Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

Question 1.
Without expanding the determinant, prove that
Solution:

Question 2.
Show that
Solution:

Question 3.
Prove that
Solution:
LHS
Taking a from C₁, b from C₂ and c from C₃ we get

Expanding along R₁ we get
(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)
1 = (abc) (4abc) = 4a²b²c²
= RHS

Question 4.

Solution:

Question 5.
Prove that
Solution:

Question 6.
Show that
Solution:

Question 7.
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.
Solution:

Question 8.

Solution:

we get – (aα² + 2bα + c) [ac – b²]
So Δ = 0 ⇒ (aα² + 2bα + c) (ac -b²) = – 0 = 0
⇒ aα² + 2bα + c = 0 or ac – b² = 0
(i.e.) a is a root of ax² + 2bx + c = 0
or ac = b²
⇒ a, b, c are in G.P.

Question 9.
Prove that
Solution:

Question 10.
If a, b, c are p^th, q^th and r^th terms of an A.P., find the value of
Solution:
We are given a = t_p,b = t_q and c = t_r
Let a be the first term and d be the common difference

Question 11.
Show that is divisible by x⁴
Solution:

Multiplying R₁ by a, R₂ by b and R₃ by c and
taking out a from C₁ b from C₂ and c from C₃ we get
==

Question 12.
If a, b, c are all positive, and are p^th, q^th and r^th terms of a G.P., show that

Solution:

Question 13.
Find the value of if x, y, z ≠ 1.
Solution:
Expanding the determinant along R₁

Question 14.

Solution:

Question 15.
Without expanding, evaluate the following determinants:

Solution:

Question 16.
If A is a square matrix and |A| = 2, find the value of |AA^T|.
Solution:
Given |A| = 2
[Property 1: The determinant of a matrix remains unaltered if its rows are changed into columns and columns. That is, |A| = |A^T|]

|A^T| = |A| = 2
∴ |A A^T| = |A| |A^T|
= 2 × 2 = 4

Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3AB|.
Solution:
Given A and B are square matrices of order 3 such that |A| = -1 and |B| = 3
[It A is a square matrix of order n then det ( kA) = |kA| = kⁿ |A|.]
A and B are square matrices of order 3. Therefore,
AB is also a square matrix of order 3.
|3 AB| = 3³ |AB|
= 27 |A| |B|
= 27 × – 1 × 3
|3 AB| = – 81

Question 18.
If λ = -2, determine the value of
Solution:
Given λ = -2
∴ 2λ = -4; λ² = (-2)²; 3λ² + 1 = 3 (4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13

expanding along R₁
0(0) + 4 (0 + 13) + 1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew-symmetric matrix is zero

Question 19.
Determine the roots of the equation
Solution:

Given the determinant value is 0
⇒ 30(1 + x) (2 – x) = 0
⇒ 1 + x = 0 or 2 – x = 0
⇒ x = -1 or x = 2
So, x = -1 or 2.

Question 20.
Verify that det (AB) = (det A) (det B) for
Solution:

{(-20)(52) (-19) + (10)(38)(—49) + (2)(64)(-17)} – {(-49)(52) (2) + (-17)(38)(-20) + (-19)(64)(10)}
= (19760 – 18620 – 2176) – (-5096 + 12920 – 12160)
= (19760 + 5096 + 12160) – (18620 + 2176 + 12920)
= 37016 – 33716 = 3300 ….(3)
Now (1) × (2) = (3)
(i.e.,) (-33) (-100) = 3300
⇒ det (AB) = (det A), (det B)