If a , b , c , x , y , z ∈ R , then prove that a - x 2 b - x 2 c - x 2 a - y 2 b - y 2 c - y 2 a - z 2 b - z 2 c - z 2 = 1 + a x 2 1 + b x 2 1 + c x 2 1 + a y 2 1 + b y 2 1 + c y 2 1 + a z 2 1 + b z 2 1 + c z 2 .

By LHS. a - x 2 b - x 2 c - x 2 a - y 2 b - y 2 c - y 2 a - z 2 b - z 2 c - z 2 = a 2 + x 2 - 2 a x b 2 + x 2 - 2 b x c 2 + x 2 - 2 c x a 2 + y 2 - 2 a y b 2 + y 2 - 2 a y c 2 + y 2 - 2 c y a 2 + z 2 - 2 a z b 2 + z 2 - 2 b z c 2 + z 2 - 2 c z = 1 x x 2 1 y y 2 1 z z 2 × a 2 - 2 a 1 b 2 - 2 b 1 c 2 - 2 c 1 (row × row) Taking - 1 common in second matrix. = 1 x x 2 1 y y 2 1 z z 2 × - 1 a 2 2 a 1 b 2 2 b 1 c 2 2 c 1 using the column interchange property C 1 ↔ C 3 = 1 x x 2 1 y y 2 1 z z 2 × - 1 - 1 1 2 a a 2 1 2 b b 2 1 2 c c 2 Now multiplying row by row then we are getting = 1 + a 2 x 2 + 2 a x 1 + b 2 x 2 + 2 b x 1 + c 2 x 2 + 2 c x 1 + a 2 y 2 + 2 a y 1 + b 2 y 2 + 2 b y 1 + c 2 y 2 + 2 c y 1 + a 2 z 2 + 2 a z 1 + b 2 z 2 + 2 b z 1 + c 2 z 2 + 2 c z = 1 + a x 2 1 + b x 2 1 + c x 2 1 + a y 2 1 + b y 2 1 + c y 2 1 + a z 2 1 + b z 2 1 + c z 2 = R . H . S .

If a 1 , a 2 , a 3 are distinct real roots of the equation p x 3 + p x 2 + q x + r = 0 such that 1 + a 1 1 1 1 1 + a 2 1 1 1 1 + a 3 = 0 , then Prove that r p < 0

It is given that a 1 , a 2 , a 3 are distinct roots of the equation p x 3 + p x 2 + q x + r = 0 So, a 1 + a 2 + a 3 = - p p , a 1 a 2 + a 2 a 3 + a 3 a 1 = q p & a 1 a 2 a 3 = - r p Now given that 1 + a 1 1 1 1 1 + a 2 1 1 1 1 + a 3 = 0 Now using the property C 2 → C 2 - C 3 ⇒ 1 + a 1 0 1 1 a 2 1 1 - a 3 1 + a 3 = 0 Now expanding determinant then ⇒ 1 + a 1 a 2 1 + a 3 + a 3 - 0 1 + a 3 - 1 + 1 - a 3 - a 2 = 0 ⇒ 1 + a 1 a 2 + a 2 a 3 + a 3 - a 3 - a 2 = 0 ⇒ 1 a 2 + a 2 a 3 + a 3 + a 1 a 2 + a 1 a 2 a 3 + a 1 a 3 - a 3 - a 2 = 0 ⇒ a 1 a 2 + a 2 a 3 + a 1 a 2 a 3 + a 1 a 3 = 0 ⇒ q p - r p = 0 ⇒ q = r So, above equation can written as p x 3 + p x 2 + r x + r = 0 ⇒ p x 2 x + 1 + r x + 1 = 0 ⇒ x = - 1 , x 2 = - r p We know that x 2 ≥ 0 for real x And for x = 0 given equation becomes r = 0 which does not gives the distinct real roots so x 2 > 0 implies - r p > 0 ⇒ r p < 0 So it is possible for real solution r p < 0 . Proved

Prove that Δ = β γ β γ ' + β ' γ β ' γ ' γ α γ α ' + γ ' α γ ' α ' α β α β ' + α ' β α ' β ' = α β ' - α ' β β γ ' - β ' γ γ α ' - γ ' α .

