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Question

Mathematics Question on Matrices

Choose the correct answer in the following questions: If [αβ γ−α]\begin{bmatrix} α & β \\\ \gamma & -\alpha \end{bmatrix} is such that A2=I then

A

1+α2+βγ=0

B

1-α2+βγ=0

C

1-α2-βγ=0

D

1+α2-βγ=0

Answer

1-α2-βγ=0

Explanation

Solution

A=[αβ γ−α]\begin{bmatrix} α & β \\\ \gamma & -\alpha \end{bmatrix}

A2=A.A =A=[αβ γ−α]\begin{bmatrix} α & β \\\ \gamma & -\alpha \end{bmatrix}A=[αβ γ−α]\begin{bmatrix} α & β \\\ \gamma & -\alpha \end{bmatrix}

= [α2+βγαβ−αβ αγ−αγβγ+α2]\begin{bmatrix} α^2+\beta\gamma & \alphaβ-\alpha\beta \\\ \alpha\gamma-\alpha\gamma & \beta\gamma+\alpha^2 \end{bmatrix}

=[α2+βγ0 0βγ+α2]\begin{bmatrix} α^2+\beta\gamma & 0 \\\ 0& \beta\gamma+\alpha^2 \end{bmatrix}

Now A2=I ⇒ [α2+βγ0 0βγ+α2]\begin{bmatrix} α^2+\beta\gamma & 0 \\\ 0& \beta\gamma+\alpha^2 \end{bmatrix}= [10 01]\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}

On comparing the corresponding elements, we have:

α2+βγ=1

⇒α2+βγ-1=0

⇒1-α2-βγ=0