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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using factor theorem

Question 1.

Show that < = (x – a)^{2} (x + 2a)

Solution:

⇒ (x + 2d) is a factor of A.

Now degree of Δ is 3 (x × x × x = x^{3}) and we have 3 factors for A

∴ There can be a constant as a factor for A.

(i.e.,) Δ = k(x – a)^{2} (x + 2d)

equating coefficient of x^{3} on either sides we get k = 1

∴ Δ = (x – a)^{2} (x + 2a)

Question 2.

Show that

Solution:

Similarly b and c are factors of Δ.

The product of the leading diagonal elements is (b + c) (c + a) (a + b)

The degree is 3. And we got 3 factors for Δ ∴ m = 3 – 3 = 0

∴ there can be a constant k as a factor for Δ.

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Question 3.

Solution:

⇒ x = 0, 0 are roots.

Now the degree of the leading diagonal elements is 3.

∴ the equation is of degree 3, so the roots are 0, 0, – (a + b + c)

Question 4.

Show that = (a + b + c) (a – b) (b – c) (c – a)

Solution:

⇒ (a – b) is a factor of Δ.

Similarly, (b – c) and (c – a) are factors of Δ.

The degree of the product of elements along the leading diagonal is 1 + 1 + 2 = 4 and we got 3 factors for Δ.

m = 4 – 3 = 1

∴ There can be one more factor symmetric with a, b, c which is of the form k (a + b + c).

Question 5.

Solve

Solution:

Question 6.

Show that = (x – y) (y – z) (z – x)

Solution:

⇒ (x – y) is a factor of Δ.

Similarly (y – z) and (z – x) are factors of Δ.

Now degree of Δ = 0 + 1 + 2 = 3 and we have 3 factors of Δ.

and so there can be a constant k as a factor of Δ.

### Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3 Additional Problems

Question 1.

Prove that = (a – b) (b – c) (c – a) (a + b + c).

Solution:

∴ (a – b) is a factor of Δ.

Similarly, we observe that Δ is symmetric in a, b, c by putting b = c, c = a, we get Δ = 0.

Hence (b – c) and (c – a) are also factors of Δ.

∴ The product (a – b) (b – c) (c – a) is a factor of Δ. The degree of this product is 3.

The product of leading diagonal elements is 1. bc^{3}. The degree of this product is 4.

∴ By cyclic and symmetric properties, the remaining symmetric factor of the first degree must be k (a + b + c), where k is any non-zero constant.

Question 2.

Using factor method show that = (a – b) (b – c) (c – a)

Solution:

⇒ (a – b) is a factor of Δ.

similarly, (b – c) and (c – a) are factors of Δ.

The product of leading diagonal elements is bc^{2}. The degree of the product is 1 + 2 = 3.

∴ there will be three factors for Δ.

We got 3 factors for Δ as (a – b), (b – c) and (c – a). Its degree = 3.

∴ m = 3 – 3 = 0

∴ there can be a constant k as a factor of Δ.

Question 3.

Solution:

⇒ (a – b) is a factor of A.

Similarly, (b – c) and (c – a) are factors of Δ.

The degree of Δ = 5 and degree of product of factors = 3.