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.

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-3>Q 36

EASY

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IMPORTANT

Earn 100

If $A$ is an idempotent matrix and $I$ is an identity matrix of the same order such that $(A + I)^{n} = I + 127 A$ , where $n \in N$ , then the value of $n$ is

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Important Questions on Matrices and Determinants

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-3>Q 37

Let

X

be the solution set of the equation

A^{x} = I,

where

A = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}]

and

I

is the corresponding unit matrix and

x \in N,

then the minimum value of

\underset{}{\sum_{x}} (\cos^{x} θ + \sin^{x} θ), θ \in (0, \frac{π}{2}),

is

MEDIUM

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-3>Q 38

If

A = [\begin{matrix} \sin α & - \cos α & 0 \\ \cos α & \sin α & 0 \\ 0 & 0 & 1 \end{matrix}]

, then the value of

|A^{T}|

is

MEDIUM

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IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 10 - Matrices and Determinants>EXERCISE-3>Q 39

If

{[\begin{array}{l} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}]}^{n} = [\begin{matrix} 1 & 18 & 2007 \\ 0 & 1 & 36 \\ 0 & 0 & 1 \end{matrix}]

then find

a + n

EASY

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IMPORTANT

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

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)

.

EASY

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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

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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 :

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