If A - begin bmatrix a - b 1b - alend bmatrix and A - 2- begin bmatrix Walpha - lalphalllendbmatrix - then A- - a 2- b 2- -2 abB- - a 2- b 2- - abC- - a 2- b 2- - a 2- b 2D- -2 ab - - a 2- b 2

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

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equality of Matrices

If A = begin ...

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

α = a^{2} + b^{2}, β = 2 a b

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

α = a^{2} + b^{2}, β = a^{2} - b^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

α = 2 a b, β = a^{2} + b^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

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

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

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}

Join BYJU'S Learning Program