Without expanding evaluate the following determinant. 1 a b + c 1 b c + a 1 c a + b

Let D = 1 a b + c 1 b c + a 1 c a + b Applying C 3 → C 2 + C 3 , we have D = 1 a a + b + c 1 b a + b + c 1 c a + b + c Now, taking a + b + c common from C 3 , we get D = a + b + c 1 a 1 1 b 1 1 c 1 Since, C 1 and C 3 are same, ∴ D = a + b + c × 0 = 0 Hence, the value of given determinant is 0 .

Prove that x + y y + z z + x z + x x + y y + z y + z z + x x + y = 2 x y z z x y y z x

To prove: x + y y + z z + x z + x x + y y + z y + z z + x x + y = 2 x y z z x y y z x L.H.S. = x + y y + z z + x z + x x + y y + z y + z z + x x + y Applying R 1 → R 1 + R 2 + R 3 , = 2 x + y + z 2 x + y + z 2 x + y + z z + x x + y y + z y + z z + x x + y Taking 2 common from R 1 , = 2 x + y + z x + y + z x + y + z z + x x + y y + z y + z z + x x + y Applying R 1 → R 1 - R 3 , = 2 x y z z + x x + y y + z y + z z + x x + y Applying R 2 → R 2 - R 1 , = 2 x y z z x y y + z z + x x + y Applying R 3 → R 3 - R 2 = 2 x y z z x y y z x = R.H.S. Hence, proved.

Solve the following equation x + 2 x + 6 x - 1 x + 6 x - 1 x + 2 x - 1 x + 2 x + 6 = 0

Given, x + 2 x + 6 x - 1 x + 6 x - 1 x + 2 x - 1 x + 2 x + 6 = 0 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , We get x + 2 x + 6 x - 1 4 - 7 3 - 3 - 4 7 = 0 Applying C 2 → C 2 - C 1 and C 3 → C 3 - C 1 x + 2 4 - 3 4 - 11 - 1 - 3 - 1 10 = 0 ∴ ( x + 2 ) ( - 110 - 1 ) - 4 ( 40 - 3 ) - 3 ( - 4 - 33 ) = 0 ⇒ - 3 ( x + 2 ) - 4 + 3 = 0 ⇒ - 3 x - 6 - 1 = 0 ⇒ - 3 x = 7 ⇒ x = - 7 3 Hence, the value of x is - 7 3

Solve the following equation. x - 1 x x - 2 0 x - 2 x - 3 0 0 x - 3 = 0

Given, x - 1 x x - 2 0 x - 2 x - 3 0 0 x - 3 = 0 Applying R 2 → R 2 - R 3 , we get x - 1 x x - 2 0 x - 2 0 0 0 x - 3 = 0 ∴ ( x - 1 ) [ ( x - 2 ) ( x - 3 ) - 0 ] - x [ 0 - 0 ] + ( x - 2 ) ( 0 - 0 ) = 0 ⇒ ( x - 1 ) ( x - 2 ) ( x - 3 ) = 0 ⇒ x - 1 = 0 or ( x - 2 ) = 0 or ( x - 3 ) = 0 ⇒ x = 1 or x = 2 or x = 3 Hence, x = 1 or x = 2 or x = 3 .

If 4 + x 4 - x 4 - x 4 - x 4 + x 4 - x 4 - x 4 - x 4 + x = 0 , then find the values of x .

Given, 4 + x 4 - x 4 - x 4 - x 4 + x 4 - x 4 - x 4 - x 4 + x = 0 Applying C 1 → C 1 + C 2 + C 2 , we get 12 - x 4 - x 4 - x 12 - x 4 + x 4 - x 12 - x 4 - x 4 + x = 0 Taking 12 - x common from 4 12 - x 1 4 - x 4 - x 1 4 + x 4 - x 1 4 - x 4 + x = 0 Applying R 1 → R 1 - R 2 , we get 12 - x 0 - 2 x 0 1 4 + x 4 - x 1 4 - x 4 + x = 0 Expanding along first row, ∴ ( 12 - x ) [ 0 + 2 x ( 4 + x - 4 + x ) + 0 ] = 0 ⇒ ( 12 - x ) 4 x 2 = 0 ⇒ 4 x 2 = 0 or 12 - x = 0 ⇒ x = 0 or x = 12 Hence, x = 0 or x = 12 .

Without expanding determinants show that 1 3 6 6 1 4 3 7 12 + 4 2 3 3 2 1 2 1 7 6 = 10 1 2 1 3 1 7 3 2 6

To prove: 1 3 6 6 1 4 3 7 12 + 4 2 3 3 2 1 2 1 7 6 = 10 1 2 1 3 1 7 3 2 6 L.H.S. : 1 3 6 6 1 4 3 7 12 + 4 2 3 3 2 1 2 1 7 6 = 2 1 3 3 6 1 2 3 7 6 + 4 2 3 3 2 1 2 1 7 6 (Taking 2 common from third column of first determinant) = 2 3 3 12 1 2 6 7 6 + 8 3 3 8 1 2 4 7 6 (Multiplying 2 and 4 to first column of their respective determinants) = 10 3 3 20 1 2 10 7 6 (Adding two determinant because row 2 and 3 are same in both determinant) = 10 1 3 3 2 1 2 1 7 6 (Taking 10 common from first column) = 10 1 2 1 3 1 7 3 2 6 (Changing rows into column doesn't affect the value of determinant) = R.H.S. Hence, proved.

E M B I B E

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2

Maharashtra Board Solutions for Chapter: Determinants and Matrices, Exercise 2: EXERCISE 4.2

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Determinants and Matrices, Exercise 2: EXERCISE 4.2

Attempt the practice questions on Chapter 4: Determinants and Matrices, Exercise 2: EXERCISE 4.2 with hints and solutions to strengthen your understanding. Mathematics and Statistics (Arts & Science) Part 1 solutions are prepared by Experienced Embibe Experts.

Questions from Maharashtra Board Solutions for Chapter: Determinants and Matrices, Exercise 2: EXERCISE 4.2 with Hints & Solutions

EASY

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 1

Without expanding evaluate the following determinant.

$|\begin{array}{lll} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{array}|$

EASY

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 1

Evaluate the following determinant.

$|\begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x \end{matrix}|$

EASY

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 1

Without expanding evaluate the following determinant.

$|\begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix}|$

MEDIUM

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 2

Prove that $|\begin{matrix} x + y & y + z & z + x \\ z + x & x + y & y + z \\ y + z & z + x & x + y \end{matrix}| = 2 |\begin{matrix} x & y & z \\ z & x & y \\ y & z & x \end{matrix}|$

MEDIUM

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 5

Solve the following equation

$|\begin{matrix} x + 2 & x + 6 & x - 1 \\ x + 6 & x - 1 & x + 2 \\ x - 1 & x + 2 & x + 6 \end{matrix}| = 0$

MEDIUM

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 5

Solve the following equation.

$|\begin{matrix} x - 1 & x & x - 2 \\ 0 & x - 2 & x - 3 \\ 0 & 0 & x - 3 \end{matrix}| = 0$

MEDIUM

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 6

If $|\begin{matrix} 4 + x & 4 - x & 4 - x \\ 4 - x & 4 + x & 4 - x \\ 4 - x & 4 - x & 4 + x \end{matrix}| = 0$ , then find the values of $x$ .

MEDIUM

11th Maharashtra Board

IMPORTANT

Mathematics and Statistics (Arts & Science) Part 1>Chapter 4 - Determinants and Matrices>EXERCISE 4.2>Q 7

Without expanding determinants show that

$|\begin{array}{r} 1 & 3 & 6 \\ 6 & 1 & 4 \\ 3 & 7 & 12 \end{array}| + 4 |\begin{array}{l} 2 & 3 & 3 \\ 2 & 1 & 2 \\ 1 & 7 & 6 \end{array}| = 10 |\begin{array}{l} 1 & 2 & 1 \\ 3 & 1 & 7 \\ 3 & 2 & 6 \end{array}|$

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