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

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

Question 1.
Construct an m × n matrix A = [a_ij], where a_ij is given by

Solution:
(i) a_ij = \(\frac{(i-2 j)^{2}}{2}\)
Here m = 2, n = 3
So we have to construct a matrix of order 2 × 3

(ii) Here m = 3 and n = 4
So we have to construct a matrix order 3 × 4
The general form of a matrix of order 3 × 4 will be

Question 2.
Find the values of p, q, r and s if

Solution:
When two matrices (of the same order) are equal then their corresponding entries are equal.

p² – 1 = 1 ………… (1)
– 31 – q³ = – 4 ……….. (2)
r + 1 = \(\frac{3}{2}\) ……… (3)
s – 1 = – π ……….. (4)
(1) ⇒ p² – 1 ⇒ p² = 1 + 1 = 2
p = ±√2
(2) ⇒ – 31 – q³ – 4
31 + q³ = 4
q³ = 4 – 31 = – 27
q³ = (-3)³
q = – 3
(3) ⇒ r + 1 = \(\frac{3}{2}\)
r = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\)
(4) ⇒ s – 1 = – π
s = 1 – π
∴ The required values are
p = ±√2,
q = 3,
r = \(\frac{1}{2}\),
s = 1 – π

Question 3.
Determine the value of x + y if
Solution:
\(\left[ \begin{matrix} 2x\quad +\quad y & 4x \\ 5x\quad -\quad 7 & 4x \end{matrix} \right] \) = \(\left[ \begin{matrix} 7 & 7y\quad -\quad 13 \\ y & x\quad +\quad 6 \end{matrix} \right]\)
Equating the corresponding entries
2x + y = 7 ………. (1)
4x = 7y – 13
4x – 7y = – 13 ………. (2)
5x – 7 = y
5x – y = 7 ………. (3)
4x = x + 6
4x – x = 6
3x = 6
x = \(\frac{6}{3}\) = 2 ………. (4)
Substituting for x in equation (1)
(1) ⇒ 2 × 2 + y = 7
y = 7 – 4 = 3
The required values are x = 2 and y = 3
x + y = 2 + 3 = 5
x + y = 5

Question 4.
Determine the matrices A and B if they satisfy

Solution:

Question 5.
If A = \(\left[\begin{array}{ll}{\mathbf{1}} & {\boldsymbol{a}} \\ {\mathbf{0}} & {\mathbf{1}}\end{array}\right]\), then compute A⁴
Solution:

Question 6.
Consider the matrix A_α = \(\left[\begin{array}{cc}{\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha}\end{array}\right]\)
(i) Show that A_αA_β = A_{α + β}.
(ii) Find all possible real values of satisfying the condition A_α + A^T_α = 1.
Solution:

General solution is α = 2nπ + \(\frac{\pi}{3}\), n ∈ Z

Question 7.
If A = \(\left[\begin{array}{rr}{4} & {2} \\ {-1} & {x}\end{array}\right]\) such that (A – 2I) (A – 3I) = 0, find the value of x.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{0}} \\ {\mathbf{0}} & {\mathbf{1}} & {\mathbf{0}} \\ {\boldsymbol{a}} & {\boldsymbol{b}} & {-\mathbf{1}}\end{array}\right]\), show that A² is a unit matrix.
Solution:

Question 9.
If A = and A³ – 6A² + 7A + KI = 0, find the value of k.
Solution:

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.
Solution:

Question 11.
Show that f(x) f(y) = f(x + y), where f(x) =
Solution:

Question 12.
If A is a square matrix such that A² = A, find the value of 7A – (I + A)³.
Solution:
Given A is a square matrix such that A² = A.
(I + A)³ = (I + A) (I + A) (I + A)
= (I . I + I . A + A . I + A . A) (I + A)
= (I + A + A + A²) (I + A)
= (I + 2A + A ) (I + A)
[Given A² = A]
= (I + 3A) (I + A)
= I . I + I . A + 3A . I + 3A . A
= I + A + 3A + 3A²
= I + 4A + 3A
[Given A² = A]
(I + A)³ = I + 7A
∴ 7A – (I + A)² = 7A – (I + 7A)
= 7A – I – 7A
7A – (I + A)² = – I

Question 13.
Verify the property A (B + C) = AB + AC, when the matrices A, B, and C are given by

Solution:

Question 14.
Find the matrix A which satisfies the matrix relation
Solution:

Question 15.

