b - alendbmatrix 1 and - A - 2- begin bmatrix Seta - lalphall lend bmatrix text- then A- -a2-b2- -a bB- - a 2- b 2- -2 abC- - a 2- b 2- - a 2- b 2D- -2 ab - - a 2- b 2

The correct option is B α = a^2 + b^2, β = 2abA^2 = [ α amp; β; nbsp; β amp; α; ] = [ a amp; b; nbsp; b amp; a ][ a amp; b; nbsp ...

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Multiplication of Matrices

b & alendbmat...

Question

$If A = [\begin{matrix} a & b b & a \end{matrix}] and A^{2} = [\begin{matrix} α & β β & α \end{matrix}], then$

α = a^{2} + b^{2}, β = a b

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α = a^{2} + b^{2}, β = 2 a b

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α = a^{2} + b^{2}, β = a^{2} - b^{2}

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α = 2 a b, β = a^{2} + b^{2}

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Similar questions

A = [\begin{matrix} a & b b & a \end{matrix}]

A^{2} = [\begin{matrix} α & β β & α \end{matrix}]

If $A = [\begin{matrix} a & b b & a \end{matrix}]$ and $A^{2} = [\begin{matrix} α & β β & α \end{matrix}]$ , then

A = [\begin{matrix} a & b b & a \end{matrix}]

A^{2} = [\begin{matrix} α & β β & α \end{matrix}]

α

β

a

cos θ + b sin θ = c

cos (α - β) = \frac{2 c^{2} - (a^{2} + b^{2})}{a^{2} + b^{2}}

cos (θ - α) = a

sin (θ - β) = b

(0 < θ - α, θ - β < \frac{π}{2})

{cos}^{2} (α - β) + 2 a b sin (α - β)

a^{2} - b^{2}

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