 4
from (1) and (2) we see that B should be a matrix of order 3 × 4

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 17.
Express the following matrices as the sum of a symmetric matrix and a skew – symmetric matrix:
(i) \(\left[ \begin{matrix} 4 & -2 \\ 3 & -5 \end{matrix} \right] \)
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 46
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 47
A = \(\frac{1}{2}\)(A + AT) + \(\frac{1}{2}\)(A – AT
Thus A is expressed as a sum of a symmetric and skew-symmetric matrix.

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

(ii) \(\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right] \)
Answer:
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 18.
Find the matrix A such that
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Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 54
Equating like entries
2a – d = -1, 2b – e = – 8, 2c – f = -10
a = 1, b = 2, c = -5

2a – d = -1 ⇒ 2 × 1 – d = – 1
⇒ 2 + 1 = d ⇒ d = 3

2b – e = – 8 ⇒ 2 × 2 – e = – 8
⇒ 4 + 8 = e ⇒ e = 12

2c – f = -10 ⇒ 2 × – 5 – f = -10
⇒ – 10 – f = -10 ⇒ f = 0
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 19.
If A = \(\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -\quad 2 \\ x & 2 & y \end{matrix} \right] \) is a matrix such that AAT = 9I, find the values of x and y.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 56
Equating the corresponding entries
x + 4 + 2y = 0 ………… (1)
2x + 2 – 2y = o ………… (2)
x + 4 + 2y = 0 ………… (3)
2x + 2 – 2y = 0 ………… (4)
x2 + 4 + y2 = 9 ………… (5)
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 57
Substituting the value of y in equation (1) we have
x + 4 + 2x – 1 = 0
x + 4 – 2 = 0 ⇒ x = – 2
Substituting x = – 2 and y = – 1 in equation(5) we have
(5) ⇒ (-2)2 + 4 + (- 1)2 = 9
4 + 4 + 1 = 9
9 = 9
∴ The required values of x and y are
x = – 2 and y = – 1

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 20.
(i) For what value of x, the matrix A = \(\left[ \begin{matrix} 0 & 1 & -\quad 2 \\ -\quad 1 & 0 & { x }^{ 3 } \\ 2 & -\quad 3 & 0 \end{matrix} \right] \) is skew – symmetric
Answer:
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The matrix A is skew-symmetric if A = – AT
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 59
Equating the corresponding entries
x3 – 3 = 0
x3 = 3 ⇒ x = 31/3

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

(ii) If \(\left[ \begin{matrix} 0 & p & 3 \\ 2 & { q }^{ 2 } & -\quad 1 \\ r & 1 & 0 \end{matrix} \right] \) is skew – symmetric find the values of p, q and r.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 60
A is skew-symmetric if A = – AT
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 61
Equating the corresponding entries.
p = – 2 , r = – 3
q2 = -q2 ⇒ q2 + q2 = 0
⇒ 2q2 = 0 ⇒ q = 0
∴ The required values are
p = – 2 , q = 0 , r = – 3

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 21.
Construct the matrix A = [aij]3×3 , where aij = 1 – j. State whether A is symmetric or skew – symmetric.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 62
aij = i – j
a11 = 1 – 1 = 1
a12 = 1 – 2 = – 1
a13 = 1 – 3 = – 2
a21 = 2 – 1= 1
a22 = 2 – 2 = 1
a23 = 2 – 3 = – 1
a31 = 3 – 1= 2
a32 = 3 – 2 = 1
a33 = 3 – 3 = 0
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 22.
Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.
Answer:
Given A and B two symmetric matrices.
∴ A = AT and B = BT
First, let us assume AB = BA.
Let us prove AB is a symmetric matrix.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 64
∴ AB is a symmetric matrix.
conversely let us assume that AB is a symmetric matrix.
we prove AB = BA
AB is symmetric then
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Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 23.
If A and B are symmetric matrices of the same order, prove that
(i) AB + BA is a symmetric matrix
(ii) AB – BA is a skew-symmetric matrix.
Answer:
Given A and B are symmetric matrices
⇒ – AT = A and BT = B
(i) To prove AB + BA is a symmetric matrix.
Proof: Now (AB + BA)T = (AB)T + (BA)T = BTAT + ATBT
= BA + AB = AB + BA
i.e. (AB + BA)T = AB + BA
⇒ (AB + BA) is a symmetric matrix.

(ii) To prove AB – BA is a skew symmetric matrix.
Proof: (AB – BA)T = (AB)T – (BA)T = BTAT – ATBT = BA – AB
i.e. (AB – BA)T = – (AB – BA)
⇒ AB – BA is a skew symmetric matrix.

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Question 24.
A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds. The pack contains 100 gm of cashew nuts, 100 gm of raisins, and 50 gm of almonds. Pack – II contains 200 gm of cashew nuts, 100 gm of raisins, and 100 gm of almonds. Pack -III contains 250 gm of cashew nuts, 250 gm of raisins, and 150 gm of almonds. The cost of 50 gm of cashew nuts is ₹ 50, 50 gm of raisins is ₹ 10, and 50 gm of almonds is ₹ 60. What is the cost of each gift pack?
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 66
Let us consider 50 gm of cashew nuts as one unit, 50 gms of raisins as one unit 50 gm of almonds as one unit.
∴ The Gift pack matrix becomes
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 67
Also given 50 gms of Cashew nuts cost = Rs. 50
50 gms of Raisins cost = Rs. 10
50 gms of Almonds cost = Rs. 60
∴ Cost matrix is B = [50 10 60]
∴ Cost of cash gift pack = BA

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.1 68
∴ Cost of I gifts Pack = Rs. 180
Cost of II gift Pack = Rs.340
Cost of III gift Pack = Rs. 480

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