(i) (A + B)^T = A^T + B^T = B^T + A^T
(ii) (A – B)^T = A^T – B^T
(iii) (B^T)^T = B.
Solution:

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?
Solution:
Given Order of A = 3 × 4
∴ Order of A^T = 4 × 3
Given that A^TB is defined.
∴ Number of columns of A^T = Number of rows of B
Number of rows of B = 3
Also given BA^T is defined.
∴ Number of columns of B = Number of rows of A^T
Number of columns of B = 4
∴ Order of B = 3 × 4

Question 17.
Express the following matrices is the sum of a symmetric matrix and a skew-symmetric matrix:

Solution:

Question 18.
Find the matrix A such that
Solution:

Question 19.
If A = is a matrix such that AA^T = 9I, find the values of x and y.
Solution:

Question 20.

Solution:

Question 21.
Construct the matrix A = [a_ij]_3×3, where a_ij = i- j. State whether A is symmetric or skew- symmetric.
Solution:
Given A is a matrix of order 3 × 3

Here A^T = -A
⇒ A is skew-symmetric

Question 22.
Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.
Solution:
Let A and B be two symmetric matrices
⇒ A^T = A and B^T = B …………….. (1)
Given that AB = BA (2)
To prove AB is symmetric:
Now (AB)^T = B^TA^T = BA
(from(1)) But (AB)^T = AB by ………….. (2)
⇒ AB is symmetric.
Conversely, let AB be a symmetric matrix.
⇒ (AB)^T = AB
i.e. B^TA^T = AB
i.e. BA = AB (from (1))
⇒ AB is symmetric

Question 23.
If A and B are symmetric matrices of the same order, prove that
(i) AB + BA is a symmetric matrix.
(li) AB – BA is a skew-symmetric matrix.
Solution:
Given A and B are symmetric matrices of the same order.
∴ A^T = A , B^T = B

(i) AB + BA is a symmetric matrix
(AB + BA)^T = (AB )^T + (BA)^T
= B^TA^T + A^TB^T
= BA + AB
= AB + BA
∴ AB + BA is a symmetric matrix.

(ii) AB – BA is a skew – symmetric matrix
(AB – BA)^T = (AB )^T (BA )^T
= B^TA^T – A^TB^T
= BA – AB
(AB – BA)^T = – (AB – BA)
∴ AB – BA is a skew symmetric matrix.

Question 24.
A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds.
The pack-I 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?
Solution:

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

Question 1.

Prove that (i) AB ≠ BA
(ii) A(BC) = (AB) C
(iii) A(B + C) = AB + AC
(iv) AI = IA = A
Solution:

Question 2.
If A = \(\left[\begin{array}{ll}{2} & {3} \\ {4} & {5}\end{array}\right]\) find A² – 7A – 2I.
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If A = \(\left[\begin{array}{rr}{3} & {-5} \\ {-4} & {2}\end{array}\right]\), show that A² – 5A – 14I = 0 where I is the unit matrix of order 2.
Solution:

Question 6.
If A = \(\left[\begin{array}{rr}{3} & {-2} \\ {4} & {-2}\end{array}\right]\), find k so that A² = kA – 2I
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}{1} & {2} & {2} \\ {2} & {1} & {2} \\ {2} & {2} & {1}\end{array}\right]\), show that A² – 4A – 5I = 0
Solution: