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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 C_{1}, b from C_{2} and c from C_{3} we get

Expanding along R_{1} we get

(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)

1 = (abc) (4abc) = 4a^{2}b^{2}c^{2}

= 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 – (aα^{2} + 2bα + c) [ac – b^{2}]

So Δ = 0 ⇒ (aα^{2} + 2bα + c) (ac -b^{2}) = – 0 = 0

⇒ aα^{2} + 2bα + c = 0 or ac – b^{2} = 0

(i.e.) a is a root of ax^{2} + 2bx + c = 0

or ac = b^{2}

⇒ 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 x^{4}

Solution:

Multiplying R_{1} by a, R_{2} by b and R_{3} by c and

taking out a from C_{1} b from C_{2} and c from C_{3} 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 R_{1}

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| = k^{n} |A|.]

A and B are square matrices of order 3. Therefore,

AB is also a square matrix of order 3.

|3 AB| = 3^{3} |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; λ^{2} = (-2)^{2}; 3λ^{2} + 1 = 3 (4) + 1 = 13

6λ – 1 = 6(-2) – 1 = -13

expanding along R_{1}

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)

Question 21.

Using cofactors of elements of the second row, evaluate |A|, where

Solution:

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

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution: