Without expanding the determinant prove that : 0 p - q p - r q - p 0 q - r r - p r - q 0 = 0

Given, 0 p - q p - r q - p 0 q - r r - p r - q 0 Apply, C 1 → C 1 − C 2 , C 2 → C 2 − C 3 ⇒ q - p r - q p - r q - p r - q q - r q - p r - q 0 ⇒ q - p r - q 1 1 p - r 1 1 q - r 1 1 0 Since two columns are identical therefore, ∆ = 0 OR According to properties, skew symmetric matrix of odd order has ∆ = 0

If u = ax 2 + 2 b xy + cy 2 , u ' = a ' x 2 + 2 b ' x y + c ' y 2 , then prove that- | y 2 − x y x 2 a b c a ' b ' c ' | = | a x + b y b x + c y a ' x + b ' y b ' x + c ' y | = − 1 y | u u ' a x + b y a ' x + b ' y |

Take y 2 - x y x 2 a b c a ' b ' c ' first and then multiply and divide it by x y , = 1 x y | x y 2 − x y 2 x 2 a x b y c a ' x b ' y c ' | C 1 → C 1 + C 2 = 1 x y 0 - x y 2 x 2 a x + b y b y c a ' x + b ' y b ' y c ' (taking common x from R 1 , y from C 2 ) = 0 - y x a x + b y b c a ' x + b ' y b ' c ' = 1 x y | 0 − x y x y a x + b y b x c y a ' x + b ' y b ' x c ' y | C 2 → C 2 + C 1 = 1 x y 0 0 x y a x + b y b x + c y c y a ' x + b ' y b ' x + c ' y c ' y Expanding along 3 r d column ⇒ a x + b y b x + c y a ' x + b ' y b ' x + c ' y Now u = a x 2 + 2 b x y - c y 2 u ' = a ' x 2 + 2 b ' x y + c ' y 2 = a x + b y a ' x + b ' y b x + c y b ' x + c ' y (Taking transpose) A T = | A | [Interchanging] = - b x + c y b ' x + c ' y a x + b y a ' x + b ' y = - 1 x y b x y + c y 2 b ' x y + c ' y 2 a x 2 + b x y a ' x 2 + b ' x y = - 1 y u u ' a x + b y a ' x + b ' y

Solve the system of equations : z + a y + a 2 x + a 3 = 0 z + b y + b 2 x + b 3 = 0 z + c y + c 2 x + c 3 = 0 , where a ≠ b ≠ c .

We have, z + a y + a 2 x + a 3 = 0 z + b y + b 2 x + b 3 = 0 z + c y + c 2 x + c 3 = 0 So, a 2 x + a y + z = - a 3 b 2 x + b y + z = - b 3 c 2 x + c y + z = - c 3 ∆ = a 2 a 1 b 2 b 1 c 2 c 1 Apply R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , then ∆ = a 2 a 1 b 2 - a 2 b - a 0 c 2 - a 2 c - a 0 ⇒ ∆ = b - a c - a a 2 a 1 b + a 1 0 c + a 1 0 ⇒ ∆ = b - a c - a b - c And, ∆ 1 = − a 3 a 1 − b 3 b 1 − c 3 c 1 Apply R 2 → R 2 - R 1 and R 3 → R 3 - R 1 ∆ 1 = − a 3 a 1 a 3 − b 3 b - a 0 a 3 − c 3 c - a 0 ⇒ ∆ 1 = a - b a - c − a 3 a 1 a 2 + b 2 + a b - 1 0 a 2 + c 2 + a c - 1 0 ⇒ ∆ 1 = a - b a - c a 2 + c 2 + a c - a 2 - b 2 - a b ⇒ ∆ 1 = a - b a - c c - b a + b + c ⇒ ∆ 1 = a - b b - c c - a a + b + c And, ∆ 2 = a 2 - a 3 1 b 2 - b 3 1 c 2 - c 3 1 Apply R 2 → R 2 - R 1 and R 3 → R 3 - R 1 ∆ 2 = a 2 - a 3 1 b 2 - a 2 a 3 - b 3 0 c 2 - a 2 a 3 - c 3 0 ⇒ ∆ 2 = a - b a - c a 2 - a 3 1 - a + b a 2 + b 2 + a b 0 - a + c a 2 + c 2 + a c 0 Apply R 3 → R 3 - R 2 ∆ 2 = a - b a - c a 2 - a 3 1 - a + b a 2 + b 2 + a b 0 b - c c 2 - b 2 + a c - b 0 ⇒ ∆ 2 = a - b a - c a 2 - a 3 1 - a + b a 2 + b 2 + a b 0 b - c a + b + c c - b 0 ⇒ ∆ 2 = a - b a - c b - c a 2 - a 3 1 - a + b a 2 + b 2 + a b 0 1 - a + b + c 0 ⇒ ∆ 2 = a - b a - c b - c a 2 + b 2 + 2 a b + a c + b c - a 2 - b 2 - a b ⇒ ∆ 2 = a - b a - c b - c a b + a c + b c And, ∆ 3 = a 2 a - a 3 b 2 b - b 3 c 2 c - c 3 Apply R 2 → R 2 - R 1 and R 3 → R 3 - R 1 ⇒ ∆ 3 = a 2 a - a 3 b 2 - a 2 b - a a 3 - b 3 c 2 - a 2 c - a a 3 - c 3 ⇒ ∆ 3 = b - a c - a a 2 a - a 3 b + a 1 - a 2 + b 2 + a b c + a 1 - a 2 + c 2 + a c Apply R 3 → R 3 - R 2 ⇒ ∆ 3 = b - a c - a a 2 a - a 3 b + a 1 - a 2 + b 2 + a b c - b 0 b - c a + b + c ⇒ ∆ 3 = b - a c - a c - b a 2 a - a 3 b + a 1 - a 2 + b 2 + a b 1 0 - a + b + c ⇒ ∆ 3 = b - a c - a c - b - a 2 a + b + c - a - a + b a + b + c + a 2 + b 2 + a b + a 3 ⇒ ∆ 3 = b - a c - a c - b - a 2 a + b + c + a a + b a + b + c - a a 2 + b 2 + a b + a 3 ⇒ ∆ 3 = b - a c - a c - b a b a + b + c - a a 2 + b 2 + a b + a 3 ⇒ ∆ 3 = b - a c - a c - b a b c - a 3 + a 3 ⇒ ∆ 3 = b - a c - a c - b a b c Now x = ∆ 1 ∆ , y = ∆ 2 ∆ , z = ∆ 3 ∆ So, x = - a + b + c y = a - b a - c b - a a b + a c + b c b - a c - a b - c = a b + a c + b c z = b - a c - a c - b a b c b - a c - a b - c = - a b c

