Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.3 Text Book Back Questions and Answers, Notes.

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

Solve the following problems by using Factor Theorem:

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 1.
Show that \(\left| \begin{matrix} x & a & a \\ a & x & a \\ a & a & x \end{matrix} \right| \) = (x – a)2 (x + 2a)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 1
By putting x = a , we have three rows of |A| are identical. Therefore (x – a)2 is a factor of |A|
Put x = – 2a in |A|
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 2
∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)2 (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 3
a3 [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k . 4a3
a3 [o + 2 + 2 ] = 4 ka3
4a3 = 4 ka3
k = 1
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 4

 

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 2.
Show that \(\left| \begin{matrix} b\quad +\quad c & a\quad -\quad c & a\quad -\quad b \\ b\quad -\quad c & c\quad +\quad a & b\quad -\quad a \\ c\quad -\quad b & c\quad -\quad a & a\quad +\quad b \end{matrix} \right| \) = 8 abc
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 5
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 6
since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 7
since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 8

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 9

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 3.
Solve that \(\left| \begin{matrix} x\quad +\quad a & b & c \\ a & x\quad +\quad b & c \\ a & b & x\quad +\quad c \end{matrix} \right| \) = 0
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 10
Put x = 0
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 11
x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 12
∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0 , – (a + b + c)

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 4.
Show that \(\left| \begin{matrix} b\quad +\quad c & a & { a }^{ 2 } \\ c\quad +\quad a & b & { b }^{ 2 } \\ a\quad +\quad b & c & { c }^{ 2 } \end{matrix} \right| \) = (a + b + c) (a – b) (b – c) (c – a)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 13
Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a , b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c2 is 4.
Therefore, the other factor is k (a + b + c).
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 14
5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 15

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 5.
Solve \(\left| \begin{matrix} 4\quad -\quad x & 4\quad +\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad -\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad +\quad x & 4\quad -\quad x \end{matrix} \right| \) = 0
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 16
Put x = 0
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 17
∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.

Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 18
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 19
∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0 , 0 , – 12

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 6.
Show that \(\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \) = (x – y) (y – z) (z – x)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 20
|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z2 is 3. Therefore, the other factor is the constant factor k.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 21
Put x = 0, y = 1, z = -1 we get
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 22
Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 23

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