If a + b + c = 0 , solve for x : a - x c b c b - x a b a c - x = 0

a - x c b c b - x a b a c - x = 0 Apply, C 1 → C 1 + C 2 + C 3 ⇒ a + b + c - x c b a + b + c - x b - x a a + b + c - x a c - x = 0 ⇒ - x c b - x b - x a - x a c - x = 0 [ ∵ a + b + c = 0 ] ⇒ - x 1 c b 1 b - x a 1 a c - x = 0 ⇒ x = 0 o r 1 c b 1 b - x a 1 a c - x = 0 R 1 → R 1 − R 2 , R 2 → R 2 − R 3 ⇒ 0 c − b + x b − a 0 b − a − x a − c + x 1 a c - x = 0 ⇒ ( c − b + x ) ( a − c + x ) − ( b − a ) ( b − a − x ) = 0 ⇒ ( c − b ) ( a − c ) + x 2 + x ( a − b ) − ( b − a ) 2 + x ( b − a ) = 0 x = 0 or x 2 = a 2 + b 2 + c 2 - a b - b c - c a x = ± a 2 + b 2 + c 2 - a b - b c - c a

If a + b + c = 0 , solve for x : a - x c b c b - x a b a c - x = 0

a - x c b c b - x a b a c - x = 0 Apply, C 1 → C 1 + C 2 + C 3 ⇒ a + b + c - x c b a + b + c - x b - x a a + b + c - x a c - x = 0 ⇒ - x c b - x b - x a - x a c - x = 0 [ ∵ a + b + c = 0 ] ⇒ - x 1 c b 1 b - x a 1 a c - x = 0 ⇒ x = 0 o r 1 c b 1 b - x a 1 a c - x = 0 R 1 → R 1 − R 2 , R 2 → R 2 − R 3 ⇒ 0 c − b + x b − a 0 b − a − x a − c + x 1 a c - x = 0 ⇒ ( c − b + x ) ( a − c + x ) − ( b − a ) ( b − a − x ) = 0 ⇒ ( c − b ) ( a − c ) + x 2 + x ( a − b ) − ( b − a ) 2 + x ( b − a ) = 0 x = 0 or x 2 = a 2 + b 2 + c 2 - a b - b c - c a x = ± a 2 + b 2 + c 2 - a b - b c - c a

Prove that b c b c ' + b ' c b ' c ' c a c a ' + c ' a c ' a ' a b a b ' + a ' b a ' b = a b ' - a ' b b c ' - b ' c c a ' - c ' a

Given, b c b c ' + b ' c b ' c ' c a c a ' + c ' a c ' a ' a b a b ' + a ' b a ' b ' Now taking ( b ′ c ′ ) , ( c ′ a ′ ) & ( a ′ b ′ ) as common from R 1 , R 2 & R 3 respectively, ⇒ Δ = ( b ' c ' ) ( c ' a ' ) ( a ' b ' ) b c b ' c ' b b ' + c c ' 1 c a c ' a ' c c ' + a a ' 1 a b a ' b ' a a ' + b b ' 1 Now, R 2 → R 2 − R 1 , R 3 → R 3 − R 1 ⇒ Δ = ( a ' b ' c ' ) 2 b c b ' c ' b b ' + c c ' 1 c c ' a a ' - b b ' a a ' - b b ' 0 b b ' a a ' - c c ' a a ' - c c ' 0 ⇒ Δ = ( a ' b ' c ' ) 2 a a ' - b b ' a a ' - c c ' b c b ' c ' b b ' + c c ' 1 c c ' 1 0 b b ' 1 0 Expanding it, ⇒ Δ = ( a ' b ' c ' ) 2 a b ' − a ' b a ' b ' a c ' − a ' c a ' c ' c c ' - b b ' ⇒ ∆ = ( a b ' − a ' b ) ( b c − b ' c ) ( c a ' − c ' a )

For a fixed positive integer n , if D = n ! ( n + 1 ) ! ( n + 2 ) ! ( n + 1 ) ! ( n + 2 ) ! ( n + 3 ) ! ( n + 2 ) ! ( n + 3 ) ! ( n + 4 ) ! then show that D ( n ! ) 3 - 4 is divisible by n .

