Let A = [ 1 − 1 2 α ] a n d B = [ β 1 1 0 ] , α , β ∈ R . Let a1 be the value of a which satisfies ( A + B ) = A 2 + [ 2 2 2 2 ] a n d α 2 be the value of a which satisfies (A + B)2 = B2. Then | α 1 − α 2 | is equal to…………….

A + B = [ β + 1 0 3 α ] ( A + B ) 2 = [ ( β + 1 ) 2 0 3 ( β + 1 ) + 3 α α 2 ] ∴ [ 1 − α + 1 2 α + 4 α 2 ] = [ ( β + 1 ) 2 0 3 ( α + β + 1 ) α 2 ] = 1 = 1 B 2 = [ β 1 1 0 ] [ β 1 1 0 ] β = 0 , α = − 1 = α 2 | α 1 − α 2 | = | 1 − ( − 1 ) | = 2

Let $A = [\begin{matrix} 1 & - 1 \\ 2 & α \end{matrix}] a n d B = [\begin{matrix} β & 1 \\ 1 & 0 \end{matrix}], α, β \in R .$ Let a₁ be the value of a which satisfies $(A + B) = A^{2} + [\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}] a n d α_{2}$ be the value of a which satisfies (A + B)² = B². Then $| α_{1} - α_{2} |$ is equal to…………….

3 Views|Posted 6 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

$A + B = [\begin{matrix} β + 1 & 0 \\ 3 & α \end{matrix}]$

${(A + B)}^{2} = [\begin{matrix} {(β + 1)}^{2} & 0 \\ 3 (β + 1) + 3 α & α^{2} \end{matrix}]$

$∴ [\begin{matrix} 1 & - α + 1 \\ 2 α + 4 & α^{2} \end{matrix}] = [\begin{matrix} {(β + 1)}^{2} & 0 \\ 3 (α + β + 1) & α^{2} \end{matrix}]$

= 1 = 1

$B^{2} = [\begin{matrix} β & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} β & 1 \\ 1 & 0 \end{matrix}]$

$β = 0, α = - 1 = α_{2}$

$| α_{1} - α_{2} | = | 1 - (- 1) | = 2$

Upvote

Similar Questions for you

If $A [\begin{matrix} 1 & 0 & 1 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 1 \end{matrix}]$ ,  $A [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = 4 [\begin{matrix} - 1 & 0 & 1 \end{matrix}]$ and $A [\begin{matrix} 0 & 1 & 0 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 0 \end{matrix}]$  where, A is a 3 × 3 matrix and (A –3I) $[\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$  then the value of (x, y, z) is

Alok Kumar Singh

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ ...(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 &nb

Vishal Baghel

$f (x) = a x^{2} + b x + c$

g (x) = px + q

$f (g (x)) = a {(p x + q)}^{2} + b (p) (+ q) + c$

$\Rightarrow 8 x^{2} - 2 x = a {(p x + q)}^{2} + b (p x + q) + c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9 (f (x)) = p (a x^{2} + b x + c) + q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

Vishal Baghel

Kindly consider the following figure

B = (I – adjA)⁵

Vishal Baghel

Kindly consider the following figure

B = (I – adjA)⁵

If the system of linear equations

2x + 3y – z = -2

x + y + z = 4

$x - y + | λ | z = 4 λ - 4$

where $λ \in R,$ has no solution, then

Raj Pandey

System of equation is

$(\begin{matrix} 2 & 3 & - 1 \\ 1 & 1 & 1 \\ 1 & - 1 & | λ | \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = [\begin{array}{l} - 2 \\ 4 \\ 4 λ - 4 \end{array}]$

R₁ – 2 R₂, R₃ – R₂

$(\begin{matrix} 0 & 1 & - 3 \\ 1 & 1 & 1 \\ 0 & - 2 & | λ | - 1 \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} - 10 \\ 4 \\ 4 λ - 8 \end{array})$