Without expanding the determinant prove that : 0 b - c - b 0 a c - a 0 = 0

Given, 0 b - c - b 0 a c - a 0 C 2 → c C 2 , C 3 → b C 3 ⇒ 1 b c 0 c b - b c - b 0 b a c - c a 0 C 2 → C 2 + C 3 ⇒ 1 b c 0 0 - b c - b b a b a c - c a 0 C 1 → a C 1 ⇒ 1 a b c 0 0 - b c - a b b a b a a c - c a 0 C 1 → C 1 + C 2 ⇒ 1 a b c 0 0 - b c 0 b a b a 0 - c a 0 = 0 [Since, one column/row elements is completely zero] OR According to properties, skew symmetric matrix of odd order has ∆ = 0 .

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

Prove that : a x b y c z x 2 y 2 z 2 1 1 1 = a b c x y z y z z x x y

To prove : a x b y c z x 2 y 2 z 2 1 1 1 = a b c x y z y z z x x y Expand. L.H.S. a x b y c z x 2 y 2 z 2 1 1 1 = a x y 2 - z 2 - b y x 2 - z 2 + c z x 2 - y 2 R.H.S. a b c x y z y z z x x y = a x y 2 - x z 2 - b x 2 y - y z 2 + c x 2 z - y 2 z = a x y 2 - z 2 - b y x 2 - z 2 + c z x 2 - y 2 Therefore, L.H.S. = R.H.S.

Prove that 1 a a 2 - b c 1 b b 2 - c a 1 c c 2 - a b = 0 .

LHS = 1 a a 2 - b c 1 b b 2 - c a 1 c c 2 - a b Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we get LHS = 1 a a 2 - b c 0 b - a b 2 - a 2 + b c - c a 0 c - a c 2 - a 2 + b c - a b LHS = 1 a a 2 - b c 0 b - a ( b - a ) ( a + b + c ] 0 c - a ( c - a ) [ a + b + c ] LHS = b - a c - a 1 a a 2 - b c 0 1 a + b + c 0 1 a + b + c LHS = ( b - a ) ( c - a ) ( 1 ) ( 0 ) ( ∵ R 2 and R 3 are identical) LHS = 0 Therefore, LHS = RHS Hence proved.

Prove that a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 = ( a - 1 ) 3 .

Given L . H . S . = a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 [to create two zeros in the third column] = a 2 + 2 a 2 a + 1 1 2 a + 1 - a 2 - 2 a a + 2 - 2 a - 1 1 - 1 3 - a 2 - 2 a 3 - 2 a - 1 1 - 1 = a 2 + 2 a 2 a + 1 1 1 - a 2 1 - a 0 3 - a 2 - 2 a 2 - 2 a 0 Expanding by third column, we have = 1 1 - a 2 1 - a 3 - a 2 - 2 a 2 - 2 a - 0 + 0 = ( 1 - a 2 ) ( 2 - 2 a ) - ( 1 - a ) ( 3 - a 2 - 2 a ) = ( 1 - a ) [ 2 - 2 a 2 - 3 + a 2 + 2 a ] = ( - 1 ) ( a - 1 ) [ - a 2 - 1 + 2 a ] = ( - 1 ) ( a - 1 ) ( - 1 ) ( a 2 + 1 - 2 a ) = ( a - 1 ) ( a - 1 ) 2 = ( a - 1 ) 3 .

Prove that 1 + a 2 - b 2 2 a b - 2 b 2 a b 1 - a 2 + b 2 2 a 2 b - 2 a 1 - a 2 - b 2 = ( 1 + a 2 + b 2 ) 3 .

Let ∆ = 1 + a 2 - b 2 2 a b - 2 b 2 a b 1 - a 2 + b 2 2 a 2 b - 2 a 1 - a 2 - b 2 Applying C 1 → C 1 - b C 3 and C 2 → C 2 + a C 3 , we get ∆ = 1 + a 2 - b 2 - b ( - 2 b ) 2 a b - 2 a b - 2 b 2 a b - 2 a b 1 - a 2 + b 2 + 2 a 2 2 a 2 b - b ( 1 - a 2 - b 2 ) - 2 a + a ( 1 - a 2 - b 2 ) 1 - a 2 - b 2 ∆ = 1 + a 2 + b 2 0 - 2 b 0 1 + a 2 + b 2 2 a b ( 2 - 1 + a 2 + b 2 ) a ( - 2 + 1 - a 2 - b 2 ) 1 - a 2 - b 2 ∆ = 1 + a 2 + b 2 0 - 2 b 0 1 + a 2 + b 2 2 a b ( 1 + a 2 + b 2 ) - a ( 1 + a 2 + b 2 ) 1 - a 2 - b 2 Taking ( 1 + a 2 + b 2 ) common from C 1 and C 2 , we get ∆ = ( 1 + a 2 + b 2 ) 2 1 0 - 2 b 0 1 2 a b - a 1 - a 2 - b 2 Applying R 3 → R 3 - b R 1 + a R 2 , we get ∆ = ( 1 + a 2 + b 2 ) 2 1 0 - 2 b 0 1 2 a b - b + 0 - a - 0 + a 1 - a 2 - b 2 + 2 b 2 + 2 a 2 ∆ = ( 1 + a 2 + b 2 ) 2 1 0 - 2 b 0 1 2 a 0 0 1 + a 2 + b 2 Expanding along R 1 , we get ∆ = ( 1 + a 2 + b 2 ) 2 [ 1 1 2 a 0 1 + a 2 + b 2 - 0 0 - 2 b 0 1 + a 2 + b 2 + 0 0 - 2 b 1 2 a ] ∆ = ( 1 + a 2 + b 2 ) 2 ( 1 × ( 1 + a 2 + b 2 ) - 0 × 2 a ) ∆ = ( 1 + a 2 + b 2 ) 2 ( ( 1 + a 2 + b 2 ) - 0 ) ∴ ∆ = 1 + a 2 - b 2 2 a b - 2 b 2 a b 1 - a 2 + b 2 2 a 2 b - 2 a 1 - a 2 - b 2 = ( 1 + a 2 + b 2 ) 3 . Hence proved.

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 )

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-3>Q 40

EASY

JEE Main/Advance

IMPORTANT

Earn 100

Let $A = [\begin{array}{l} a & b \\ c & d \end{array}]$ and $P = [\begin{array}{l} p \\ q \end{array}] \neq [\begin{array}{l} 0 \\ 0 \end{array}] .$ Such that $AP = P$ and $a + d = 5050$ . Find the value of $(a d - b c)$ .

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Important Questions on Matrices and Determinants

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 1

Without expanding the determinant prove that :

$|\begin{matrix} 0 & b & - c \\ - b & 0 & a \\ c & - a & 0 \end{matrix}| = 0$

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 :

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

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 3

Prove that :

$|\begin{matrix} a x & b y & c z \\ x^{2} & y^{2} & z^{2} \\ 1 & 1 & 1 \end{matrix}| = |\begin{matrix} a & b & c \\ x & y & z \\ y z & z x & x y \end{matrix}|$

MEDIUM

JEE Main/Advance

IMPORTANT

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

Prove that

|\begin{matrix} 1 & a & a^{2} - b c \\ 1 & b & b^{2} - c a \\ 1 & c & c^{2} - a b \end{matrix}| = 0 .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 5

Prove that

|\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = (a - 1)^{3}

.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-4>Q 6

Prove that $|\begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b \\ 2 a b & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix}| = (1 + a^{2} + b^{2})^{3} .$