Cos-alpha -sinlalpha -0011-sinlalpha - coslalpha -0U 0 - 0 -e - beta 1

The correct option is A A( α , β ) A( ,β)=[ cos amp; sin amp; 0; sin amp; cos amp; 0; 0 amp;0 amp;e^β nbsp; ] A^ 1=[ cos amp ...

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Byju's Answer

Standard XII

Mathematics

Multiplication of Matrices

cosłalpha &si...

Question

$A (α, β) = ⎛ ⎜ ⎝ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & e^{β} \end{matrix} ⎞ ⎟ ⎠ \Rightarrow {[A (α, β)]}^{- 1} =$

$A (- α, β)$

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$A (- α, - β)$

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$A (α, - β)$

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$A (α, β)$

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Solution

The correct option is B
$A (- α, - β)$

$A (\propto, β) = ⎡ ⎢ ⎣ \begin{matrix} c o s \propto & s i n \propto & 0 - s i n \propto & c o s \propto & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦ A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} c o s \propto & - s i n \propto & 0 s i n \propto & c o s \propto & 0 0 & 0 & e^{- β} \end{matrix} ⎤ ⎥ ⎦ = A (- \propto, - β)$

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Similar questions

Using properties of determinants prove the following questions.

$∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n α & c o s α & c o s (α + δ) s i n β & c o s β & c o s (β + δ) s i n γ & c o s γ & c o s (γ + δ) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

Evaluate $∣ ∣ ∣ ∣ \begin{matrix} c o s α c o s β & c o s α s i n β & - s i n α - s i n β & c o s β & 0 s i n α c o s β & s i n α s i n β & c o s α \end{matrix} ∣ ∣ ∣ ∣$

If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then the value of determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & c o s (α - β) & c o s α c o s (α - β) & 1 & c o s β c o s α & c o s β & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

Q. Let

f (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

, then

(f (α))^{- 1}

is equal to

Q. If

α, β, γ

are real numbers, then

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & c o s (β - α) & c o s (γ - α) c o s (α - β) & 1 & c o s (γ - β) c o s (α - γ) & c o s (β - γ) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

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A(α,β)=⎛⎜⎝cosαsinα0−sinαcosα000eβ⎞⎟⎠⇒[A(α,β)]−1=

The correct option is B A(−α,−β) A(∝,β)=⎡⎢⎣cos∝sin∝0−sin∝cos∝000eβ⎤⎥⎦A−1=⎡⎢⎣cos∝−sin∝0sin∝cos∝000e−β⎤⎥⎦=A(−∝,−β)

$A (α, β) = ⎛ ⎜ ⎝ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & e^{β} \end{matrix} ⎞ ⎟ ⎠ \Rightarrow {[A (α, β)]}^{- 1} =$