Given: D = n ! n + 1 ! n + 2 ! n + 1 ! n + 2 ! n + 3 ! n + 2 ! n + 3 ! n + 4 ! We know that n ! = n n - 1 ! = n n - 1 n - 2 ! . . . . i Taking n ! , ( n + 1 ) ! and ( n + 2 ) ! common from R 1 , R 2 and R 3 , respectively. ⇒ D = n ! n + 1 ! n + 2 ! 1 n + 1 n + 1 n + 2 1 n + 2 n + 2 n + 3 1 n + 3 n + 3 n + 4 ⇒ D = n ! 3 n + 1 2 n + 2 1 n + 1 n + 1 n + 2 1 n + 2 n + 2 n + 3 1 n + 3 n + 3 n + 4 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 2 D = n ! 3 n + 1 2 n + 2 1 n + 1 n + 1 n + 2 0 1 2 n + 4 0 1 2 n + 6 Expanding along C 1 D = n ! 3 n + 1 2 n + 2 2 n + 6 - 2 n + 4 ⇒ D = 2 n ! 3 n + 1 2 n + 2 On dividing both sides by n ! 3 ⇒ D n ! 3 = 2 n ! 3 n + 1 2 n + 2 n ! 3 ⇒ D n ! 3 = 2 n 3 + 4 n 2 + 5 n + 2 ⇒ D n ! 3 = 2 n n 2 + 4 n + 5 + 4 ⇒ D n ! 3 - 4 = 2 n n 2 + 4 n + 5 Now we can say that D n ! 3 - 4 is divisible by n

Find the value of the determinant 1 z 1 z - ( x + y ) z 2 - ( y + z ) x 2 1 x 1 x - y ( y + z ) x 2 z x + 2 y + z x z - y ( x + y ) x z 2

Let D = 1 z 1 z - ( x + y ) z 2 - ( y + z ) x 2 1 x 1 x - y ( y + z ) x 2 z x + 2 y + z x z - y ( x + y ) x z 2 ⇒ D = 1 x 4 z 4 z z - x + y - y + z x x - y z y + z x z x + 2 y + z - x y x + y R 2 → R 2 + R 1 ⇒ D = 1 x 4 z 4 z z - x + y - y x + z - y - y z y + z x z x + 2 y + z - x y x + y Expanding along R 2 , we get D = 1 x 4 z 4 y - x y z x + y + x z x + y x + 2 y + z + x + z - x y z x + y - y z x + y y + z + y x z 2 x + 2 y + z + y z 2 y + z ⇒ D = 1 x 4 z 4 y x z x + y - y + x + 2 y + z + y z x + y x + z - x - y + z + y z 2 x 2 + 2 x y + x z + y 2 + y z ⇒ D = 1 x 4 z 4 y x z x + y x + y + z - y z x + y x + z x + y + z + y z 2 x + y x + y + z ⇒ D = x + y x + y + z x 4 z 4 x y z - y z x + z + y z 2 ⇒ D = x + y x + y + z x 4 z 4 x y z - x y z - y z 2 + y z 2 ⇒ D = 0

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 7

EASY

JEE Main/Advance

IMPORTANT

Earn 100

If $a + b + c = 0$ , solve for $x : |\begin{matrix} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{matrix}| = 0$

Important Questions on Matrices and Determinants

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 8

Prove that

|\begin{array}{l} b c & b c^{'} + b^{'} c & b^{'} c^{'} \\ c a & c a^{'} + c^{'} a & c^{'} a^{'} \\ a b & a b^{'} + a^{'} b & a^{'} b \end{array}| = (a b^{'} - a^{'} b) (b c^{'} - b^{'} c) (c a^{'} - c^{'} a)

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 9

Let the three-digit numbers

A 28, 3 B 9

and

62 C

, where

A, B, C

are integers between

0 and 9

be divisible by a fixed integer

k

. Show that the determinant

|\begin{array}{lll} A 3 6 \\ 8 9 C \\ 2 B 2 \end{array}|

is divisible by

k

.

HARD

JEE Main/Advance

IMPORTANT

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

For a fixed positive integer $n$ , if

$D = |\begin{matrix} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{matrix}|$ then show that $[\frac{D}{(n!)^{3}} - 4]$ is divisible by $n$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 11

Find the value of the determinant

|\begin{matrix} \frac{1}{z} & \frac{1}{z} & - \frac{(x + y)}{z^{2}} \\ - \frac{(y + z)}{x^{2}} & \frac{1}{x} & \frac{1}{x} \\ - \frac{y (y + z)}{x^{2} z} & \frac{x + 2 y + z}{x z} & - \frac{y (x + y)}{x z^{2}} \end{matrix}|

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 12

Solve the following sets of equations using Cramer's rule and remark about their consistency.

$x - 3 y + z = 2 3 x + y + z = 6 5 x + y + 3 z = 3$

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 $c$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 15

Let

a, b, c, d

be real numbers in G.P. If

u, v, w

satisfy the system of equations

u + 2 v + 3 w = 6

4 u + 5 v + 6 w = 12, 6 u + 9 v = 4

, then show that the roots of the equations

(\frac{1}{u} + \frac{1}{v} + \frac{1}{w}) x^{2} + [(b - c)^{2} + (c - a)^{2} + (d - b)^{2}] x + u + v + w = 0

and

20 x^{2} + 10 (a - d)^{2} x - 9 = 0

are reciprocals of each other.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 16

Prove the identity

$|\begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ - b c + c a + a b & b c - c a + a b & b c + c a - a b \\ (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix}|$