Use matrix to solve the following system of equations (c) x + y + z = 3 x + 2 y + 3 z = 4 2 x + 3 y + 4 z = 7

(a) Given, x + y + z = 3 … … … 1 x + 2 y + 3 z = 4 … … … 2 2 x + 3 y + 4 z = 7 … … … 3 A X = B Where A = 1 1 1 1 2 3 2 3 4 , X = x y z & B = 3 4 7 A = 1 8 - 9 - 1 4 - 6 + 1 3 - 4 = 0 Now, a d j . A = - 1 2 - 1 - 1 2 - 1 1 - 2 1 T = - 1 - 1 1 2 2 - 2 - 1 - 1 1 a d j . A B = - 1 - 1 1 2 2 - 2 - 1 - 1 1 3 4 7 = - 3 - 4 + 7 6 + 8 - 14 - 3 - 3 + 7 = 0 0 0 Thus the system of equation has infinitely many solution. Let z = k where k ∈ R , Substituting value in equation 1 , x + y = 3 − k Substituting value of z in equation 2 we get, x + 2 y = 4 - 3 k Solving these two equations we get, x = 2 + k , y = 1 − 2 k , z = k

Use matrix to solve the following system of equations (d) x + y + z = 3 x + 2 y + 3 z = 4 2 x + 3 y + 4 z = 9

(d) x + y + z = 3 x + 2 y + 3 z = 4 2 x + 3 y + 4 z = 9 ⇒ ∆ = 1 1 1 1 2 3 2 3 4 = 1 ( 8 − 9 ) − 1 ( 4 − 6 ) + 1 ( 3 − 4 ) = 0 Since, ∆ = 0 hence an entire row or column is zero.Therefore the system of equations is inconsistent, hence no solution. (An inconsistant system of equations has no solution).

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4

Embibe Experts Solutions for Chapter: Matrices and Determinants, Exercise 4: EXERCISE-4

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Matrices and Determinants, Exercise 4: EXERCISE-4

Attempt the free practice questions on Chapter 10: Matrices and Determinants, Exercise 4: EXERCISE-4 with hints and solutions to strengthen your understanding. Beta Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Matrices and Determinants, Exercise 4: EXERCISE-4 with Hints & Solutions

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 2

Without expanding the determinant prove that :

$|\begin{matrix} 0 & p - q & p - r \\ q - p & 0 & q - r \\ r - p & r - q & 0 \end{matrix}| = 0$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 14

Find the values of $c$ for which the equations $2 x + 3 y = 0$ , $(c + 2) x + (c + 4) y = c + 6$ , and $(c + 2)^{2} x + (c + 4)^{2} y = (c + 6)^{2}$ are consistent. Also solve above equations for these values of $c$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 21

If $u = {ax}^{2} + 2 b xy + {cy}^{2}, u' = a' x^{2} + 2 b' x y + c' y^{2}$ , then prove that-

$| \begin{matrix} y^{2} & - x y & x^{2} \\ a & b & c \\ a' & b' & c' \end{matrix} | = | \begin{matrix} a x + b y & b x + c y \\ a' x + b' y & b' x + c' y \end{matrix} | = - \frac{1}{y} | \begin{matrix} u & u' \\ a x + b y & a' x + b' y \end{matrix} |$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 22

Solve the system of equations : $\{\begin{cases} \begin{matrix} z + a y + a^{2} x + a^{3} = 0 \\ z + b y + b^{2} x + b^{3} = 0 \end{matrix} \\ z + c y + c^{2} x + c^{3} = 0 \end{cases}$ , where $a \neq b \neq c$ .

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 23

If x, y, z are not all zero and if $a x + b y + c z = 0; b x + c y + a z = 0; c x + a y + b z = 0$ Prove that $x : y : z = 1 : 1 : 1$ or $1 : ω : ω^{2}$ or $1 : ω^{2} : ω$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 24

Prove that the system of equations in $x$ and $y; a x + h y + g = 0, h x + b y + f = 0, a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = t$ is consistent if $t = |\begin{array}{l} a & h & g \\ h & b & f \\ g & f & c \end{array}| \div |\begin{array}{l} a & h \\ h & b \end{array}|$