Taking ∆ = β γ β γ ' + β ' γ β ' γ ' γ α γ α ' + γ ' α γ ' α ' α β α β ' + α ' β α ' β ' Now multiplying R 1 by α , R 2 by β and R 3 by γ ∆ = α β γ α β γ ' + α β ' γ α β ' γ ' β γ α β γ α ' + β γ ' α β γ ' α ' γ α β γ α β ' + γ α ' β γ α ' β ' Now dividing by α β γ then we are getting ∆ = 1 α β γ α β γ 1 α β γ ' + α β ' γ α β ' γ ' 1 β γ α ' + β γ ' α β γ ' α ' 1 γ α β ' + γ α ' β γ α ' β ' Now use the property R 3 = R 3 - R 1 & R 2 = R 2 - R 1 ∆ = 1 γ ' α β - α β ' γ γ ' α β ' 0 γ α ' β - α β ' γ ' α ' β - α β ' 0 β α ' γ - α γ β ' α ' γ - α γ ∆ = α ' β - α β ' α ' γ - α γ 1 γ ' α β - α β ' γ γ ' α β ' 0 γ γ ' 0 β β ' Now expanding determinant, then we are getting ∆ = α ' β - α β ' α ' γ - α γ ' 1 γ β ' - β γ ' - 0 + 0 ∆ = α ' β - α β ' α ' γ - α γ ' β ' γ - β γ '

Let M be a 2 × 2 symmetric matrix with integer entries. Then M is invertible if

the product of entries in the main diagonal of M is not the square of an integer

the first column of M is the transpose of the second row of M

the second row of M is the transpose of first column of M

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-3>Q 44

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Earn 100

If $A = [\begin{matrix} 5 a & - b \\ 3 & 2 \end{matrix}]$ and $A$ $adj A = A A^{T}$ , then $5 a + b$ is equal to

50% studentsanswered this correctly

Important Questions on Matrices and Determinants

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-3>Q 45

It

S

is the set of distinct values of

' b'

for which the following system of linear equations

$x + y + z = 1$

$x + a y + z = 1$

$a x + b y + z = 0$

has no solution, then

S

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-3>Q 46

If

A = [\begin{matrix} 2 & - 3 \\ - 4 & 1 \end{matrix}]

, then

adj (3 A^{2} + 12 A)

is equal to

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-3>Q 47

The number of $3 \times 3$ matrices $A$ whose entries are either $0$ or $1$ and for which the system $A [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 0 \\ 0 \end{array}]$ has exactly two distinct solutions, is

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 1

If

a^{2} + b^{2} + c^{2} = 1,

then prove that

|\begin{matrix} a^{2} + (b^{2} + c^{2}) \cos ϕ & a b (1 - \cos ϕ) & a c (1 - \cos ϕ) \\ b a (1 - \cos ϕ) & b^{2} + (c^{2} + a^{2}) \cos ϕ & b c (1 - \cos ϕ) \\ ca (1 - \cos ϕ) & cb (1 - \cos ϕ) & c^{2} + (a^{2} + b^{2}) \cos ϕ \end{matrix}|

is independent of

a, b, c

.

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 2

If

a, b, c, x, y, z \in R,

then prove that

|\begin{matrix} {(a - x)}^{2} & {(b - x)}^{2} & {(c - x)}^{2} \\ {(a - y)}^{2} & {(b - y)}^{2} & {(c - y)}^{2} \\ {(a - z)}^{2} & {(b - z)}^{2} & {(c - z)}^{2} \end{matrix}| = |\begin{matrix} {(1 + a x)}^{2} & {(1 + b x)}^{2} & {(1 + c x)}^{2} \\ {(1 + a y)}^{2} & {(1 + b y)}^{2} & {(1 + c y)}^{2} \\ {(1 + a z)}^{2} & {(1 + b z)}^{2} & {(1 + c z)}^{2} \end{matrix}|

.

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 3

If

a_{1}, a_{2}, a_{3}

are distinct real roots of the equation

p x^{3} + p x^{2} + q x + r = 0

such that

|\begin{matrix} 1 + a_{1} & 1 & 1 \\ 1 & 1 + a_{2} & 1 \\ 1 & 1 & 1 + a_{3} \end{matrix}| = 0,

then Prove that

\frac{r}{p} < 0

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 4

Prove that

Δ =

|\begin{matrix} β γ & β γ' + β' γ & β' γ' \\ γ α & γ α' + γ' α & γ' α' \\ α β & α β' + α' β & α' β' \end{matrix}| =

(α β' - α' β) (β γ' - β' γ) (γ α' - γ' α)

.

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Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 5

If $a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = d a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = d a x_{3}^{2} + b y_{3}^{2} + c z_{3}^{2} = d$ and $a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = f a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = f a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f$

then prove that ${|\begin{array}{l} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{array}|}^{2} = {(d - f)}^{2} \frac{d + 2 f}{a b c}$ where $a, b, c \neq 0$